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-   -   69660 and 92020 (https://www.mersenneforum.org/showthread.php?t=25410)

 enzocreti 2019-11-06 13:46

binary form of the exponents 69660, 92020, 541456

pg(69660), pg(92020) and pg(541456) are probable primes with 69660, 92020 and 541456 multiple of 86

69660 in binary is 10001000000011100
92020 in binary is 10110011101110100
541456 in binary is 10000100001100010000

you can see that the number of the 1's is always a multiple of 5

a chance?

 enzocreti 2019-12-18 08:26

69660, 92020, 541456

69660, 92020 and 541456 are 6 mod 13 (and 10^m mod 41)

69660, 92020 and 541456 are multiple of 43

is there a reason why

(69660-6)/26 is congruent to 13 mod 43
(92020-6)/26 is congruent to 13 mod 43
(541456-6)/26 is congruent to 13 mod 43?

215, 69660, 92020, 541456 are multiple of 43

let be log the log base 10

int(x) let be the integer part of x so for example int(5.43)=5

A=10^2*log(215)-215/41

int(A)=227=B

215 (which is odd) is congruent to B+1 mod 13
69660 which is even is congruent to B mod 13
92020 is congruent to B mod 13
and 541456 is also congruent to B mod 13

215 (odd) is congruent to -1215 mod 13
69660 (even) is congruent to 1215 mod 13
92020 (even) is congruent to 1215 mod 13
541456 (even) is congruent to 1215 mod 13

215, 69660, 92020, 541456 can be written as 13x+1763y+769

(541456-769-13*93)/1763=306

(69660-769-13*824)/1763=33

(92020-769-13*781)/1763=46

as you can see 306,33 and 46 are all 7 mod 13

769+13*824
769+13*781
769+13*93 are multiples of 43

(69660-(10^3+215))/13=5265 which is 19 mod 43
(92020-(10^3+215))/13=6985 which is 19 mod 43
(541456-(10^3+215))/13=41557 which is 19 mod 43

10^3+215 is 6 mod 13

now
69660/13=5358,4615384...
92020/13=7078,4615384...
541456/13=41650,4615384...

the repeating term 4615384 is the same...so that numbers must have some form 13s+k?

215 (odd) is congruent to 307*2^2-10^3 or equivalently to - (19*2^6-1) mod 13
69660 (even) is congruent to 307*2^2-1001 or equivalenly to (19*2^6-1) mod 13
92020 (even) the same
541456 (even) the same

(19*2^6*(541456-92020)/(13*43)+1)/13+10=75215 is a multiple of 307=(215*10-1)/7=(5414560-1)/17637

pg(51456) is another probable prime with 51456 congruent to 10^n mod 41

75215=(51456*19+1)/13+10=(19*2^6*(541456-92020)/(13*43)+1)/13+10

215 is congruent to -1215=5*3^5 mod 13

69660 is congruent to 1215 mod 13 and so also 92020 and 541456

1215=(51456/2-2*13*10^2-43)/19

...so summing up...

215 (odd) is congruent to 3*19*2^2 mod((41*43-307)/(7*2^4)=13) where 41*43-307 is congruent to 10^3+3*19*2^2 mod (3*19*2^2=228)
69660 (even) is congruent to (3*19*2^2-1) mod 13 ...
the same for 92020 and 541456

another way is 215 (odd) is congruent to -15*81 mod ((41*43-307)/(307-15*13))
69660 (even) is congruent to 15*81 mod ((41*43-307)/(307-15*13))
the same for 92020 and 541456

215 is congruent to (41*43-307-10^3)/2 mod ((41*43-307)/(7*2^4))
69660 is congruent to (41*43-307-10^3)/2-1 mod ((41*43-307)/(7*2^4))
92020 is congruent to (41*43-307-10^3)/2-1 mod ((41*43-307)/(7*2^4))
541456 is congruent to (41*43-307-10^3)/2-1 mod ((41*43-307)/(7*2^4))

215 is congruent to 3*19*2^2 mod ((41*43-307)/(2*(19*3-1)))
69660 is congruent to 3*19*2^2-1 mod ((41*43-307)/(2*(19*3-1)))
and so 92020 and 541456

215 is congruent to 3*19*2^2 mod ((3^6-1)/(3*19-1))
69660 is congruent to 3*19*2^2-1 mod((3^6-1)/(3*19-1)) and so 92020 and 541456

51456 (pg(51456) is probable prime and 51456 is 10^n mod 41) is congruent to 19*3*2^4 mod 13

Pg(2131) is probable prime
2131 is prime
227=2131-307*4-26^2
So 215 is congruent also to 2131-307*4-26^2 mod 13
And 69660 is congruent to 2131-307*4-26^2+1 mod 13 and so also 92020 and 541456

69660 is congruent to 1763-307*5-1 mod ((1763-307)/112)=13)

And so 92020 and 541456

215 which is odd is congruent to - 1763+307*5+1 mod 13
215+1763-307*5-1 is divisible by 17 and by 13
And also 541456-1763+307*5+1 is divisible by 13 and 17
215 and 541456 have the same residue 10 mod 41

((541456-1763+307*5+1)/(13*17)+1) *200+51456=541456

69660 92020 541456 are congruent to 7*2^6 mod 26
7*2^6-(1763-307*5-1)=221=13*17
215+7*2^6 is a multiple of 221
541456-7*2^6 is a multiple of 221

 enzocreti 2020-01-24 15:31

215 , 51456, 69660, 92020, 541456

51456, 69660, 92020, 541456 are even and congruent to 10^n mod 41

pg(51456), pg(69660), pg(92020) and pg(541456) are prp

51456, 69660, 92020, 541456 are all congruent to 7*2^q-1 mod 13 with q a nonnegative integer

215 is odd and pg(215) is prp

215 is congruent to 7*2^q mod 13

 enzocreti 2020-03-30 11:52

69660 and 92020

69660 and 92020 are multiple of 215 and congruent to 344 mod 559
92020=lcm(215,344,559)+69660

| denotes concatenation in base 10

2^69660-1 | 2^69559-1 is prime
2^92020-1 | 2^92019-1 is prime!!!

 LaurV 2020-03-30 12:19

please show us a proof that they are prime

 enzocreti 2020-03-30 12:50

...

Well...
actually they are only probable primes... maybe in future they will be proven primes

 enzocreti 2020-03-30 17:39

[QUOTE=LaurV;541319]please show us a proof that they are prime[/QUOTE]

[url]http://factordb.com/index.php?id=1100000001110801143[/url]

[url]http://factordb.com/index.php?query=%282%5E92020-1%29*10%5E27701%2B2%5E92019-1[/url]

 enzocreti 2020-03-30 23:06

... I note also...

69660 I note also that

(lcm(215,344,559))^2=4999*10^5+69660-60

I notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)

I note that the polynomial

X^2-X*429^2+7967780460=0 has the solution x=69660

If you see the discriminant of such polynomial you can see interesting things about pg primes with exponent multiple of 43

I note that 429^2 is congruent to 1 mod 215 and to 1 mod 344.

I note that 92020*2+1=429^2

The discriminant of the polynomial is 429^4-4*7967780460 which is a perfect square and lcm(215,344,559) divides 429^4-4*7967780460-1

Pg(331259) is prime and pg(92020) is prime.
92020+(92020/215-1)*559+546=331259

Again magic numbers 559 and 546 strike!

So 331259 is a number of the form
215*(13s-1)+(13*b-2)*559+546

For some s, b positive integers

So there are pg primes pg(75894) and pg(56238) with 75894 and 56238 multiple of 546 and pg(331259) with 331259 of the form 215m+559g+546 for some positive m and g.

Pg(69660) is prime. 69660=(3067*8-546)*3-11# where # is the primorial and 3067 is a prime of the form 787+456s

notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)

So we have pg(215) is prime pg(69660) is prime pg(92020) is prime

With 215 69660 and 92020 multiple of 43

69660=215+(18*18-1)*215

92020=69660+18*18*69+4

-215 is congruent to 108 mod (18*18-1=323)
541456 is congruent to 108 mod (18*18-1)
92020 is congruent to (17*17-1) mod (18*18-1)
69660 is congruent to 215 mod (18*18-1)

108=6^3-18^2

17*17-1+36-216=17*17-1+6^2-6^3=108

215=6^3-1

To make it easier

215, 69660, 541456 are congruent to plus or minus 215 mod 323

92020 is congruent to (17*17-1) mod (18*18-1)

curious that 289/215 is about 1.(344)...

and 541456 92020 69660 215 are congruent to plus or minus 344 mod 559

215, 69660, 541456 are congruent to plus or minus 6^3-1 mod 323

92020 is congruent to (12/9)*6^3 mod 323

92020*9/12 is congruent to 6^3 mod 323

92020 is congruent to (17^2-1) mod 323 and to - (6^2-1) mod 323

92020 is a number of the form 8686+13889s

13889=(6^3+1)*64

215 69660 92020 541456 are + or - 344 mod 559

lcm(215,344,559)-86*(10^2+1)+6^3-1=(6^3+1)*2^6+1

92020=69660+lcm(215,344,559) so you can substitute

92020=69660+86*(10^2+1)-6^3+1+(6^3+1)*2^6+1

86*(10^2+1) mod 323 is 17^2-1

215, 69660, 541456 are multiple of 43 and congruent to 10 and 1 mod 41

They are congruent to plus or minus 215 mod 323

92020 is congruent to 2^4 (not a power of 10) mod 41

92020 is congruent to (2^4+1)^2-1 mod 323

288 is 17^2-1

288 in base 16 is 120

120=11^2-1

also 323=18^2-1 in base 16 is 143=12^2-1

344*((1444456-1763*2^9) /344-1)=541456

1444456=lcm(13,323,344)

541456=lcm(13,323,344)-344*(41*2^6+1)

215 69660 92020 541456 are congruent to plus or minus (3^a*2^b) mod 323

215 is congruent to - 108 mod 323
541456 is congruent to 108 mod 323
69660 is congruent to - 108 mod 323
92020 is congruent to 288 mod 323

108 and 288 are numbers of the form 3^a*2^b

So exponents multiple of 43 are congruent to plus or minus 344 mod 559 and to plus or minus a 3-smooth number mod 323

108 and 288 are both divisible by 36

 enzocreti 2020-08-19 05:48

Q77I [QUOTE=enzocreti;541360]69660 I note also that

(lcm(215,344,559))^2=4999*10^5+69660-60

I notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)

I note that the polynomial

X^2-X*429^2+7967780460=0 has the solution x=69660

If you see the discriminant of such polynomial you can see interesting things about pg primes with exponent multiple of 43

I note that 429^2 is congruent to 1 mod 215 and to 1 mod 344.

I note that 92020*2+1=429^2

The discriminant of the polynomial is 429^4-4*7967780460 which is a perfect square and lcm(215,344,559) divides 429^4-4*7967780460-1

Pg(331259) is prime and pg(92020) is prime.
92020+(92020/215-1)*559+546=331259

Again magic numbers 559 and 546 strike!

So 331259 is a number of the form
215*(13s-1)+(13*b-2)*559+546

For some s, b positive integers

So there are pg primes pg(75894) and pg(56238) with 75894 and 56238 multiple of 546 and pg(331259) with 331259 of the form 215m+559g+546 for some positive m and g.

Pg(69660) is prime. 69660=(3067*8-546)*3-11# where # is the primorial and 3067 is a prime of the form 787+456s

notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)

So we have pg(215) is prime pg(69660) is prime pg(92020) is prime

With 215 69660 and 92020 multiple of 43

69660=215+(18*18-1)*215

92020=69660+18*18*69+4

-215 is congruent to 108 mod (18*18-1=323)
541456 is congruent to 108 mod (18*18-1)
92020 is congruent to (17*17-1) mod (18*18-1)
69660 is congruent to 215 mod (18*18-1)

108=6^3-18^2

17*17-1+36-216=17*17-1+6^2-6^3=108

215=6^3-1

To make it easier

215, 69660, 541456 are congruent to plus or minus 215 mod 323

92020 is congruent to (17*17-1) mod (18*18-1)

curious that 289/215 is about 1.(344)...

and 541456 92020 69660 215 are congruent to plus or minus 344 mod 559

215, 69660, 541456 are congruent to plus or minus 6^3-1 mod 323

92020 is congruent to (12/9)*6^3 mod 323

92020*9/12 is congruent to 6^3 mod 323

92020 is congruent to (17^2-1) mod 323 and to - (6^2-1) mod 323

92020 is a number of the form 8686+13889s

13889=(6^3+1)*64

215 69660 92020 541456 are + or - 344 mod 559

lcm(215,344,559)-86*(10^2+1)+6^3-1=(6^3+1)*2^6+1

92020=69660+lcm(215,344,559) so you can substitute

92020=69660+86*(10^2+1)-6^3+1+(6^3+1)*2^6+1

86*(10^2+1) mod 323 is 17^2-1

215, 69660, 541456 are multiple of 43 and congruent to 10 and 1 mod 41

They are congruent to plus or minus 215 mod 323

92020 is congruent to 2^4 (not a power of 10) mod 41

92020 is congruent to (2^4+1)^2-1 mod 323

288 is 17^2-1

288 in base 16 is 120

120=11^2-1

also 323=18^2-1 in base 16 is 143=12^2-1

344*((1444456-1763*2^9) /344-1)=541456

1444456=lcm(13,323,344)

541456=lcm(13,323,344)-344*(41*2^6+1)

215 69660 92020 541456 are congruent to plus or minus (3^a*2^b) mod 323

215 is congruent to - 108 mod 323
541456 is congruent to 108 mod 323
69660 is congruent to - 108 mod 323
92020 is congruent to 288 mod 323

108 and 288 are numbers of the form 3^a*2^b

So exponents multiple of 43 are congruent to plus or minus 344 mod 559 and to plus or minus a 3-smooth number mod 323

108 and 288 are both divisible by 36[/QUOTE]

215 69660 92020 541456 are congruent to plus or minus (6^k-1) mod 323 for k=3,2

69660=(2^5*3^7)-(2^3*3^4) so it is the difference of two 3 smooth numbers (2^a*3^b)-(2^(a-3)*3^(b-3))

69660 is multiple of 3 and congruent to 0 mod (6^2-1)

215, 92020, 541456 are not multiple of 3 and multiple of 43 and are congruent to plus or minus 2^k mod 36 for some k

1763*323-(6^3+1)*(2^7+1)=541456 or

(42^2-1)*(18^2-1)-(6^3+1)*(2^7+1)=541456

I also note that

69660=(2^7+1)*(6^3+1)+(2^7+1)*(18^2-1)

And by the way 215=(42^2-1)*4^2-(2^7+1)*(6^3+1)

27993=(217*129)=(2^7+1)*(6^3+1)

I notice that - 541456 mod 27993=9202*2

92020=10*9202

And 9202*2*10+1=429^2

27993=3/5*(6^6-1)

27993 in base 6 is 333333

27993 has also the representation:

(42^2-1)*4^2-(6^3-1)=27993=2*43*(18^2-1)+(6^3-1)

x/5+(42^2-1)*(18^2-1)-3/5*(6^6-1)=20*3/5*(6^6-1)

The solution of this equation x=92020

this identity:

(42^2-1)*(18^2-1)=(10^3+18^2)*430+43*3

Maybe it is not a chance that pg(10^3+18^2-1=1323) is prime

pg(1323), pg(215), pg(69660), pg(92020), pg(541456) are primes

1323, 215, 69660, 92020, 541456 are congruent to plus or minus (2^a*3^b-1) mod 323 where 2^a*3^b is a 3 smooth number with 2^a*3^b<323

or 1323, 215, 69660, 92020, 541456 are congruent to plus or minus (p-1) mod 323 where p is a perfect power

92020 has the factorization 2*(6^3-1)*(6^3-2)

lcm(215,344,559)=(6^3-1)*(2*(6^3-2)-18^2)=(2*(6^3-2)-18^2)*(2*(6^3-2)-18^2+111)

69660=92020-lcm(215,344,559)

215=2*(6^3-2)+2*(6^3+1)-2*18^2+1

92020=(2*(6^3-2)+2*(6^3+1)-2*18^2+1)*(2*(6^3-2)+2*(6^3+1)-2*18^2)*2

so 215=2*(6^3-2)+2*(6^3+1)-2*18^2+1

69660==(2*(6^3-2)+2*(6^3+1)-2*18^2+1)*(2*(6^3-2)+2*(6^3+1)-2*18^2)*2-((6^3-1)*(2*(6^3-2)-18^2))

92020=(2*(6^3-2)+2*(6^3+1)-2*18^2+1)*(2*(6^3-2)+2*(6^3+1)-2*18^2)*2

541456=(42^2-1)*(18^2-1)-(7^3-6^3+2)*(6^3+1)

69660=111111-5*11111+8*(42^2-1)

344=7^3+1=2*(111111-5*11111)/(18^2-1)=A

so lcm(215,344,559)=lcm(215, A, A+215)

541456=5*111112-(42^2-1)*8

(12^2-1) divides (111111-5*11111-6^3+1)

559=(111111-5*11111-6^3+1)/(12^2-1)+(111111-5*11111)/(18^2-1)

Quite clear that 215, 69660, 92020,541456 multiple of 43

are congruent to plus or minus (18^2-6^k) mod (18^2-1) where k is 2 or 3 and 3 indeed is the maximum exponent such that 18^2-6^k i 1s positive

(541456-18^2+6^3)/323=41^2-5=1676
(92020-18^2+6^2)/323=17^2-5=284
(69660+18^2-6^3)/323=6^3
(215+18^2-6^3)/323=1

There is clearly a pattern!!!

I notice also that (1+5)=6 is a semiprime
(6^3+5)=13*17 is a semiprime
(1676+5)=41*41 is a semiprime
(284+5)=17*17 is a semiprime

(6,221,1681,289) are either squares of primes or product of two consecutive primes

so when pg(k) is prime and k is a multiple of 43, then k can be expressed in this way:

(p*q-5)*(18^2-1) plus or minus (18^2-6^k) with 6^k<18^2

p and q are primes

when pg(k) is prime and k is a multiple of 546, then

k is congruent to 78 mod (18^2-6^3)

as in the cases k=56238 and k=75894

Curio of the curios:

pg(331259) is probable prime and 331259 is prime

magic:

71*6^6-331*(10^4-3*331)=331259

71*6^6 and 331259 have in common the first five digits 33125=182^2+1

If you consider 71*6^n for n>3

For n even

71*6^n is congruent to (11^2-1) mod (13^2-1)

and for n odd

Is congruent to (7^2-1) mod (13^2-1)

For n=4

71*6^4=92016

for n=6

71*6^6=3312576

As you can see

Pg(92020) is prime and pg(331259) is prime

92020 has the last two digits 20 different from 92016

The same for 3312576 and 331259

Moreover 331259 mod (71*6^3) =9203

Both 331259 and 92020 are 5 mod 239

Is there something connected with the fact

ord (71*6^k)=4

I mean the smallest value k for which 71*6^k is congruent to 1 mod 239 is k=4???

71 and 6 are both quadratic residues mod 239

92020 and 331259 are congruent to 71*3^3 mod (239*13)

I wonder if this concept could be generalized

pg(51456), pg(92020), pg(331259) are probable primes

51456 is congruent to 71 mod (239*(6^3-1))
92020 is congruent to 71*3^3 mod (239*13)
331259 is congruent to 71*3^3 mod (239*13)

92020 and 331259 are congruent to 6 mod 13

51456 and 331259 are congruent to 2^3 mod 109

71 and 71*3^3 are both congruent to 6 mod 13

so

51456, 92020, 331259 are congruent to 13*(5+71*k)+6 either mod (239*215) or mod (239*13)

are these primes infinite?

the odd thing is that

51456, 92020, 331259 are either congruent to 10^m mod 41 or prime (331259 infact is prime)

So I think it is no chance that 51456, 92020, 331259 are either congruent to 2^j mod 71 or to 13*2^i mod 71

No chance at all!

There is a file Rouge!
Pg(541456) is probable prime as pg(51456)

And 541456 mod (239*215)=9202*3, that is 3*( 92020/10 )and pg(92020) is prime

541456 is congruent to (3/20)*(429^2-1) mod (239*215)

541456 is congruent to 14*71*3^3+3*2^8 (mod (239*215))

Maybe there is some connection to the fact that

215, 69660, 92020 and 541456 are plus or minus 344 mod 559

lcm(344,559)=4472=71*(2^6-1)-1

I suspect that something in field F(239) is in action!

331259^(-1) mod (215*239)=49999=(10^5-2)/2

((71*(2^6-1))-1)/2=lcm(344,559)=9202-6966

Multiplying both sides by 10 you have

92020=69660+lcm(215,344,559)

I would suggest to study these exponents in field F(51385=239*215)

(239*10*215+3*(429^2-1)/10-541456) /3=9202

So

10*
(239*10*215+3*(429^2-1)/10-541456) /3=92020

In this equation we have 215, 92020, 541456 multiple of 43 and not of 3

The other multiple of 43 is 69660 which is multiple of 3

And 92020=69660+lcm(344,215,559)

239*10*215+3*(429^2-1)/10 is a multiple of 559

239*10*215+3*(429^2-1)/10=(42^2-1)*(18^2-2)+43*2^5

This identity

(239*10*215+(3/10)*(429^2-1)-541456+92020)/559=92020/(215*2)

One can play around with this expression containing 215 and 559

And substitute for example 92020 with (429^2-1)/2

Pg(331259) is prime and also 331259 is prime

The inverse modulo (215*239) of 331259 is the prime 5*10^4-1=49999

(331259*49999-1)/(215*239)=7*(18^2-1)*(12^2-1)-10^3+1

So 331259=((((18^2-1)*(10^3+1)-10^3+1)*239*215)+1)/(5*10^4-1)

Using Wolphram Alpha i considered this equation:

((323*(10^x+1)-10^x+1)*239*215+1)/(5*10^(x+1)-1)=y

wolphram say that the integer solution is x=3, y=331259

wolphram gives an alternate form:

84898302/(5*(2^(x+1)*5^(x+2)-1))+1654597/5=y

the number 1654597=69660+(30^2-1)*(42^2-1)

so 69660 and 331259 (both 6 mod 13 and pg(69660) pg(331259) primes) are linked by this equation:

84898302/(5*(2^(3+1)*5^(3+2)-1))+(69660+(30^2-1)*(42^2-1))/5=331259

541456 in field F(239*215) and in field F(239*323)

541456 mod (239*215)=9202*3
541456 mod (239*323)=359*3

Pg(359) is prime, pg(9202*10) is prime

9202-359 is a multiple of 239

Pg(92020+239239=331259) is prime

I note also

331259=71*6^6-331*(10^4-3*331)=6^2*9202-13=92020+239239

239239 is congruent to -13 mod 107 and mod 43

22360=(10*(239239+13))/107

92020=69660+22360

92020=331259-239239

92020 is a multiple of 107

lcm(215,344,559)*10=22360

multiple of 86, that is 69660, 92020 and 541456 are congruent to a square mod 428

so the muliple of 86=k for which pg(k) is prime are numbers for which there is a solution to this modular equation:

y is congruent to 36*x^2 mod 428

infact

92020 is congruent to 36*0^2 mod 428
69660 is congruent to 36*3^2 mod 428
541456 is congruent to 36*1^2 mod 428

There are pg(k) primes with k multiple of 215 and k multiple of 546

probably that numbers 215 and 546 are not random at all

look at this equality:

71*6^6-182^2-3^2-331*(10^4-3*331)-10=546^2

546-215=331=546-(6^3-1)

by the way 182^2+3^2+546^2 is prime

546^2 is congruent to 6^3 mod 331

541456-((3*239) ^2-239+(429^2-1)/20)=(429^2-1)/10

541456=(18/5)*(429^2-1)/2+13*((429^2-11)/10-2235)

where 2235=lcm(215,344,559)-1

541456*10=3*331259+239*(136^2+1)

92020 and 331259 are both 5 mod 239

71*6^k is congruent to 1 mod 239 for k=4

But 71*6^4 is 920...
And 71*6^6 is 33125...

92020 mod 71 is 4 and mod 331 is 3

71*6^4 is congruent to 1 mod (385*239)

541456 is congruent to - 239 mod 385

I notice that 92020 and 331259 are congruent to 5 mod 239 but also to - 72 mod (1001)

I notice that 92020 is congruent to 146 mod 71

541456 and 331259 are congruent to 146 mod 703

I notice that 215, 69660, 541456 are plus or minus 215 mod 323
92020 which is congruent to 16 mod 41, is congruent to 288 mod 323 and mod 71
288 and 92020 are both 4 mod 71

so 92020 is congruent to 4 mod 71
and is congruent to 4+284 (mod 323)

where 284 is the residue mod 323 of 71*6^4

215, 69660, 541456 are congruent to plus or minus 215 mod 323

where 215=4+211

211 is the residue of 71*6^4 mod 215

in particular 92020 is congruent to 288 mod 323 and mod 284, infact 92020=71*6^4+4

I think that also 331259 has something to do with 71*6^6

so multiple of 43, that is 215, 69660, 92020, 541456 are either of the form 323k+108, or 323k-108, or 323k-288

I notice that (323-108)=6^3-1 and (323-288)=6^2-1

211 and 284 are also the residues of 71*6^4 and 71*6^6 mod 323

the 18-th pg prime is pg(1323),
the 36-th is pg(360787)

modulo (18^2-1=323), 1323 is 31 and 360787 is 319

the difference between 319 and 31 is again 288

1323 is congruent to 31 mod (6^4-4)
360787 is congruent to 6^2*2^3+31 mod (6^4-4)

319 is congruent to 31 mod 6^2 infact

here:

215, 69660, 541456 are congruent to plus or minus 108 mod 323
92020 is congruent to (6^4-1008=288) mod 323

so multiple of 43 are either congruent to plus or minus (18+90) mod (18^2-1) or to 6^4-(18+90+900) mod (18^2-1)

92020 is congruent to (11*6^2-108) mod (18^2-1)
and to - 11*6^2 mod (304^2)

so great fact 215, 69660, 92020, 541456 are either congruent to plus or minus (108*2-1) mod (108*3-1) or to plus/minus (108/3-1) mod (108*3-1)

[B]pg(331259) and pg(92020) are probable primes.[/B]

[B]331259=92020+239239 as said[/B]

[B]331259 and 92020 are congruent to 5 mod 239[/B]
[B]331259 and 92020 are congruent to 6 mod 13[/B]

[B]using chinese remainder theorem it yields[/B]
[B]something like[/B]

[B]239x(1)+13y(1)=1[/B]

[B]using Euclidean algor you have the solution y(1)=92[/B]

[B]92 are the first two digits of 92020[/B]

[B]now 331259 can be rewritten as 239*10^3+92*10^3+331-71-1[/B]

[B]331259 and 92020 have the property that they are congruent to the last two decimal digits modulo 9200[/B]

[B]331259 is congruent to 59 mod 9200[/B]
[B]92020 is congruent to 20 mod 9200[/B]

(239*77+1)*5=92020

239*77+1 is a multiple of 215

541456 is congruento to 3/2*(239*77+1) mod (239*215)

331259 is congruent to 5 mod (239*77) and also 92020

(541456-9202*3)/215/239=10

(13*331-1)*7*11+5-7*11*13*239=92020

541456 is congruent to (9203+(71*6^4-1)/5) (mod (239*5*43))

9203 is prime

(541456-9203-(71*6^4-1)/5)/215/239=10

331259-13*(9202*2-1)=92020

92020-13*1720=69660

lcm(215,344,559)=13*1720

67*(16683/67+22360/43-1)=51456

where 16683=9202*2-1-1720 and 22360=lcm(215,344,559)

331259 has the representation

331259=36*(7/180+(92020-4)/10)

Let be Floor(x) the floor function floor(5.5)=5 for example

Floor(239*(331259/(71*6^4))/4)=215

So i suspect that there are infinitely many pg(k) primes with k multiple either of 215 or multiple of something transformed by the Floor function in 215 and in these Cases k is alway 6 mod 13

215=(239*6^2-2^2)/40=(239*6^2-2^2)/(6^2+2^2)

331259=(3*71*6^5+7)/5

71*6^6-(3^3*71*6^5-7)/5=331259

(3^3*71*6^5-7)/5=(10^4-3*331)*331

so we have 92020=71*6^6-(3^3*71*6^5-7)/5-13*(71*6^4-1)/5

(69660/3-860)/13=1720

I see where 215 comes from

(331259-23666)/(12^2-1)=9*239-1=215*10

23666 I used CRT x is congruent to 6 mod 13 and to 5 mod (239*7)

As you can see Both 331259 and 92020 are congruent to 23666 mod 143

But this simply means that both 331259 and 92020 are congruent to 71 mod 143

By the way also 215 is congruent to -71 mod 143

so 215, 92020, 331259 are congruent to plus or minus 71 mod 143

215 is 2 mod (71)
92020 is 4 mod (71*6^3)
331259 is 9203 (prime) mod (71*6^3)

there is clearly a pattern

note that also 541456 is congruent to -(9203*3) mod (7*71)

i have no idea how to develop these ideas but i strongly suspect that there is a structure

It should be clear that

541456 215 69660 331259 92020 are congruent to plus or minus (19+13s) mod 143 for some non negative s.

331259 is also congruent to sqrt(239*9-215) mod 71

215*9=44^2-1

So 331259 is congruent to sqrt(215*9+1) mod 71

another way to see the same thing is that 215 92020 and 331259 are congruent to plus or minus 72 (mod 143)

and 92020 and 331259 are even congruent to - 72 (mod 143*7)

infact 92020+72=92092 and 331259+72=331331

It's easy to see that 9203 conguent to 44 mod 71 and 331259 is as well congruent to 44 mod 71

and 9203 has the same residue 259 mod 344 and mod 559

((541456/921)-((41*43*10+7)/30))^(-1)-7=9203

note that 921=3*307

9210=30*307=((541456/921)-((41*43*10+7)/30))^(-1)

2150=307*7+1

92020*(5879)/999-71.107107107=541456

or 92020*(1763*10+7)/2997-71.107107...=541456

541456-(92020*(1763*10+7)/58790)=10*215*239

331259 is congruent to (71*6^3+307*10)/2 (mod (71*7*6^4))

9203=(331259*2-7*71*6^4)/2

331259=(71*6^3*43)/2+307*5

331259=(2^10+1)*(18^2-1)+184

I would conjecture that if pg(k) is prime and k is multiple of 43,

then k is a multiple of (72+143s) for some positive integer s

the multiple of 215=72+143 are 215 itself, 69660, 92020

the other multiple of 43 is 541456 which is a multiple of 787=72+143*5

multiple of 43 are 215, 69660, 92020, 541456

they are either congruent to (plus/minus) 215 mod 323 or to 288 mod 323

215 and 288 are integers for which a integer solution exists for the equation x^2+71*y^2=z (x,y,z positive integers)

multiple of 43 are 215 69660 92020 541456

Now consider the equations:

x^2+71*y^2=215
x^2+71*y^2=69660
x^2+71*y^2=92020
x^2+71*y^2=541456

there are Always non zero integer solutions x and y

541456 is about 271*999*2 i wonder Why???

so multiple of 86, 92020 69660 and 541456 are congruent to (271*999-1) mod 172

this implies (92020 69660 541456 are also 6 mod 13) that 92020 69660 and 541456 are congruent to 344 mod 2236

and I remember that 92020=69660+22360

this could suggest why multiple of 43 are congruent to plus or minus 344 mod 559

infact (271*999-1) is congruent to 172 mod 559

(271*999-1) is congruent to -172 mod 215

pg(51456) and pg(541456) are probable primes

observing that 51456 is congruent to 508 mod 542 and to 507 mod 999 using CRT:

solutions are 51456+541458n

infact lcm(542,999)=541458=541456+2

331259= 11*2^10+39*2^13+507

51456*9 is congruent to 700^2 mod (164^2)

541456=51456+700^2

(700^2-164^2)/51456=3^2

(10*700^2-164^2)/3^2=541456

So i wonder inf there are infinitely many primes pg(k) with k multiple of 86 with the property that k is congruent to 164^2 mod (268) as pg(541456) and pg(92020)
More generally if k is multiple of 86 and k is multiple of 3, 69660 is the example

k/3 is congruent to - 164^2 mod 268

if k is not multiple of 3 and k is multiple of 86, then

Examples are 92020 and 541456 are congruent to 164^2 mod 268

541456=271*9*999*2-18

271*9=2439

Maybe this explain the fact that 541456/41 and 51456/41 have a repenting term 2439 and so they are 10^m mod (700^2+239)*2^8/(271*9)=51456

(2439*111-1)*2-(2439*201-239)=51456

From above reasoning it results that

541456 and 51456 are both congruent to -239 mod 245=7*25

In subastance here the formulas

541456=245*2211-239
51456=245*211-239

Where 211 is prime and 2211 is a multiple of 67 as 51456

(541456-92020) is a multiple of 559 and 67

(541456-92020)/(67*3)=2236

69660+22360=92020

But ed can substitute 67*3 with 2211/11

So (541456-92020)*11/2211=2236

i note also that (541456-210*1001+13)=331259

92020 is so congruent to (245*2211-239) mod (3*67*43*13)

and (245*2211-239=541456) is congruent to (331259+13) mod 559

i think that it shuould be possible to prove that when pg(k) is prime and k is congruent to 6 mod 13, (examples known 215, 69660, 92020, 331259, 541456) then k is congruent to plus or minus (344-13s) mod 559
with s=1 in the case of 331259 which infact is congruent to 331 mod 559 and s=0 in all the other cases where k is a multiple of 43

i think that something interesting could be found examining this formula:

f(x)=((71*6^x+4)/10)*6^2-13 for x integer

x=4 f(4)=331259
331259 congruent to -13 mod 9202

i think that the study of this function could shed light on these numbers

331259 is congruent for example to -7 mod 55211=f(3)

331259=92020+239239

note that also 239239 is congruent to -13 mod 9202

239239 is congruent to 546 mod 559

215*(239239-546)/559+215=92020

I suspect that 546 is not a random residue

infact there are primes pg(k) with k multiple of 546

for example pg(75894) is prime and 75894 is multiple of 546

modulo 559 75894 is 429 (92020=(429^2-1)/2)

(239239-546)/559+2=429

so( ((239239-546)/559+2)^2-1)/2=92020

(239239+(12^2-1)*4)=429

Given the attention this question has received, it is disappointing that it is closed. I have a more concise answer that I cannot post other than as a comment: 245⋅22…11−239=245(2000*(10^m−1)/9+211)−239. Using the congruences: 245≡−1, 2000≡−9, 451≡0mod41 we get (−1)(−9)10m−19+(−1)(211)−239≡10m−1−211−239=10m−451≡10m

245 is congruent to -1 mod 41

modulo 41 we have 245*211=-211

-239-211-1 =451 congruent to 10^m mod 41

pg(451) is prime

((71*6^4+4)-344)/559=164

but we saw before that

51456=(700^2-164^2)/9
541456=(10*700^2-164^2)/9

so we arrived to this:

(700^2-(((71*6^4+4)-344)/559)^2)/9=51456
(10*700^2-(((71*6^4+4)-344)/559)^2)/9=541456

So multiple of 43

215 69660 92020 and 541456 are of the form

s(71*6^4-340)/52 +r with r being the residue 1763s+r such that is congruent to 10^m mod 41

340-52=288 maybe this explain why 92020 is congruent to 288 mod 323???

I think yes because 92020=52*1763+344

(71*6^4+4)
is congruent to 344 both modulo 1763 and modulo 559

(92020-344)/559+(92020-344)/1763=6^3

92020=(71*6^4+4)
331259=(71*6^6+4+10)/10

71*6^6 is congruent to -12^2 mod (71*6^4+4)

2236=(331*1001-81*143)/143

Anyway we have a fact:

92020 and 331259 are congruent to 71*3^3 mod (239*13)..so I wonder if there are infinitely many of such exponents

By the way I note that

541456 is congruent to 71*3^3 mod (1001)

541456 is congruent to 331*1001+71*3^3 mod (1001*13)

Or 541456 is congruent to 71*3^3 mod (11*1001)

so 541456=71*3^3+11011n where 11011 is 3^3 in base 2

71*3^3 is congruent to 344 mod 143

I think that this could be useful

It holds 541456-(331259)+13-210*1001=0

So this could be useful if you have in mind that 331259 is congruent to 71*3^3 mod (239*13)

92020=541456-13*(449*77-1)
331259=92020+239*1001

it is clear that there is a relationship

541456 is congruent to (331259+71*3^3+111) mod (239*13)

331259=92020 mod(239*13)

so 541456 is congruent to 2*71*3^3+111 mod (239*13)

92020 and 331259 are congruent to 71*3^3 mod (239*13) and to 71*3^2+13 mod (1001)

92020 and 331259 are congruent to 929 mod (1001)
541456 is congruent to (929-13) mod 1001

69660*559 is congruent to (331259-559-331) mod (1001)

69660*559 is so also congruent to (92020-559-331) mod (1001)

so 69660*559 is either congruent to (x-890) mod 1001 or to (x-903) mod (1001)

where x=541456, 331259, 92020 numbers congruent to 6 mod 13

331259 and 92020 are -72 (mod 1001) and so 69660*559 is congruent to (x-890) mod (1001) with x=331259, 92020
if x=541456 not congruent to - 72 mod 1001, then

69660*559 is congruent to (x-903) mod 1001

but this is equivalent to say

69660*559 is congruent to either (x +111) or (x+111+13) mod (1001)

x=331259, 541456, 92020

541456, 331259, 92020 are congruent to either (13*2-111) or to (13*3-111) mod 1001

69660 is congruent to 591 mod 1001

591*559 is congruent to 39 mod 1001

69660/3 is congruent to (14^2+1) mod 1001

92020 and 331259 are congruent to -(14^2-111) mod 1001
541456 is congruent to -(14^2-111-13) mod 1001

so

69660 is congruent to 591 mod (1001)
331259 is congruent to 559*591-111 mod (1001)
541456 is congruent to 559*591-111-13 mod (1001)

92020 the same as 331259 if i am not wrong

so 69660 is (14^2+1)*3 mod 1001

331259 is congruent to 559*3*(14^2+1)-111 mod 1001...

-(69660/81) is congruent to 141 mod 1001

so

541456 is congruent to (4*(14^2+1)-860-13) mod (1001)

92020 and 331259 to (4*(14^2+1)-860) mod 1001

69660 is congruent to 3*(14^2+1) mod 1001

14^2+1=197 is a prime

69660/81=860

541456 is congruent to 3*860-41*5*13 mod (1001)
331259 and 92020 are congruent to 3*860-41*5*13+13 mod (1001)
69660 is congruent to -2*41*5 mod 1001

541456 is congruent to 344-13*2^5-13 mod 1001
331259 and 92020 are congruent to 344-13*2^5 mod 1001

541456 is congruent to (344+13*2^5-13*2^6-13) mod 1001
92020 and 331259 to (344+13*2^5-13*2^6) mod 1001
69660 to (344+13*2^5+13*2^6) mod 1001

-(344+3*13*2^5) is congruent to 410 mod 1001
-(344-13*2^5) is congruent to 72 mod 1001
-(344-13*2^5-13) is congruent to 85 mod 1001

69660 is -410 mod 1001
331259 and 92020 are -72 mod 1001
541456 is -85 mod 1001

pg(2131) pg(2131*9=19179) and pg(92020) are probable primes

2131 is congruent to -1^2 mod 164
19179 is congruent to - 3^2 mod 164
92020 is congruent to 4^2 mod 164

92020 is congruent to 43*2^3 mod (2132) and congruent to 43*3^2 mod 2131

19179 is congruent to -81 mod (2131+9)
2131 is congruent to -9 mod (2131+9)
92020 is a multiple of (2131+9)

This implies that

(92020-2131) is congruent to 9 mod 4280
(92020-2131*9) is congruent to 81 mod 4280

92020 is congruent to 2140 mod 4280

92020 s congruent to 2140 mod 8988

2131=p is prime
pg(2131) is prime and pg(2131*9=19179) is prime

pg(92020) is prime

pg(69660) is prime
92020=69660+2236*10

92020 can be written both as (2236*43+344) and (2132*41+344)

so 92020 leaves the same residue 344 mod (41*2236) and mod (2132*43) so 92020 is also congruent to 344 mod 1763

Also 541456 and 69660 multiple of 86 are 344 mod 2236

So the exponents multiple of 86 are 69660 92020 and 541456

They are congruent to 344 mod 2236

But only 92020 is congruent to 344 modulo 2236, mod 1763 and modulo 2132=p+1=2131+1 where this prime p 2131 gives the other two probable prime pg(2131) and pg(19179)

19179 is congruent to -214 mod 1763

92020 is a multiple of 214

-2131*3^2*430 is congruent to 344 mod 1763

92020=214*430

so

modulo 1763

-2131*3^2=214

92020=214*430 congruent to 344 mod 1763

pg(331259) is prime
331259 is congruent to 6 mod 13

331259 is congruent to 344-559*3 mod (2132)

pg(56238) and pg(331259) are probable primes

the difference 331259-56238 is congruent to - 7 mod (2132*43)

331259 is -7 mod 13, 56238 is 0 mod 13

331259+7-56238=43*2132*3

this is equivalent to 92020+239*1001+7-56238=43*2132*3

92020+239239=331259

239239 is congruent to (56238-344-7) mod (2132*43)

92020 is congruent to 344 mod (2132*43)

331259 is congruent to (56238-7) mod (2132*43)

92020 is congruent to 344 mod 2132

331259 is congruent to 344*2+111 mod 2132

56238-7 is congruent to 344*2+111 mod 2132

pg(451=11*41) is prime pg(2131) is prime pg(92020) is prime

92020 is congruent to -2131*9*430 both mod 1763 and mod 451 -2131*9 is 214 mod 451 and mod 1763

92020 is congruent to -215 mod (6149=11*43*41)

541456 is congruent to 344 mod 6149

(215+2131*9-1)/43=451

11*43*41 mod 2131 is 214

92020=214*430

Mod 2131

214*430 is 387

So 92020 is congruent to 387 mod 2131

-(1763*11) mod 2131 is 71*3^3

331259 and 92020 have something tondo with 71*3^3

In fact 92020 and 331259 are congruent to 71*3^3 mod (239*13)

71*3^3=2131*10-41*43*11

This means that 331259 and 92020 are congruent to (2131*10-451*43) mod (239*13)

and to (21310-451*43+13) mod (1001)

541456 is congruent to (21310-451*43) mod (1001)

Consider this set of congruences:

x==-13 mod 214
x==6 mod 13
x+13=344 mod 43

Using CRT calulator I found the solution

x=92007+119626k

92007+13=92020 and pg(92020) is prime

92007+119626*2=331259 and pg(331259) is prime

another curio:

pg(1323) is prime

pg((1323*10+3)*3=39699) is prime

1323 is congruent to -215*3 mod 984

39699 is congruent to -215*3 mod 984

Pg(69660) is prime

69660 is multiple of 215*3=645

1323/3=441

69660 is congruent to -215*3+441 mod 984

Or to -215*3+21^2 mod 984

modulo 984, the number 69660 is a perfect square (42*9)^2
modulo 984

69660 is -1323*108=-(42*9)^2

as you can see 69660 is -1323*108=-(42*9)^2 mod (6^3)

now i consider this modular equation

1323x is congruent to -6^3 mod 984

the soluton wolphram gives me is (40+328n)

for n=-1 you get -288

92020 is congruent to 288 mod 323
and 215, 69660 and 541456 are + or - 108 mod 323

modulo 328:

21^2 is -215

-69660=21^2*42^2 modulo 328

-92020=107*42^2 modulo 328

-215 and -69660 in a certain sense are perfect squares (21^2 and 378^2)

378^2 is congruent to 21^2 mod (42*323)

92020=428*215

428 is congruent to -13 mod 21^2

92020 is congruent to 215*10^2 mod (215*328)

Ah ah this is weird

Also 92020 is a perfect square

-210^2 mod 328

Infact 428 is 10^ 2 mod 328

So mod 328 92020 is -210^2

1323 and 39669 are 11 mod 328

69660 modulo 328 is 11*108

Modulo 328 92020 is 11*4*(108^2-1)

-69660 is congruent to 204 mod 984

204=645-21^2

69660 is a multiple of 645

1323=42^2-21^2

1323*108 mod 984 is 204 as -69660 mod 984

69660 is congruent to 21^2-645 mod 984
92020 is congruent to 13^2-645 mod 984

1323 is 11 mod 328
39699 is 11 mod 328
69660 is (11+21^2) mod 328

1323 is 339 mod 984
39699 is 339 mod 984
69660 is (339+21^2) mod 984

69660 mod 984 is 780

42^2 also is 780 mod 984

this because
1323 is congruent to 339 mod 984

69660 is congruent to (339+21^2) mod 984

so
69660 is congruent to (3*21^2+21^2=42^2) mod 984

239*7*11 is -1 mod 172

69660 92020 and 541456 are congruent to (239*7*11-171) mod 2236

I notice that 92020 and 331259 are both congruent to 5 mod (239*7*11)

331259 is congruent to (239*7*11-171-13) mod 2236

modulo 2236 infact

239*7*11-171 is 344

239*7*11 is 515 mod 2236

so

239*7*11+1 so is 516 mod 2236

69660 is divisible by 516

92020, 541456 are 172 mod 516

331259 is -13 mod 516

-69660 is congruent to 1323 mod (239*11)
69660 is congruent to 6 mod 13
using wolphram alpha solution is
:
69660=1306+34177*2

92020=1306+22360+34177*2
331259=1306+22360+34177*9

what is quite clear is that

there are k such that pg(k) is prime with k congruent to 6 mod 13 and k of the form 215+1001s-287 (case 92020 and 331259) and 541456 which is of the form 215+1001s-287-13

so k is of the form -72+1001s or -85+1001s

215 is congruent to -(929-11*13) mod (7*11*13)
92020 is congruent to (929) mod (7*11*13)
331259 is congruent to (929) mod (7*11*13)
541456 is congruent to (929-13) mod (7*11*13)

929 and 331259 are primes congruent to -72 mod (7*11*13)

so mod 143

215 331259 92020 ...are plus or minus 71 mod 143
541456 is (71-13=58) mod 143..

215 is congruent to -786 mod (11*13*7)

786 is a number of the form 71+143*s

331259 and 92020 are congruent to 929 mod (11*13*7)

929 is of the form 71+143s

so these exponents leading to a prime with k congruent to 6 mod 13 have this property

215 is congruent to (71+143s) mod 1001
541456 is congruent to (58+143s) mod 1001
331259 is congruent to (71+143s) mod 1001
92020 also
I dont know 69660??? maybe not but 69660 is the only one multiple of 3

i note that 215 541456 331259 92020 69660 are congruent to plus or minus (71+r) mod 1001

where r is a 13-smooth number which is not a 11 smooth number

r infact can be one of these numbers: 845, 858 or 520

520=71+18^2+14^2
845=71+18^2+14^2+18^2+1
858=71+18^2+14^2+18^2+14

multiple of 43 (215, 69660, 92020, 541456) are congruent either to phi(323)=288 mod 323 or to +/- phi(324)=108 mod 323

phi is the Euler function

215-eulerphi(324)=107

92020 is a multiple of 107, 69660 is 3 mod 107...the thing becomes too difficult

multiple of 215=6^3-1 are 69660 and 92020

69660=324*215
92020=428*215

324 and 428 are numbers n such that

n-eulerphi(n)=6^3

is it a chance that the other multiple of 43 (and not multiple of 215), that is 541456 is congruent to 6^2 mod 428?

Another way to see the problem :

215 69660 92020 541456 are congruent to plus or minus K mod 323

Where K is either 215 or 288

215 and 288 are numbers of the form 41s+r where r is in the set (1,10,16,18,37)

So 215 69660 92020 541456 are of the form 41a+b (b in the set 1,10,16,18,37)and congruent to +- (41s+r) mod 323

215 and 288 are numbers of the form (3+n)*71+2^(n+1) for n non negative integer

note that 12345679*323 divides (10^288-1)

12345679*323*81 divides (10^(phi(323)-1)

phi(323)=288

92020/324 gives a repeating decimal period of 012345679

92020 is congruent to 288 mod 323

215, 69660, 541456 are of the form k*(41s+r)*323 + or minus 215, where r is in the set (1,10,16,18,37) and k some nonnegative integer

541456=323*1677-215
215=0*(41s+r)+215
69660=323*215-215

the only exception is 92020 which is congruent to 16 mod 41

92020 is of the form 323*284+phi(323)

now pg(1323) and pg(39699) are probable primes

1323=441*3
39699=4411*3^2

numbers of the form 441...4411...are

(397*10^n-1)/9

1323 and 39699 are congruent to -645 mod (41*3*2^3)

(397*10^n-1) is divisible by 9209 for a certain n...

397*10^5-((397*10-1)/9-126)*1001-5^3 is a multiple of 92020

39699 and 1323 are of the form 328n^2+11

maybe there are infinitely many of such exponents

541456 is a multiple of 787

69660 is a multple od (787-13=774)

541456-(222*2021+787-13)=92020

331259-118*2021-(787-13*2)=92020

from the equation above

541456-(222*2021+787-13)=92020

if we work mod 787

0+715+13=728

so 92020 is congruent to (3^6-1) mod 787

92020 is also congruent to 43*2^n mod (787-13)

92020-69660=22360 is also congruent to 43*2^n=688 mod (787-13)

69660 is congruent to (3^6-1-18^2) mod 787

22360 is congruent to 18^2 mod 787

92020=69660+22360

92020 is congruent to 3^6-1 mod 787

69660 is divisible by 18^2 and it is of the form (2^j*3^k)-18^2 with (2^j*3^k) congruent to 3^6-1 mod 787

69660 is also congruent to 18^2 mod (107*2)

92020 instead is a multiple of (107*2)

92020 is a multiple of 428

69660 is congruent to 18^2 mod 428

541456 is 0 mod 787

curious that pg(787-428=359) is prime

215 69660 92020 541456 are +/- 344 mod 559

lcm(215,559,344)=22360

69660+22360=92020

22360 is of the form 787n+324

maybe working in the field F(787):

the inverse mod 787 of 107 is 559

92020 is congruent to (107*73) mod (107*787)

541456=107*2942+787*288+6

541456=107*(17*19*10-phi(17*19))+787*phi(17*19)+6

-215 (because odd so signus -), 69660, 92020, 541456=G are solutions of the diophantine equation:

107x+787y+6=G

I think that we think that the inverse mod 239 of 428 is 43...

43*428-1=7*11*239

and that lcm(428,559)=239*7*11*13+13

then maybe we could explain why

541456+(72)+13 is congruent to 0 mod 1001

331259+72 is congruent to 0 mod 1001=7*11*13

92020+72 is congruent to 0 mod 1001
...

A strategy could be working in the field F(239)

107*172 is congruent to 1 mod (239*7*11)

so the inverse of 172 mod 239 is 107

172=344/2

215 69660 92020 and 541456 are all congruent to +-344 mod 559

92020 is a multiple of 107

331259=92020+239239

69660 92020 and 541456 are multiple of 172...but 172 in the field 239 is (107^(-1))

multiple of 86 are 69660 92020 541456...

69660*(1-1/172) is congruent to 0 mod (3^2*5*19)
92020*(1-1/172) is congruent to 0 mod (3^2*5*19=855)
541456*(1-1/172) is congruent to 513=19*3^3 mod (3^2*5*19)

855 is congruent to -1 mod 107

69660*(1-1/172) is congruent to -9^2 mod 107
92020*(1-1/172) is congruent to 0^2 mod 107
541456*(1-1/172) is congruent to -3^2 mod 107

-2^9-7^3 is congruent to 6^3-1 mod 107

-2^9=-19*3^3+1

6^3+7^3 is the famous 559

344=7^3+1

lcm((6^3+7^3),(7^3+1),(6^3-1))=22360

69660+22360=92020!

curious that 559 is congruent to (1-2^9) mod 107

and to (1+2^9) mod 67

maybe this is not chance because there are primes pg(67s) like pg(51456) but i dont know

92020 for example is congruent to (6^3+2^9) mod 787
541456 is a multiple of 787
69660 is congruent to (6^3+2^9-18^2) mod 787

6^3+2^9=3^6-1

so 3^6 congruent to 2-7^3 mod 107

541456=344*1574=-2^9*1574 mod 107

2^9 mod 107 is 84

1574 mod 107 is 76

so 541456=-84*76 mod 107

but incredibly

541456=-84*76 mod (107*10*2^9)

and the difference 541456-107*10*2^9 is divisible by 456

in other words:

541456=(6^3+7^3+2^9-1)*2^9-84*76

if you reduce the right side mod 107

6^3=2
7^3=22
2^9-1=83

2+22+83=107

69660 is congruent to (6^3+1)*(6^3-1) mod (5*43*107)

infact 324 is congruent to (6^3+1) congruent to 3 mod 107

69660=324*215

so 69660 is congruent to (6^6-1) congruent to 3 mod 107

69660 is congruent to (6^6-1) mod (215*107)

215*107 is congruent to 1 mod (71*6^2)

is this connected to the fact that for example 92020 is about 71*6^4???

69660-(6^6-1) divides 92020

92020=69660+22360

69660 is a multiple of 645

22360=215*107-215*3

69660 is congruent to 645 mod (215*107)
69660 is congruent to -22360 mod (215*107)

using wolphram (chinese remainder theorem) this leads to numbers of the form:

645+23005n

6^6-1 and 69660 are so nummbers of the form 645+23005n

92020 is divisible by 23005

pg(331259) is probable prime

331259 is congruent to 9203 (prime of the for 107n+1) mod (107*215-1)

((215*107-1)*12^2+12^2)/10-(239239+13)=92020

and 541456 is congruent to the curious 12341 mod (215*107), 12341=7*41*43, this remainds me the famous 1763

69660 is congruent to 6^6 mod (71*6) congruent to 222 mod (71*6)

69660/3-222/3-786=22360

a rapid calculation from above leads

69660+1 congruent to 2^6*(6^3+1)+2^15 mod (215*107)...
this yields 69660+1 is congruent to (13888+9763) mod (215*107)

69661-13888-9763=46010

46010 is a divisor of 92020

69660=3^2*(71^2+71*36+71*2+1)

so this leads to a polynomial x^2+36x+2x+1

71+6^2=107 so I arrived to

69660=3^2*(71^2+71*36+71*2+1)

feom this I found this polynomial:

(x+19)^2-9^2*10^2=0

whcih has solutions:

x=71

x=-109

71*109+1=6^2*(6^3-1)

from this we obtain

(9^2*10^2-19^2+1)=6^2*(6^3-1)=7740, which divides 69660

69660 has also the factorization 6^2*(44+1)*(44-1)

pg(331259) is probable prime

331259 is congruent to (71*18^2-1)=215*107-2 mod (456) and mod (26^2)

maybe there is a link 513=2^9+1=19*3^3 and 456=2^3*3*19, the difference is 15*19 congruent to 1 mod 71

215*107-15*19+6^2*(6^3-1)-90^2=22360

where 215*107 15*19 and 6^2*(6^3-1) are 1 mod 71

pg(331259) is probable prime

331259=(13*(71*11*18^2+1)+71*18^2+1)/10

71*18^2+1=23005 which divides 92020

69660-71*18^2=6^6

and pg(69660) is prime

(13*(71*11*18^2+1)+71*18^2+1) is congruent to -13*10 mod (215*107)

maybe it has something to do with the fact that

23005*143 is congruent to 1 mod 71

infact 23005 is 1 mod 71 and so 23005*143 is 1 mod 71

I don't know how to connect things but

107*73 is congruent to 1 mod 71

and so all the numbers of the form

(73+71s)*107 are congruent to 1 mod 71...

numbers of the form 73+71s are 215 and 428

maybe it is not a chance that there 92020 is divisible by 428 (and also 215)

92020 is also divisible by 23005 which is of the form 73+71s+1

215 and 428 in the field Z 71 have 36=6^2 as inverse

maybe it is not chance that 541456 is 36 mod 107

and 92020 is (324-36=288) mod 323

pg(359) is prime

359-71=288 the famous 288=phi(323)

pg(359) comes after pg(215) which is probable prime

359 and 215 are numbers of the form 71+144s

69660=71*324+6^6 if we rearrange

324*(71+144)=69660

I note that 359=19^2-2

pg(359) is prime

215, 69660, 92020, 541456 are congruent to plus or minus (19^2-73s) mod 323

infact 215=19^2-73*2 and 288=19^2-73

pg(51456) is prime

2^4*(73*49-361)=51456

pg(67) is prime

215 is a number of the form 67+2+73s

also 288 is a number of the form 67+2+73s

maybe it is not a chance that also 1456 the final digits of 51456 and 541456 is of the form 69+73s

(1456-69-1)=1386

1386*239+5=331259 and pg(331259) is prime
(1386-1001)*239+5=92020 and pg(92020) is prime

215 288 and 1456 in the field Z 73 have 18 as inverse...curious that 1456*18-1=73*359 and pg(359) is prime

by the way 541456 is congruent to 1456 mod 288

108 and 288 are numbers of the form 54+18s

215 69660 92020 541456 are congruent to +/- (108 or 288) mod (18^2-1)

also 774 is a number of the form 54+18s and 774 divides 69660

69660 is divisible by 18^2 which is a number of the form 54+18s, by 774 which is again a number of the form 54+18s, by 90 which is a number of the form 54+18s and by 108 which is of the form 54+18s

288 and 108 are two smooth numbers congruent to 18 mod 90

71*18+224^2+700^2+2=541456
71*18+224^2+2=51456

224^2+700^2 is congruent to 215*2=43*5*2 mod (71*18)

224^2=(67*3-5)*2^8

51456=67*2^8*3

curious that 224^2+1 and 700^2+1 are both prime

69660=(9*71+6^4)*6^2

645 divides 69660

92020=69660+22360

22360=23005-645

23005=215*107 divides 92020

23005 is congruent to 1 mod 71

curious that there are pg(k) primes with k multiple of the reverse of 645=546

645=(18^2-1)*2-1

92020=215*428

Both 215 and 428 are congruent to 2 mod 71

The inverse of 215 in the ring F426 is 107

curious that pg(7) is prime , 7 is prime...the inverse of 7 in the ring F426 is 61...(61*7-1)/426=1

541456 is a multiple of 787

in the ring F426 the inverse is 367

367*787=288828

288828=6^2*(8*1003-1)

(359*89-1)/426=75=(8*1003+1)/107

8*1003 is 1 mod 71

pg(359) is prime and the inverse mod 426 of 359 is 89

remark:

in the ring F426

a^2 is congruent to 1 mod 426, for a=143

(143*143-1)/426=288

lcm(426,288)=143^2-1

maybe there is a link to the fact that

541456-429*10^3-143^2+13=92020

in the ring Z428

(215*215-1)/428=108

(211*71-1)/428=35

215 69660 92020 and 541456 are congruent to + or - (108 or 35) mod 323

pg(3371) is probable prime with 3371 prime
pg(331259) is proabble prime with 331259 prime

3371=59+46*72
331259=59+46*10^2*72

I would conjecture that there are infinitely many prime pg(k) with k a prime of the form 59+72s.

I note that 46*10^2+1=4601 which divides 92020

this is equivalent to primes of the form (72b-13), as Peter Scholze pointed out

I note that 331259 equiv (congruent) 999 mod (3371-1)

3371 and 331259 leave also the same resiude 13 mod 46

so they are primes of the form 59+1656x

1656 is a multiple of 46 naturally

I think it is not a chance that 4601 divides 92020

72*4601-13 is 331259

so 3371 and 331259 are primes of the form 72*k-13 with k congruent to 1 mod 46

because 331259=239239+92020 follows that

(239239+13) is a multiple of 428 and (239239+13)/428=the famous 559

3371 and 331259 are primes of the form 59+3312x

331259=(4601*4-1)*13+4601*20 where 4601*20=92020

because 59 and 3312 are coprime we expect infinitely many primes of the form 59+3312x so there could be infinitely many pg(59+3312x) primes with 59+3312x prime

I note that 3371, 331259, 92020 are all congruent to 5 mod 11

so 3371 and 331259 are primes of the form -589+792s

curious that 589*792+1=683^2

putting all together we have that

3371 and 331259 are primes of the form -14845+18216s.

infact 3371 and 331259 are 13 mod 46, -13 mod 72 and 5 mod 11

3371 and 331259 leave the same remainder 82 (mod 23*11)

so they are also primes of the form 23*11*s+82

3371=82+253*(13)
331259=82+253*(13+6^4)

3371 and 331259 are primes of the form 59+414s where 414 is 1 mod 59 and 0 mod 23

4601 which divides 92020 is 1 mod 23 and -1 mod 59

so 3371 and 331259 are primes congruent to 59 mod (46*72)

3371 and 331259 are congruent to 13 mod 46

(3371-13)/46=73 which is congruent to 1 mod 72

(331259-13)/46=19*379 which is congruent to 1 mod 72

19*379 is congruent to 51^2 mod 46

51^2 is congruent to 1 mod 26

4601 which divides 92020 is congruent to -1 mod 26

19*379+1 is a multiple of 26

4602 is a multiple of 26 and 59

19*379+1-4602+1=51^2

51^2 is congruent to 1 mod 26
51^2 is congruent to 2 mod 23

4601 is congruent to 1 mod 200 4601 divides 92020

19*379 is congruent to 1 mod 200

we can say

3371 and 331259 are congruent to 13 mod 46

3371 and 331259 are congruent to -13 mod ((46s+1)*72)

3371 infact is congruent to -13 mod (47*72)
331259 is congruent to -13 mod 4601*72 where 4601=46s+1

92020 is 0 mod 4601

72 is the least integer such that

4601*x is congruent to 1 mod 337

3371 is 1 mod 337

69660 (divisible by 215*3) is congruent to 215*3 mod 4601

4601 is congruent to -1 mod 59

22360 is congruent to -1 mod 59

4601 divides 92020

69660+22360=92020

774 divides 69660, 428 divides 92020

331259 is congruent to -13 mod (774*428)

774*428 is congruent to 1 mod 337

3371 is 1 mod 337

91*100 is congruent to 3371 mod 337

331259 is -12 mod 337 and -13 mod (428*774)

331259 and 3371 are primes of the form 59+3312s

3312 is congruent to 2 mod 331

92020=428*215 is 2 mod 331

428*774 is congruent to -59 mod 331

curious that pg(1323) is prime and 1323 is congruent to -1 mod 331

69660 is (6^4+4) mod 230

so 69660 is 200 mod 230

the inverse mod 230 of 43 is 107

so 5*324 is 10 mod 230

5*94 is 10 mod 230

5*47*2=5*94

the inverse mod 230 of 47 is 93

so 10 is congruent to 930 mod 230

92020 is congruent to 20 mod 920

43*107*72 is congruent to 1 mod 337

43*107*72 is congruent to 3371 mod 337

43*107*72-1=331271, the number is 3371 with a 12 sandwiched 33(12)71

the difference between 331271 and 3371 is 1093*300 where 1093 is a Wieferich prime

300 is the least integer such that 1093x is congruent to -1 mod 337

331259 is congruent to -12 mod 331271 and to -13 mod (43*107*72)

3371 is congruent to -1001 mod 1093
331259 is congruent to -1001-12 mod 1093

working mod (-1093)

3371 is -1001

331259=3371+327888k

working mod (-1093)

331259=-1001-12s

3371 is congruent to -12*28 mod 337

the inverse mod 337 of 28 is 325

3371*325 is congruent to -12 mod 337

so 3371*325 is congruent to 331259 mod 337

and so 331259 is congruent to -12 mod 337

Se know that 69660 is congruent to 6^6 mod 71

23004 (23005 divides 92020) has the form 288*3^x-18^2

Also 69660 should have the form 288*3^x-18^2

6^6 is 648 mod (71*324)

(69660+648)/(288*3^2-324)=31

69660 is divisible by 540=288*3-324

given 69660 is congruent to 2^3*3^4 mod 213

after some passage :

111 is congruent to 2^7*(2^7+1) mod 213

this implies

324 is congruent to 2^7*(2^7+1) mod 213

and 69660=324*215 so is congruent to 2*(2^7+1)*2^7 mod 213

(2^7+1)=129 is a divisor of 69660

I notice that 128*129=2^7*(2^7+1) is congruent to -1 mod 337

so 3371 is congruent to (128*129+2) mod 337

I notice that 128*129+2 is a multiple of 359 and pg(359) is prime

I notice that 128*129 is congruent to -7^3 mod 3371

3371 is congruent to 336*128*129 mod 337

336*128*129-3371=5544661=10*(2^10-1)*271*2

331259 is congruent to 5544649 congruent to 325 mod 337

so 331259 is congruent to 11*83*6073 mod 337

and so
331259 is congruent to 239*7 mod 337

if we multiply by 143

143*331259 is congruent to 239239 mod 337

331259=92020+239239

because 239239 is -(2^5+1) mod 337, then

331259-92020 is conruent to -(2^5+1) mod 337

92020 is -10 mod 9203 (this strange prime is 1 mod 43*107)

331259 is 9203 mod 23004=215*107-1=71*...
331259 is -7^2 mod 9203

71*6^6+1 is congruent to 2131*1399 mod 9203

pg(2131) is prime

71*6^6 is congruent to 8699 mod 9203

71*6^6 is congruent to 8699 mod (9203*359)

pg(359) is prime

331259 and 71*6^6 are both congruent to -72 mod 331 or (259) mod 331

(71*6^6+72)/331-(331259+72)/331=9007=10^4-3*331

-71*6^4 is congruent to 2 mod 331 and 92020 is congruent to 2 mod 331

in particular (71*6^6+72) is divisible by (92020-2=92018, which is multiple of 331)
(331*139-1)/2=23004=71*...
23005 divides 92020

139*331 is -1 mod (43*107)

curious that 331259 is congruent to -(9203*2^2+1) mod (331*139)
(331*139-215*107)+6^6=69660

331259 is congruent to -2000 mod 9007=10^4-3*331

71*6^6 also is -2000 mod 9007

curiously 92020 is congruent to -(84^2+1) mod 9007

after a post on mathexchange I had a proof that:

6^(6+35j) for some j, has the form 648+23004s

69660 has the form 648+23004s

69660 is divisible by (6^5-6^2)

6^5+1=7777

6^5 is congruent to (6^2+1) mod 71

i didn't realized that

331259 is congruent to 9203 mod (71*324)

331259 and 9203 are primes

331259 and 9203 are both congruent to 44 mod 71

(331259-9203)/71=4665

6^6=46656

4665 has in common with 46656 the first four digits

6^6 is congruent to 44^2 mod 2236

I remember that

69660=(44+1)*(44-1)*2

and 69660 is congruent to 6^6 mod 23004

331259=(2236*2+193)*71+44

193 is 1936=44^2 with a six truncated

mod 2236 331259 is 331

69660=(44^2-1)*36=(193*10+6)*36

(6^6-1) is congruent to 1935 mod 2236

1935=(44^2-1) divides 69660

I looked at the factorization of 215, 69660, 92020, 541456 (multiple of 43 leading to a pg prime)

215=43*5 for example

every prime in the factorization of one of these numbers is a quadratic residue mod 71. This is quite surprising.
for example 43 is a quadratic residue mod 71
also 5
but also 787 which divides 541456

331259 is congruent to (4667) mod 6^6

4667 is a multiple of 359 (pg(359) is prime)

-69660 is congruent to 345 mod 359 as 6^6 is congruent to 345 mod 359

69660+6^6 infact is 359*324=359*18^2

consider

71x is congruent to -1 mod 215

359x is congruent to 1 mod 215

x=109+215n

71*(215+109)=23004
71*109=7739

7739+1 divides 69660
23004+1 divides 92020

curious that (359*109-1)/215=182

and 541456 is congruent to 182 mod 2131, pg(2131) is prime

182^2+1=33125 and pg(331259) is prime

359*(109+215x)-1

for x=2 we have 193500

1935 divides 69660

(359*(109+215x)-1)/10^x is an integer for x=2

69660 is congruent to 6^6 mod (71*324) and to -6^6 mod (359*324)

pg(359) is prime

the difference between 359 and 71 is the famous 288=phi(323)

I think it's not chance that 92020 is congruent to 288 mod 323

288=2*6^6/324

in the ring F116316 -69660 is congruent to 6^6 mod (116316)
in the ring F116316 (116316=359*324)
:
-331259 is congruent to 133^2 mod (116316)
pg(331259) is prime

pg(92020) is prime.

92020=215*428

428=21^2-13

pg(1323) is prime. 1323 is a multiple of 21^2

pg(331259) is prime

331259+13 is multiple of (43*107)

331259+13+6^2 is multiple of 9203

9203 is a prime of the form 96^2-13

69660 and 92020 can be written in the form 3+215x+23004y, for some x,y

92020=3*215^2-6^6+1

(6^6-1) is congruent to 3 mod 107

69660 is congruent to 3 mod 107

92020=69660+22360

-22360 is congruent to -324*213 congruent to 3 congruent to (6^6-1) mod 107

22360*36 congruent to 23004 congruent to -3*36 congruent to -1 mod 107

22360*36 is congruent to -5*36 mod (71*324*35)

Divisors of 69660 are 36 and 7740, Numbers of the form 36+107s. 7740=36+107*72 and is congruent to 1 mod 71

4472*36 congruent to -36 mod 23004

22360=4472*5

4472 is congruent to -1 mod 71

And 4472=lcm(344,559) the famous 344 and 559

pg(331259) is prime

331259 is prime congruent to 6 mod 13

331259 is congruent to (2^8+1=257 prime) mod (4473)

22360=(22622-262)

331259 is congruent to 257 congruent to (22622-262)+262 mod (4473)

subtracting 5 we have

1386*239 congruent to 252 congruent to 359*63 mod (4473)

dividing by 63

22*239 congruent to 4 congruent to 359 mod 71

so 22*239 congruent to 4 congruent to 359 congruent to 92020 mod 71

331259=92020+239239

92020=4601*20

4601 congruent to - 5 mofd 71

69660*63 congruent to -(4601+5)*63 congruent to 9*63 congruent to -1 mod 4473

so

69660*63 congruent to -92020-4601*43-5*63 congruent to -1 mod 4473

so

69660*63 congruent to -92020+76*10^2 mod 4473

11701*5 congruent to 43*324 mod 23004, 43*324 congruent to -7*6^4 mod 23004

11701*2 is congruent to 43 mod 71

so 69660 is congruent to -(6^2-1)*6^4 mod 23004

pg(51456) and pg(92020) are primes
51456 and 92020 are congruent to 10^m mod 41

51456 is divisible by 4288
92020 is divisible by 428

4288 and 428 are numbers of the form (386*10^n-8)/9

pg(51456) is prime

51456 is divisible by 2^8 and by 67

51456 congruent to -12*2^8 mod (71*3*2^8)

51456 is congruent to -12*43 mod (71*3)

Pg(67) is prime pg(359) is prime pg(92020) is prime

-67 congruent to 4 congruent to 359 congruent to 92020 mod 71

Remarkably 92020+67=71*1297 where 1297 is a prime of the form 6^k+1

you can start from

(6^4+1) congruent to 5 mod 1292

multiplying both sides by 71

71*(6^4+1) congruent to (359-4) mod (71*323)

71*(6^4+1) +355 is congruent to -288 mod (43*107)

69660 is congruent to 17^2+359=648 mod 23004

-17^2-359 mod 23004 is -22715-22645=-45360

69660 is congruent to 9 congruent to -45360 mod 71

-45360 is even congruent to 9 mod (213^2)

7740 divides 69660

-(71^2-1)=5040 is congruent to 7740 mod (71*3)

from here we have

-1008=-10^3-2^3 is congruent to 43*6^2 mod (71*36)

it holds:

((71^2-1)+71*109+1)/71+108=288

where 71*109+1=7740 divides 69660

maybe there is some connection to the fact that 215 69660 92020 541456 are congruen to +- 108 (or 288) mod 323

probably in the ring Z426 is hidden something special about these numbers

69660 is congruent to 222 mod 426

204*24 is congruent to -6^3 mod 5112

69660 has the curious factorization (387/28)*(71^2-1)

pg(51456) is prime

noting that 359 is congruent to -67 mod 426

we have

359*2^8*3 is congruent to -51456

so 51456 is congruent to 336 mod 426 and mod 5112

336 is congruent to 1 mod 67 and 51456 is multiple of 67

in the ring Z426 the multiplicative inverse of 67 is (336+1=337)

pg(6231=67*93) is prime

6231 is congruent to -159 mod 426

curious that 337*67 is congruent to 1 mod 159

pg(331259) is prime

331259 is congruent to -13 mod (43*107)

331259 is congruent to -13^2 mod (426) and mod 777

541456 is congruent to 10 mod 426 and 10 is also the residue mod 41

my sensation is that the exponents of these primes follow some ver complex logic

331259*5 is congruent to 7 mod (426)

(2^8+1)*5 is congruent to (2^3-1) mod 426

331259*5-6 is a semi-prime congruent to 1 mod 426 and congruent to 17^2 mod 4600 (4601 divides 92020)

331259*5-6=1151*1439 and the difference is 1439-1151=288=17^2-1

pg(2131) is prime

7740 divides 69660 (7740 is congruent to 1 mod 71)

2131 is congruent to 1 mod (71*3)

7740 is congruent to 2131 mod (71*79)

so 69660 is congruent to 19179 mod (71*79)

pg(19179) is prime

curiously also pg(79) is prime

69660 is congruent to 3*28^2 mod (71*79)

i have the vague idea that these exponents are not random at all

92020 is congruent to -3333 mod (71*79)

in conclusion

69660 is congruent to 19179 which is congruent to 6^6 which is congruent to 648 which is congruent to 9 mod (71*9)

92020 is congruent to 4 mod (639)

92020 is congruent to 6^6-5 mod (213^2)

6^6 infact mod 639 is 9

92020 is congruent to 2*641 mod (213^2)

641 is a prime I think it has something special (Euler? Fermat?)

641 is congruent to 2 mod 639 anyway

72^2-21^2*10=774

69660=72^2*90-(21*30)^2

(21*30)^2=396900

pg(39699) is prime and 39699 congruent to 6^6 mod (773)

(2130^2) is congruent to 1 mod 2131 , pg(2131) is prime

69660 is congruent to (60^2-1) mod 2131

pg(2131) is prime
pg(69660) is prime
pg(2131*9) is prime

2131, 69660, 19179 are congruent to (214*3^j) mod 639, for some nonnegative j

2131 mod 639 is 214=107*2

92020 is a multiple of 107

pg(331259) is prime

331259 is congruent to -6*2^6+2 mod 23004 (6*2^6+1 is Woodal prime???)

the inverse of 2^6 mod 23004 is 10

so
331259*10 is congruent to 14 mod (23004*12^2=71*6^6)

if we work mod 639

331259*10 is congruent to -5^4 mod 639

(331259*10+625)/639=72^2+1

so 331259*10 is congruent to -5^4 mod (71*9*(72^2+1))

331259*10 is congruent to 456*10 mod ((72^2+1)*(71*3^2-1))

pg(541456) is prime

541456 is congruent to 2*6^3 mod 638
69660 is congruent to 3*6^3 mod 639

331259*10 is congruent to -5^4 mod (71*(6^6+3^2))

69660 is congruent to 3 mod 107 pg(69660) is prime

107 is the multiplicative inverse of 215 mod 23004

but 69660 is also congruent to 3 mod 651.

651=(2^2-1)*(2^3-1)*(2^5-1) so is the product of (2^p-1) taken over the first three primes 2,3,5

in other words
69660 is congruent to 3 modulo the product of the first three Mersenne primes

92020=2*6^6-(6^4-4)

2*6^6 is congruent to 6^4 mod 23004

2*12^2 is congruent to 4 mod 71

i think that something is involved related to p-adic numbers

69660 is congruent to (17^2+359) mod 71

so 69660 is congruent to (5+4=9) mod 71

92020 is congruent to 359 which is congruent to (17^2-1) mod 71

so 92020 is congruent to 4 mod 71

i think that is in some way related to the fact that 92020 is congruent to (17^2-1)=2*12^2 mod 323

69660 is congruent to (17^2+18^2+(6^2-1))=80=9 mod 71

92020 is congruent to (18^2+(6^2-1))=75=4 mod 71

69660 is congruent to 2*18^2 mod 71

92020 is congruent to (2*18^2-17^2) mod 71

so 69660 is congruent to 9 mod 71

92020 is congruent to (9+66=75=4) mod 71

pg(67) is prime, 67 is congruent to -(17^2-1) mod 71

69660 is congruent to 2*17^2+2*(6^2-1) mod 71

92020 is congruen to 17^2+2*(6^2-1) mod 71

359=17^2+2*(6^2-1)

69660 is congruent to 3*(6^3+71) mod (323*71)

92020 is congruent to (6^3+72) mod (323*71)

69660 is multiple of 18^2

69660 is congruent to 18^2=3 mod 107

maybe 36*2^k is involved...

36*2^8 for example is 9216 which is 92016=71*6^4 without the 0

pg(359) is prime and 359 is -1 mod 36

absolute value of (92020/359) is a power of 2

pg(331259) is prime. 331259 is prime
curiously (maybe it is not chance) 71*648^2/90 is near to 331259

((71/10)*6^2)*(19^2-1)+4=92020

92020*359/100+(7*6^4)/10=331259

359 is of the form 323k+36

6^6+1 is divisibile by 3589 which is 323*11+36

331259 is congruent to 4667 mod 6^6

4667=6^6-323k for some k

23005*4=92020

23005 is congruent to 5 which is congruent to -18 mod 23 (23005 is a multiple of 107 and 5)

541456 is congruent to 331259 congruent to 13 mod 23

331259=92020+239239

so mod 23:

541456 is congruent to 20+16=6^2 mod 23

or equivalently

541456 is congruent to -7*8 mod (23*6^3)

541456 is even congruent to 6^2 mod (23*107*5)

I recall that 69660 is congruent to 6^6 or 3*6^3 mod (23004)

69660 is congruent to 4^2 mod 23

using CRT 541456 is so a number of the form 36+12305s

pg(36) is prime

I could conjecture that there are infinitely many pg(k) primes with k of the form 36+12305*s

(1/18)*(331259*(2^2+1)+(2^6+1))=92020=(5/18)*(331259+13)

the curious 71*72=5112=512+4600 (4601 divides 92020)

92020/5112 is very near to 18

(331259/71)=6^6/10+7/355

(331259/6^6)=71/10+7/(2^5*3^6)

-104*215 congruent to 69660 congruent to 3 mod 107

dividing both sides by 215

-104 is congruent to 18^2 congruent to 3 mod 107

104+18^2=428 which divides 92020

-104*215 is congruent to 69660 which is congruent to 3 mod 107

the inverse mod 107 of 104 is 71

so

-215 is congruent to 69660*71 which is congruent to 213 mod 107

69660*71 is even congruent to -215 mod (107*215^2)

69660*71 is congruent to -1 mod 107

equivalent to:

-(6^3-1) congruent to -69660*6^2 congruent to 213 mod 107

213+215=428=107*4 which divides 92020

324*71 is congruent to -1 mod 107

324*71=23005 which divides 92020

(6^6-1)*215 congruent to 69660 mod (43*107*5*433)

6^6 is congruent to 325=18^2+1 mod (107*433)

(6^6-1)*(6^3-1)-(x/2+3*215^2)=107x

soluion to this equation is x=92020

curious that 331259 (pg(331259) is prime) is congruent to 44 mod (71*4665)

4665 is 6^6 (46656) with the last 6 truncated

(71/10)*6^6 is congruent to (71/10)*6^4 mod 23004

so mod 23004, 331259 is congruent to 9203 mod 23004

9203=(71/10)*6^4+1.4

consider

71*6^k+14

71*6^k+14 is divisible by 10

71*6^2+14 is divisible by 257

331259 is congruent to 257 mod (71*9)

71*6^3+14 is divisible by 1535

331259 is congruent to 1535 mod (71*9)...
and so on

331259 is congruent to 71*6/10+1.4=44 mod (71*3=213)

pg(2131) is prime as well as pg(2131*9) ...2131 is prime and congruent to 1 mod 213

I note that 71*6+14=440 which is 1 less than 21^2=441

pg(441*3=1323) is prime curious

curious that 1323 is congruent to -(13^2-1) mod 213

331259 is congruent to -13^2 mod 213

becasue (71*6^k+14)/10 is congruent to -27 mod 71

331259 is congruent to -3^3 mod 71

1323 is congruent to -(3^3-1) mod 71

for example (331259+27)/71=4666

331259*(6^3-1) is congruent to 7*43 mod (71*43*18)

1535*(6^3-1) is congruent to 7*43 mod (71*43*18)

331259 is so congruent to 1535 mod (71*43*18)

1535=71/10*6^3+1.4

23005 is congruent to 559 mod 774

dividing by 43

535 is conguent to 13 mod 18

so

535*172=92020 is congruent to 4 mod 18 which is congruent to 2236 mod 18

92020=69660+2236*10

92020 is so even congruent to 2236 mod (43*18)

-7*11*239*20-20 is congruent to 92020 mod 23005

331259 is so congruent to -539*239 which is congruent to 9203 mod 23004

and 331259 is congruent to -539*239-20 which is congruent to 9189 mod 23005

331259=92020+239*1001

(331259+539*239+20)/92020=5

43*107*k-1 i think this is the key , for k integre

331259 for example:

(43*107*6-1)*6*2-1=331259

43*107*9-1=(3*6^3-1)*2^6

331259=(3*6^3-1)*2^9-5

so for example 331259 is congruent to 7*43*107 congruent to 9203 mod 23004

43*107*10 is congruent to 1 mod (139*331)

92020 is so congruent to 2 mod (139*331)

331259 is congruent to -72 which is congruent to 71*6^6 which is congruent to 259 mod (331)

71*6^6 is even congruent to -72 mod (139*331)

by the way (331259 is 331 and 331-72=259)

71*6^6 is congruent to -215+71 mod (43*107*5)

From here

71*(6^6-1) is congruent to -215 mod (43*107*5)

From here

71*217 is congruent to -1 mod (43*107*5)

From here

-69660 is congruent to 4600*(6^6-1) mod (43*5*107)

4601 divides 92020

4601*72-13=331259

71*(6^6-1) is congruent to -215 mod 23005

(6^6-1) is divisible by 5

so

71*9331 is congruent to -43 mod 4601

so

71*129 is congruent to -43 mod 4601

71*129+43=9202

331259 is congruent to (71*129+44) mod 23004

331259 is so congruent to 44 mod 71

71*9331 is congruent to -43 mod 4601

from here

9331 is conguent to -43*4277 mod 4601

so

9331 is congruent to -18^2*43 mod 4601

from here

69660=18^2*43*5 is congruent to 9331*5 mod 4601

from here 69660 is congruent to 645=(6^3-1)*3 mod 4601

69660 is congruent to -4600*(6^6-1) mod (9332)

from here

69660 is congruent to 5*4600 mod 9332=2333*2*2

43*18^2 is congruent to 129 mod 4601

331259=(43*324-129)*3*2^3-13

331259 is congruent to -3^3 mod (71*2*2333) 9332/2=2333*2

From 71*6^6 is congruent to 6^2 mod (239*5)

I derive

92020=71*6^4+4 is congruent to 5 mod (239*5)

331259 is congruent to 244=1+3^5 mod (239*5)

3*331259 is congruent to 71^2*2^5 whcih is congruent to 3*71*6^6 mod 331

-3*331259 is congruent to -71^2*2^5 which is congruent to - 3*71*6^6 mod 331

-3*331259 is congruent to 6^3 mod 331

so
6^3 is congruent to -71^2*2^5 which is congruent to -3*71*6^6 mod 331

this leads to 1 is congruent to -3*71*6^3 mod 331

-3*331259 is congruent to 6^3 mod (331*1001)

71*6^2 is congruent to 239 mod 331

71*6^2*1001 is congruent to 257 mod 331

so

239*1001 is congruent to 239*6^4-2 which is congruent to (331259-2) whcih is congruent to 257 mod 331

so

239*1001 is congruent to 3*239*101-2 which is congruent to 331259-2 whcih is congruent to 257 mod 331

331259=92020+239*1001

so

0 is congruent to 239*3*101-2-239*1001 which is congruent to 92020-2 whcih is congruent to 257-239*1001 mod 331

so

0 is congruent to 239*(-698)-2 which is congruent to 92020-2 mod 331

or

0 is congruent to -239*6^2-2 which is congruent to 92020-2 mod 331

so 92020 is congruent to 2 mod 331

so

0 is congruent to 92*6^2-2 which is congruent to 92020-2 mod 331

so

0 is congruent to 3312-2 which is congruent to 92020-2 mod 331

331259 and 3371 are primes of the form 59+3312s

pg(331259) and pg(3371) are primes

92020 is so congruent to 3312 mod 331

so

23005 is congruent to 23*6^2 mod 331

from this the curious thing:

92020 is congruent to 92*6^2 mod 331

92020=4601*20

4601 is congruent to 298 mod 331

541456 is congruent to 298*72-1=21455 mod 331

21455 is congruent to 271 mod 331

it holds:

331259+85 is congruent to 72 which is congruent to -71*6^6-72 mod 4601

so

4673 is congruent to 72 which is congruent to -4457-72 mod 4601

4673+4457+72=9202

331259 is congruent to 9203 mod (4601*5-1=23004=71*324)

-71*6^6 is congruent to 1656*72 which is congruent to 43*107*10*72 which is congruent to -331259 mod 331

92020 is congruent to 2 mod (139*331)

71*6^6 is congruent to -46010*72 mod (139*331)

71*6^6 is congruent to -92020*36 mod (139*331)

259 is congruent to -46010*72 which is congruent to -92020*36 which is congruent to 71*6^6 which is congruent to 331259 which is congruent to 590 mod 331

92020*36+590 for example is 3313310=(331259+72)*10

92020*6^2 is congruent to -259 which is congruent to -331259 which is congruent to -71*6^6 which is congruent to 72 mod 331

so

2*6^2=72 is congruent to -259... mod 331

in the ring Z331 the multiplicative inverse of 215 is 214

so 92020=215*214*2 is 2 mod 331

in Z139 the multiplicative inverse of 215 is 214

so 92020 is also 2 mod 139

for example 331259 is congruent to -(4*9203+1) mod (139*331)

curious that 69660+22360=92020

22360 are the first five digits of the decimal expansion of 1/(5*sqrt(2))

so 92020=69660+10^n/(2*sqrt(5)) for some n

92020=2*6^6-6^4+4

(215*107*9-1)*2^4-2^7=71*6^6

(215*107*9-1) *2 is congruent to 7 mod (331*139)

so

92020 is congruent to 7*((215*107*9-1)^(-1)) which is congruent to 2 mod (139*331)

 enzocreti 2021-07-11 10:02

Now i consider

215*107*x is congruent to 1 mod (331*139)

the solution is x=2+46009*s, for some s

when s=8

(215*107*(2+46009*8)-1)/(331*139)=429^2=2*92020+1

so

92020=(215*107*(2+46009*8)-1)/(331*139*2)-1/2

starting from

215*107*(2+46009*(6^6-1)) is congruent to 1 mod (331*139)

i arrived to

69660 is congruent to 107*2^6*3^7 mod (331*139)

form this

2*430 is congruent to (430)^(-1)*12^3 mod (331*139)

 rudy235 2021-07-11 17:32

[QUOTE=enzocreti;529796]pg(69660), pg(92020) and pg(541456) are probable primes with 69660, 92020 and 541456 multiple of 86

69660 in binary is 10001000000011100
92020 in binary is 10110011101110100
541456 in binary is 10000100001100010000

you can see that the number of the 1's is always a multiple of 5

a chance?[/QUOTE]

I never quite understand what you are trying to say. Can you explain to the unwashed masses the function [B]pg[/B]? What is it? Paying guest? Picogram?

 enzocreti 2021-07-11 19:07

...

[QUOTE=rudy235;583019]I never quite understand what you are trying to say. Can you explain to the unwashed masses the function [B]pg[/B]? What is it? Paying guest? Picogram?[/QUOTE]

the last you said

 enzocreti 2021-07-11 20:35

215*107*12^2 is congruent to -71*6^6 which is congruent to 72 mod (331*139)

331*139*8 is congruent to -8 mod (215*107)

s^2 is congruent to 1 mod 23005

the first not trivial solution is

s=429

the second is s=9201

the third is s=13374

(13374^2-1)/23005 is congruent to -1 mod 6^5

215*107*(2+331*139*8)-1 is a multiple of 331*139 and 429^2

mod 429^2 we have

23005*184033-1 which is a multiple of 429^2 and 71

((184033*23005-1)/71-429^2)/429^2=18^2-1

92020 is congruent to 4*(2+331*139*8)^(-1) mod 429^2 and mod (331*139) so

92020 is congruent to 4*(368074)^(-1) mod 429^2 and mod (331*139)

the inverse of 368074 so is 23005

92020 is congruent to 4*(429^2-2^2)^(-1) mod (331*139*429^2)

maybe it is useful

431=(427)^(-1) mod 46009???

331259 for example is congruent to -(9203*4+1) mod (331*139)

and 331259 is congruent to 9203 mod 23004

(92020)^(-1)=23005 mod (331*139)

92020*23005=(46010)^2

92021 divides 215*107*(2+331*139*4)-1

pg(69660), pg(19179) are primes

maybe something useful can be derived from this:

69660 is congruent to 19179 which is congruent to 429^2 which is congruent to 9 mod (71)

69660 and 19179 are of the form 648+213s

probably there are infinitely many pg(648+213s) which are primes

in particular

6^6 is congruent to 19179 which is congruent to 429^2 which is congruent to 861 mod (71*43)

6^6 is congruent to 860 mod (214^2)

92020 is congruent to -2^0 mod (17*5413)

92020 is congruent to -2^1 mod (3*313*7)

92020 is congruent to -2^2 mod 11503

23005*(2+331*139*2^0)-1 is a multiple of 11503
23005*(2+331*139*2^1)-1 is a multiple of 3*313*7
23005*(2+331*139*2^2)-1 is a multiple of 17*5413

The inverse of 9203 mod (331*139) is x

9203*x is congruent to 1 mod 429^2

331259 is congruent to 9203 mod 23004

23005 Is binomiale(214+1,2) so 23005 Is the 214th triangolare Number

92020=(858^2*139*331+4)/(2+331*139*8)

3371 and 331259 are primes pg(3371) and pg(331259) are primes

3371 and 331259 leave the same remainder 59 mod 3312

3371 and 331259 are primes of the form 59+ ((71*6^6-24^2)/10^3)*10^m, for some nonnegative integer m

(46009x-1)/(y^3-1)+y^3=(x^2-1)/2 over positive integers

x=429 y=6

(23005*(2+46009*(4))-1)/92021=46009

215^2 is congruent to 46009 mod 216

11503=71*2*3^4+1

92020 is congruent to -4 mod 11503

69660 is congruen to 3*6^3 mod 11502

92020 is congruent to 2^2 mod 11502

92020*23005 is congruent to 4 mod (71*11503)

23005*(2+331*139*(8+184041*(8+184041*(8+184041... is congruent to 1 mod (429^n)

23005*(2+139*331) is congruent to (429^2-1)/10 mod 230051=31*41*181

92020 Is 4 mod 11502 and -4 mod 11503

92020 Is congruent to (71*6^6/11502=288=17^2-1) mod 323=18^2-1

6^6/162=288

162 divides 69660

(2*(46009*6)^2+144*(2+46009*2))/313/49/3-6^3=71*6^6

46009+2 Is a multiple of 313*49*3

2*46009+2=92020

69660=3/2*(6^6-6^3)

92020-69660=22360 which is divisible by 860

22360=860*(3^3-1)

Z46009 is isomorphic to Z331XZ139

429^2 Is congruent to 2^2 which Is congruent. To (139*331*8)^2 mod (431*427)

71*6^3 is congruent to -23005*(2+46009*2) mod (7^2*313)

71*107*6 is congruent to -430 mod 11503

69660 mod 11503=642=107*6

642=6*(3*6^2-1)=6*107

11476*2*860+428 is congruent to 429 mod (3*313*49)

Consider

71*6^6 is congruent to -72k mod j

for k=1 j=46009
for k=2 j=4601
for k=3 j=(313*7^2*3)
for k=4 j=11503
for k=7 j=9203

71*6^6 is congruent to -72*7 mod 9203

46008 (=331*139-1) is congruent to -7 mod 9203

331259 is congruent to -7^2 mod 9203

or 9203=331259-46008*7

(139*331)^2 is congruent to 1 mod (92020) and mod (23004)

69660 is congruent to 92020-(15229*(2+46009*2)-216)/313/49=648=3*6^3 mod (23004)

23005*(2+46009*2) is congruent to 1 mod (313*7^2)

71*6^6 is congruent to -6^3 mod (313*7^2)

23005*6^3 mod (313*7^2)=15229

(6^3/2)*(2+46009*2) is congruent to -6^3 mod (313*7^2)

69660*313*7^2 mod 23004=3*6^3

6^6-(69660*49*313-648)/23004=3*71

71*6^6 is congruent to 67 mod 139 and 259 mod 331

chinese remainder theorem to the rescue:

45937+46009k...allowing negative k, you have -92090 which is -2 mod 1001 and to 92020

331259 is 259 mod 331
331259 is 4588 mod 4601
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Using the Chinese Remainder Theorem, solve the following system of modulo equations: x=259 mod 331 x=4588 mod 4601
Enter modulo statements
x=259 mod 331
x=4588 mod 4601

Using the Chinese Remainder Theorem, solve the following system of modulo equations
x ≡ 259 mod 331
x ≡ 4588 mod 4601

We first check to see if each ni is pairwise coprime
Take the GCF of 331 compared to the other numbers
Using our GCF Calculator, we see that GCF(331,4601) = 1

Since all 1 GCF calculation equal 1, the ni's are pairwise coprime, so we can use the regular formula for the CRT

Calculate the moduli product N
We do this by taking the product of each ni in each moduli equation above where x ≡ ai mod ni
N = n1 x n2
N = 331 x 4601
N = 1522931

Determine Equation Coefficients denoted as ci
ci = N
ni

Calculate c1
c1 = 1522931
331

c1 = 4601

Calculate c2
c2 = 1522931
4601

c2 = 331

Our equation becomes:
x = a1(c1y1) + a2(c2y2)
x = a1(4601y1) + a2(331y2)
Note: The ai piece is factored out for now and will be used down below

Use Euclid's Extended Algorithm to determine each yi
Using our equation 1 modulus of 331 and our coefficient c1 of 4601, calculate y1 in the equation below:
331x1 + 4601y1 = 1
Using the Euclid Extended Algorithm Calculator, we get our y1 = 10

Using our equation 2 modulus of 4601 and our coefficient c2 of 331, calculate y2 in the equation below:
4601x2 + 331y2 = 1
Using the Euclid Extended Algorithm Calculator, we get our y2 = -139

Plug in y values and solve our eqation
x = a1(4601y1) + a2(331y2)
x = 259 x 4601 x 10 + 4588 x 331 x -139
x = 11916590 - 211089292
x = -199172702

Now plug in -199172702 into our 2 modulus equations and confirm our answer
Equation 1:
-199172702 ≡ 259 mod 331
We see from our multiplication lesson that 331 x -601731 = -199172961
Adding our remainder of 259 to -199172961 gives us -199172702

Equation 2:
-199172702 ≡ 4588 mod 4601
We see from our multiplication lesson that 4601 x -43290 = -199177290
Adding our remainder of 4588 to -199177290 gives us -199172702

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71*6^6 is congruent to - 12^2 mod 4601 and mod 774

331259=11502*(17^2-1)-9007*331

-9007*331 cogruent to 9203 congrue
nt to 331259 mod 11502

(9007*331+9203)/11502=259+1

331259*11502 is congruent to 4473*666 mod (23004*331)

4473=4472+1

maybe something useful can be derived by this:

139^(-1)=331 mod 23004

for example

331259*139 is - 9007 mod (1001*23004)

(6^6)^(-1)=22 mod 331

331259 is congruent to 22 mod 139

331259 is -72 mod 1001
92020 is -72 mod 1001

71*6^6 is -72 mod 331

331259, 92020 and 71*6^6 are numbers y that satisfy this congruence equation

y is congruent to (-72+331*(10^x-1)) mod 1001 for some nonnegative integer x

(71*6^6+72-331*999)/9-72=331259=(71*6^6+72-331*999-3*6^3)/9

(359+71)*(6^6+11502)/359=69660

pg(359) is prime

69660 is congruent to 14 mod 359
6^6 is congruent to -14 mod 359

-23004 is congruent to 331 mod 359

92020 and 331259 are 5 mod 239

92020 is congruent to -331259 which is congruent to -3^5 mod 257

92020+3^5=359*257

1001-((331259-243*2-92020)/257)=72

331259 and 92020 are -72 mod 1001

28 is congruent to (429^2-1) mod 257

-239239 is congruent to 28 mod 257

92020+239239=331259

14 is congruent to 129*(429^2-1) mod 257

so

107 is congruent to 69660*92020*2 mod 257

from this follows

331259 is congruent to -14 mod 257

69660 is 13 mod 257

92020 is 14 mod 257

92020 Is congruente to 1 mod 829 and mod 37

331259=92020+239239 Is congeuent to - 3*2^11 mod 829 and mod 37

541456 Is congeuent to -2 mod 37 (and also 331259 Is -2 mod 37)

541456+3*2^11 Is a perfect square

PG(359) Is prime

331259*5 Is congruente (23004+331=multiple of 359)=4667 mod 6^6

71*6^6-(331259*5-(23004+331))=6^8

541456=43*(10^4+2^5*3^4)
280 Is congeuent to -2592=2^5*3^4 mod 359

69660 Is 28/2 mod 359 23004 Is 28 or -331 mod 359

The inverse of 10 mod 359 Is 36

69660*10^3 Is -1 mod 359 so 69660 Is -6^6 mod 359

-6^6 =14 =69660=-2592*18 mod 359

From here

3870=-2592 mod (359x18)

Dividing by18

215=-12^2 mod 359

12^2=(71^2-1)/(6^2-1) mod 359

So (6^3-1)=-(71^2-1)/(6^2-1) mod 359

PG(541456) PG(331259) and PG(92020) are primes

541456 92020 and 331259 are Numbers of the form

-a+1001*s where a is a Number congruente to 7 mod 13
a=72 and a=85

Because a=13d+7 and 1001=7*11*13

541456 92020 and 331259 are of the form -13d-7(1-143f) for some dnand f

So

(541456+7)/13 Is 71 mod 77

(92020+7)/13 and (331259+7)/13 are 72 mod 77

((X^2-1)*(46009+1/4)-1)/46009-x^2=0

This Is a parabola for x=+ or - 429 this goes to zero...

429^2-1=2*92020

I dont know of from that equation One can derive something more general

parabola
focus | (0, -33870353513/736148)≈(0, -46010.2)
vertex | (0, -184041/4) = (0, -46010.3)
semi-axis length | 1/184037≈5.43369×10^-6
focal parameter | 2/184037≈0.0000108674
eccentricity | 1
directrix | y = -33870353521/736148

This Is the parabola

((X^2-1)*(46009+1/4)-1)/46009=y

429^2 Is congruente to 6^6 which Is congruente to 69660 which Is congruente to 9 mod 71

(429^2-9)/71=2592=2^5*3^4

Curious that 541456 Is divisible by (10^4+2592)

43*2592 Is 111456...the last digits 1456 are the same as in 541456

92020/2592/5-1/3240=71/10

324 divides 69660....i think there Is something involving 18^2

2*92020/2592-1/324=71

(429^2-1)/2592-1/324=71

1/69660=(1/215)*((184040/2592-71))

2592=72^2/2

215/10*(20000+72^2)=541456

1/(1/(71*6^6/331259-10)/99+239)=-277.199999...

The inverse of 5 mod 46009 Is 9202

429^2-5 Is a multiple of 46009

(429^2-5) Is then congruent to 6 mod (239*7*11)

92020 Is 10 mod 3067

239239 Is 13 mod 3067

So 331259=92020+239239 Is 23 mod 3067

71*6^6 Is 6^3 mod 3067...

71*6^6=429^2=3=-239239=3*6^4 mod 37

92020 for example =1 mod (37*3*829)

429^2=3 mod (37*829)

331259=92020+239*1001

so

331259 is -2 mod 37

331259 is congruent to 9203 mod 23004

9203 mod 71=44

(331259-44)/71=4665

4665 are the first four digits of 6^6=46656

4665 in base 6 is 33333 a repdigit

331259 is congruent to 9203 mod 23004

9203 is 5 mod 7
331259 is 5 mod 7

9203 is 44 mod 71
331259 is 44 mod 71

so 331259 and 9203 are numbers of the form 19143+497k

curious that 19143+6^2=19179 and pg(19179) is prime

curious that allowing negative numbers k -1234 is a number of the form 19143+497k

9203 and 331259 are also congruent to 131 mod 648=3*6^3

so using CRT they are numbers of the form 9203+322056k

71*6^6 is congruent to -(9203-131)/648 mod 331259

(331259-131)/648=2^9-1

71*6^6/648-4601=2^9-1

4601 divides 92020

Numbers of the form 512, 5112, 511...12,...

The difference 5112-512, 51112-5112,...Is a multiple of 46

(331259-9203)/5112=2^6-1

71*6^6=5112*3*6^3

331259/648=511,20216...

1/216=46/10^4+1/(10*15^3)

from above

370=92020X648 mod 511

370=92020X137 mod 511

138010=92020 mod 511

6^6=(138010-92020) mod 666

(20*71*6^6-4601*648*20)/511=10*6^4

370=92020X648 mod 511

370=40*648=10*2^5*3^4 mod 511

138010=92020 mod 511
69005=46010 mod 511
pg(69660) is prime

69660-69005 is a multiple of 131

pg(331259) is prime

331259=(6^2-1) mod 13801

370=92020X((69660-9-155)/511+1)=92020X137=92020X(70007)x511^(-1) mod 511^2

155 is 6^6 reduced mod 511

so

370X511=92020X70007 mod 511^2....92020 reduced mod 511 is 40

(40*70007-370*511)/511^2=10

9203-131=7*6^4

5112=71*72

(331259-131)/14-6^4+4=22360=92020-69660

92020+(6^4-4)=0 mod 6^6

9203=5=331259 mod 14

774*(1301-131)/13=69660

71*6^6=-14 mod 331259

71*6^6=-15 mod 43

(71*6^6+15)=4472 mod (43*107)

4472 divides (92020-69660)

(71*6^6+15-4472)/43/107=719

719=-1 mod 72

The inverse of 15 mod 4601 Is 1227
15*1227=18404

18404/2=9202

71*6^6 Is also =-15 mod 129

4472+129=4601

maybe It Is for that readon that

71*6^6=-144=-129-15 mod (4601*5)

-15*1227=(331259+13)/2 mod 429^2

19179=2131*3^2=-6=-429^2=-71*6^6=331259*6=9 mod 15

from 23005*(2+331*139*8)=1 mod (429^2*331*139) we obtain:

23005X2X431X7X61=1 mod (429^2*331*139)

7X61=427
so
431X7X61 is the factorization of (429^2-2^2)=(429+2)x(429-2)

71*6^6=-90 mod (431X7X61)

curious that

-69660=-90X774=71X6^6*774=(7^3-2)x7^3 mod 431

69660+(7^3-2)x7^3=431X433=432^2-1

pg(92020), pg(331259) are probable primes

92020=2^5=215=71x6^6=-331259 mod 61

69660=-2 mod 61 i think it is not chance

there are two probable primes pg(56238) and pg(75894) where 56238 and 75894 are multiple of 546
75894=139(again this 139!)*546 and the other 56238=103*546

103=139-6^2

71*6^6*429=-lcm(429,546)=-6006 mod 331259

69660=-2 mod 61

46009X8=-2 mod 61 (23005X(2+46009X8))=1 mod 429^2

so

69660=46009X8 mod 61

71*6^6=-6^3 mod (46011)

69660=2^8X3^9 mod (46011)

92020=-2 mod 46011

71*6^6=259=331259 mod 331

71*6^6=-72 mod (46009=331*139)

331259=9203 mod (23004)

331259=(9203-7) mod (46009x7)

331259=(9196+7) mod 23004x7

331259=(9196-7) mod (4601*7)

4601 divides 92020 and 4601x5=23005

(331259-9196)=7*139*331

331x7=2317, whose last digits are 317

71*6^6-331259=2981317, whose last three digits are 317

71*6^6-331259-(331*7)=331*3^2*10^3

331259-9196 is a multiple of 331*139

pg(69660) and pg(2131*3^2=19179) are primes, pg(92020) is prime

69660=19179 mod 639

(69660-19179)=4472 mod 46009
4472 divides (92020-69660)

curious that pg(2131), pg(19179=2131*9) and pg(69660=19179 mod 213) have this property: pg(2131) is the 19-th pg prime, pg(19179) is the (19+4)=23th pg prime and pg(69660) is the (19+4+4)=27th pg prime

we know that

23005*(2+46009*8)=1 mod 429^2
i noticed that

23005*(2+46009*8)=-1 mod 359

pg(359) is prime

it is not clear yet but maybe there is a connection to the fact that 6^6=-14 mod (359x13)

infact 17x13^2=1 mod 359

23005*(2+46009*8)=359*17^(-1)-1 mod 359^2

follows

29*(2+97)=359x17^(-1)-1 mod 359
99=(359*13^2-1)x260 mod 359

331259=-98 mod (359x71x13)

331259=(260-1) mod 331 by the way

331259=-46009x8-1 mod 359

331259=-(260x358^(-1)-1) mod 359
because 358x(10^2-1)=260 mod 359

71x6^6=83 mod 359

358x(10^2-1)=1 mod 83

so 331259=-98 mod (71*13*359)

71*6^6=-98-7x2^7=-994 mod (71*13*359)

71*6^6=-14*71 mod 359

331259=-98=+7*6^6 mod 13x359

359-99=260

331259=261 mod 359

99*(331-1)=1 mod (359x13)

331*99=1 mod (2^15)

331259=(1-2^15) mod (359x13)

7*6^6+2^15-1=359*1001 so 71x6^6=1-2^15 mod (359x13)

2^15=-1 mod 331
((2^(15+330*(1+138*k))+1)) is a multiple of 139x331

(71*6^6+2^15-1-359359)=12^6

12^6=(71*6^6-331259) mod (359x13)

12^6=-7*2^7 mod (359x13)

(71*6^6-331259)=-7x2^7 mod (359x13x71)

6^6=-14 mod 359 pg(359) is prime
71*6^6=-14 mod 331259 pg(331259) is prime
the difference 71*6^6-6^6=2^7x3^6x(6^2-1) where 2^7x3^6 is a 3 smooth number 3^(n+1)x2^n. 92020=2^7x3^6-(6^4-4)

(6^4-4)=215=6^3-1 mod 359

2^7x3^6=331 mod 359

331+215=546 and it is curious (but I think it is not a chance) that there are two pg(k) probable primes ( and perhaps infinitely many) with k multiple of 546

the multiplicative inverse mod 359 of 3^6 is 98

331259=-98 mod 359

so 331259=-(3^6)^(-1) mod 359

I think that there is a giant structure under these exponents but it is so complicated that no simple tool can shed even the slightest light on it

after few calculaions I found:

331259=-98=-(123*111)^(-1) mod 359

form here I could find that

3^6x324=333 mod 359

and so

333x98=324=18^2 mod 359

and

69660=324*215=333*98*215=14=-6^6 mod (359)

331259=-98 mod 359

69660=6^6 mod 23004 and -6^6 mod 359 infact

prime 359 is generating something, but I have no tools except some modular procedure to catch something

I could notice that

6^6=-14 mod (359x13)

331259=98 mod (359x13)

(69660-14)/359-(6^6+14)/359=2^6

infact 23004=28 mod 359 or equivalently 23004=-331 mod 359

in particular 23004=-331 mod (359x13)

and -23004=6^2 mod (2^6)

(331259+98)/359-(69660-14)/359=3^6

(69660-14)/359-(6^6+14)/359=2^6

some other possible ideas:

3*6^3=-70=17^2 mod 359

648x72=6^6=-14=17^2*72=-69660 mod 359

17*13^2=1 mod 359

92020=-3^5 mod 359 but also mod 257

331259=-(-3^6)^(-1) mod 359
and 331259=3^5 mod 257

257-14=3^5

i think that there should be an explanation why this number 14 appears so often

6^6=-14 mod 359
69660=14 mod 359
71x6^6=-14 mod 331259

i think that mod (23x5x3) something interesting can be found

so for example

331259=59 mod (23x5x3)

-3371=79 mod (23x5x3) pg(79) is prime

-359=331 mod (23x5x3) pg(359) is prime ...

but these are just ideas they have to be developed

mod 69 for example

331259=-79=-10=3371 mod 69

pg(331259) pg(79) pg(3371) are primes...

pg(359) is prime

359=-55 mod (23x3)

331 reduced mod 69 is 55

I could conjecture that there are infinitely many pg(k) primes with k=+/- 10 mod 69 and when it happens k is prime

-79=59=331259=3371 mod 138

i think this could be connected in some way to the fact that

ord (71*6^k) mod 23 =6

71*6^6 infact=1 mod 23

curious that 71x6^6=83 mod 359

and 359=83 mod 138

331259-3371 is divisible by 138 and by 6^3

71*6^6=(331259-59)/13800=24 mod 138

6^6+14 is a multiple of 359

69660-14 is a multiple of 359

6^6+15 is a multiple of 331x3

6966-15 is a multiple of 331x3

6966=69660/10

curious that 69660+6^6=-14^2 mod 331

6^6=-14=-69660 mod 359

The inverse of 14 mod 331259 Is a multiple of 359...why?

92020=5=331259 mod (239x7x11x13)

-92020=29=331259 mod (61x3)

648=-14 mod 331

the inverse mod 331 of 648 is 71

-(6^6+14)=1 mod 331 (6^6+14) is a multiple of 359

-(69660-14)=(14^2-1) mod 331 (69660-14) is a multiple of 359

pg(69660) and pg(19179=2131*3^2) are primes

69660=19179=-18 mod 79

69660=9=19179 mod 71

curious that 69660=-18 mod 79 and mod (21^2)

pg(3*21^2=1323) is prime and also pg(79) is prime

on the other hand 19179=6^3=-15^2 mod 21^2

(69660+18)/79+21^2=1323

From 23004=-331 mod (359x13x5)

I derived 23004x141=-1 mod (359x13x5)

10011x18^2=-1 mod (359x13x5)

18^2 divides 69660

10011^(-1)=23011 mod (359x13x5)

Curious that pg(10011/3-1=3336) Is prime

-3336=19999 mod (359x5x13)

14*10^3=-1 mod (359*13)

6^6=-14 mod (359*13)

The inverse of 1000 mod 359 Is 345=359-14

From

(10^4+1667)x3335=10^4 mod (359x13x5)

I arrived After some steps

-10^4x5x667+8191x5=-10^4 mod 359

And so

3810+8191=-2*10^3 mod 359

71*6^6=-14 mod 331259

71x6^6x359x725=-1 mod 331259

From this follows

70984x71x6^6=1 mod 331259

70984=261 mod 359

331259=70984=261 mod 359

In particolare (70984-261) Is divisible by 359 and (14^2+1)

so 70984=-14^(-1) mod 331259
70984=261 mod 359

so you can apply CRT here

and find the form of 70984

70984x5=92020=331259 mod (239x11)

Curious that pg(1323) PG(69660) are primes 1323=-1 mod 331

69660=150=70984 mod 331 and 70984-69660=1324 PG(1323) Is prime

92020-69660=22360

(92020=4) mod 71 (92016=71*6^4). 92016-69660=22356

-331259=22356 mod (197*359)

22356=18^2*69

197*359 Is 70984-261

curoius that -331259=22356 mod (359*197) and -331259=22357 mod 139

maybe something to do with 46009=331*139???

maybe...(331259+22357)=0 mod 139 (331259+22357)=108 mod 331

i have no tools anyway no advanced skills to make progress

(331259+22357)=7^3 mod 541

probably this is connected to the fact that 541456=-85 mod (541x1001)

(331259+22356)/359=444 mod 541

(116315+1)/359=18^2

maybe 331259+22356=107 mod 331 is connected in sojme way to the fact that 23005 is divisible by 107

maybe there is something in F46009 and F23005

331259+22356=-107 mod (3337x106)

pg(3336) is prime

3337=47x71

216x104=2x11x111=22356+108=22360+104=-331258
mod 3337

Curious that pg(75894) Is prime and -75894=(2^16+16) mod (359x197)

I notice that 22356=-23 mod (23x139)...

i think that something very complicated is under these exponents...
331259=59+23k

3371=59+23k pg(3371) and pg(331259) are primes...the numbers 23 and 139 are in some way involved but I have no idea how it happens

69660=648 mod 972

22356=-23 mod 973=139x7

i think that something very very hard to understand is happening in Z139 and in Z107

69660=648 mod 23004=71x972

2^2x3^5=-1 mod (139x7)

22356=0 mod 972 (22356=972x23)

973 divides 75894 and pg(75894) is prime

69660=16=239239=71x6^4 mod 23

(239239-71*6^4)/23-1=80^2 by the way

92020=69660+239239

22360=92020-69660=4 mod 23

4 is the square root of 16 by the way

(239239-16)/23+3=102^2 by the way

22360^2=16=69660=71x6^4=239239 mod 23

PG(359) Is prime PG(3336) Is prime PG(92020) are primes

92020=-(3336-359) mod 331

92020=21297 mod (359*197)

21297=21296+1
21297=11*44^2+1

44^2-1=1935 divides 69660

mod (359x197=359x(14^2+1)) I can see:

239239=27070 mod (359x197)

27070=541456/20-14^2/70

331259=92020+239239

(6^3-1)*(6^2-1)+1 divides (331259+22356+107) which is also divisible by 3337

pg(3336) is prime

this because 215x35=-1 mod (2x53x71)

(22356+107+3337)+331259 is a multiple of 71x107

215x106=-1 mod (71x107)

maybe is not chance that 22356+107+331259+3337=-1 mod 541

(107+22356-3337+331259)/10011=6^2-1

(107+22356-3337+331259)=1 mod 359

22357=22360-3 is a multiple of 139

(331259+22357)/106=3336 and pg(3336) is prime

22357=-8=79 mod 71

from here I have (because 9 is the multiplicative inverse of 79 mod 71)

201213=-72=-711 mod 71

in particular

201213=-711 mod (71x79)

but 71x79 =5609 divides (69660-19179) where 69660=9=19179 mod 71

201213+711=71x79x6^2

I don't understand why powers of 6 are involved in these numbers!

pg(3336) and pg(75894) are primes and 3336 and 75894 are multiple of 139

it holds:

-(75894-3336)=4=92020 mod 71

curiously

((75894-3336)+4)/(72^2-1)=14

3336=139x24

75894=139x546

I suspect that when pg(k) is prime and k is a multiple of 139 then k is of the form 139x(29xs-5) with s some integer.

An observation:

3336 and 75894 are multiple of 139

PG(3336) and PG(75894) are primes

3336=1=75894 mod 29

Maybe all PG(k) primes with k multiple of 139 k=1 mod 29?

Are there infinitely many PG(k) with k of the form 3336+139x29s?

92020-69660=1=75894=3336 mod 29

I think that there Is some ccomplicated connection among these exponents

In particolare 92020=2 mod 139
92020=3 mod 29

So 92020 Is a Number of the form 3338 (=71x47+1)+139x29s

Whereas 3336 and 75894 are of the form (71x47-1)+139x29s

139x29x2-3338/2=2131x3

pg(2131) is prime and also pg(2131x9)

indeed

2131=(46x139-1)/3

3336=-1 mod 71
75894=-5 mod 71

75894=3336x5 mod (71x139)

-75894=17^2 mod (71x29)

3336=-1 mod 71
3336=1 mod 29
75894=1 mod 29

so 75894=3336=92020-69660=1=-17^2=-(3x6^3-359) mod 29

75894=22360=-17^2=-(3x6^3-359) mod (29x71)

I think that that is the rub

because 69660=3x6^3 mod 23004

22360=92020-69660

359 mod 71=4

92020=4 mod 71

because 359=-1700 mod (71x29) and because the inverse of 1700 mod (17x29) is (14^2-1)=195

75894=22360=-3x6^3+195^(-1) mod (71x29)

75894x195=6^4+1 mod (71x29)

92020=2 mod 139
92020=3 mod 29

331259=21 mod 29
331259=22 mod 139

3336=0 mod 139
3336=1 mod 29

75894=0 mod 139
75894=1 mod 29

as you can see there are classes of exponents that are 1 unit far away mod 29 and mod 139 (21,22-0,1-2,3)

92020 (not multiple of 139)=2 mod 139
92020=2x2 mod 71

331259 (not multiple of 139)=22 mod 139
331259=22x2 mod 71

the not multiple of 139 (92020 and 331259)=d mod 139 and =2xd mod 71 for some d

pg(19179=2131*9) is prime and also pg(2131)

19179=9 mod 71
19179=8 mod (19x1009)

but 19x1009 is a divisor of 23005*(2+331x139*5)-1

it seems that there is a connection between the exponents leading to a prime and the divisors of 23005*(2+139x331xs)-1 for some s

but for more general results it takes a deep knowledge of field theory that I don't have

92020=71x6^3 mod (19x1009)
92020=71x6^3=-11503x2=-3835 mod (19x1009)

after some steps (dividing 3835 by 5 and 92020 by 5) I came to

7x11x239=-3x2^8 mod (19x1009)

13x7x11x239=239239

331259=92020+239239

...

331259=53x101-1 mod (19x1009)

92020+2=6^6-69660 mod (19x1009)

92020+69660=6^6-6 mod 11503

331259=44 mod (311x71)

curious that (23005*(2+46009*5)-1)=276054 is a number in Oeis sequence A007275, walks on hexagonal lattices

276054x12=3312648=3x6^3 mod 3312

3312648=72 mod (23004)
3312648=-72 mod (23005)

In my opinion something interesting should be found studyng
Z23005xZ46009

(71x6^3+1) for example divides (23005*(2+46009*2)-1) and 92020+2

23005x46009=3 mod (3539x11503x13) (by the way 3539 is a Wagstaff prime)

92020=-4 mod 11503
92020=6 mod (13*3539)

curious that 331259=2131+1 mod 3539 a chance???

92020=71x6^3 mod 19171

i think that theory of ideals should help 46009Z

curious that 541456=(13*359+1) mod 19171

23005x92015=-1 mod 92019

92015 is multiple of 239

curious that 92020=15336 mod 19171 and 92020=15335 mod (313x49)

-75894 (pg(75894) is prime))=790 mod 19171 and -75894=791 mod (313x49)

75894=-790 mod 19171 75894=-791 mod (313x49)

there are pg(k) primes (as pg(92020) and pg(75894) I have not yet checked if there are others) such that k=a mod 19171 and k=a-1 mod (313x49) where a is a certain integer belonging to the set Z

i checked also pg(69660) is one of these

69660=-7024 mod 19171
69660=-7025 mod (313x49)

14377x92020=1 mod 19171

14377x71x6^3=23005x(2+46009x5) mod 19171

follows:

14377x15336=3834x19166 mod (1917x19171)

(23005*(2+46009*20)-1)/138027-71*6^3*10=7

138027 divides also

(23005*(2+46009*8)-1) and 23005*(2+46009*5)-1

71x6^3x10=-5 mod (829x37)

829x37 divides (92020-1)

71x6^3x10=-8 mod (19171x8)

92020-19171x4=71x6^3

92020-19171x4=-1 mod (313x49)
92020-19171x4=1 mod (3067)

92020=10 mod 3067

153367-(92020-19171*4+1)=138030

consider 429^2-19171x

the maximum value of x such that 429^2-19171x >0 is x=9

429^2-19171x9=11502
429^2-19171x8=37x829

37x829 divides (92020-1)

429^2-19171x6=69015
69015+645=69660
645 divides 69660

92020x2=429^2-1

19171x6=1=429^2 mod 23005

69660=645 mod 23005

69660=6^2*71*3^3+3*6^3

46011=19171*12-429^2

i observe that 19171=3^9-2^9

541456+13-449449=92020=46010x2=331259-239239

(46010x2-19171x2-2222)=51456
Pg(51456) and pg(541456) are primes

46010-1111+1=449x10^2

-19171x2=51456+2 mod 449

51456+2+19171x2+1 is a multiple of 1009

1111=71x3 mod 449

46010=71x3-1 mod 449

4601=111 mod 449

so 4601x20=92020=111x20=2222-2 mod 449

92020+2=2222 mod 449

(92020+2) is divisible by (71x6^3+1)

92020=-5^2=2220 mod 449

331259=-(4601x3-2) mod 23004

331259+2 is a multiple of 37 whereas 92020=331259-239239=1 mod 37

71x6^3=1 mod 3067 (3067x13=9201)

71x6^3=-1 mod (313x7^2)

92020=10 mod 3067
92020=-2 mod (313x7^2)

239239=13 mod (3067x13)

429*46009=-1 mod (214x92233)

92233 is a prime. 92233-92020=71x3

 enzocreti 2022-01-03 13:06

92233=427x6^3+1

92233=6^3=51456 mod 427

((92233-51456)-1)/3-12592=10^3

12592 divides 541456

PG(51456) and PG(541456) are primes

92233-51456=1=69660 mod 1699

13592=1699*8

43*(((51456+(3^6-1)*4)/4)-10^3)=541456

69660-(487-51456)=69660-(92233-51456)=0 mod (71x1699)

(69660-487)=0 mod 313
(69660-486)=0 mod 427

486=162x3

162 divides 69660

-449449-13=64=541456+13 mod 139

so

541456=51 mod 139 (449x13=-1 mod 139)

92020=541456+13-449449

331259=71*6^6=259 mod 331

331259=44 mod 71
331249-44=331215 that Is the concatenation of 331 and 215

331259-44-215 Is a multiple of 331

44+215=259

71x6^6-331259+92020=6^2-6^5 mod 311

(6^2-6^5)=7740 which divides 69660 (7740=1 mod 71)

multyplying by 9 both sides

71x6^6x9-331259x9+92020x9=-69660 mod 311

331215 is the concatenation of 331 and 215

331+215=546

546 divides 75894 and 56238

pg(75894) and pg(56238) are primes

75894-92020/2=215x139-1

546=215+331

215x139 can be cancelled in both sides

this leaves

46009-46010=-1

56238 can be factorized as

(139-6^2)*(331+215)=56238

probably there is something in 79*3^j

pg(79) is prime by the way

79*3^2=711 and 69660=71x711+2131x3^2

79*3^3=2131+1

71x79+2131=7740 which divides 69660

79x3^2=1 mod 71

69660-19179=71x79x3^2

239239=-9 mod 12592

541456=239239+9 mod 12592

92020=3867+9 mod 12592

23004*46009-1-(92020-6)=3539x13x23003

i think that there should be some group in action...

pg(75894) and pg(56238) are primes with 75894 and 56238 multiple of 139

75894+56238=132132=1 mod (1861x71)

331259=1 mod 1861

75894+56238=331259 mod (1861x107)

331259-132132=107x1861

75894+56238-1=71x1861

107=71+6^2

56238 and 75894 are congruent to +/- 6 mod 132

in particular (75894+6)/132+1=24^2

so 331259=56238+75894+92020+107x1001

pg(394) is prime

1323=-1 mod 331 pg(1323) is prime

1323=72 mod 139

1323=(1001-72) mod 394

i suspect that this is connected to the fact that

331259=-72 mod (1001x331)

331259=-1323 mod (1001-72=929)

331259=-394=-1323 mod 929=1001-72

3x6^3-3=394x(72)^(-1) mod 1001
In particolar 394x431+3-3x6^3=169169

92020=1323-394=929 mod 1001

541456+13=1323-394-13=916 mod 1001

(92020-1323+1)=0 mod 449

(541456+13-1323+1)=0 mod 449

541456-449449=92020-13

i forgot to see that 92020=43*2132+344

344 is the residue of the multiples of 43 (69660, 541456, 92020) mod 559

43*2132 is infact divisible by 559

69660-(46009*23005-1)/(313x49)=2^3x3^4=3x6^3

92020=-2 mod (313*49)

313x49x3x6^3-1=215^3

23005*(2+46009x8)=1 mod 429^2

23005x(2+46009x8)=-1 mod (359xp) where p is a prime p=23586469...i wonder if it has some special property

the logic behind these primes requires tools that are far beyond current knowledge...

331259=259=71x6^6 mod 331

6^k=1 mod 259

the order mod 259 of 6 is 4 ord(6)=4 infact 6^4=1 mod 259

92020=2 mod 331
92020=4 mod 6^4

92020=71+4=75 mod 259

(92020-4)=71x6^4

239239=-77=-75-2 mod 259
239239=-78 mod 449

541456-77x5837+13=92020

curious that

449449-(239239+78)=210132=13*6^2x449 which is congruent to (6^4-2) mod 2131

239239+78 is 449x41x...
449x41 is 1840...anything to do with 429^2-1???

541456=7740 mod 18404

7740 divides 69660

18404=(429^2-1)/10=449x41-5

pg(3336) is prime

3336=24x139=-1 mod 71

(24+71x9)x139=-1 mod 71

(24+71x9)x139+1-138=92020

(24x71x9)x139+1)/71=6^4+2

so 92020=(6^4+2)x71-138

or equivalently

92020=71x6^4+142-138=71x6^4+4

((24+71*15)*139+1)/71=2132

so for example 19179=3^2*2131 can be rewritten as

3^2x(((24+71*15)*139-70)/71)=19179

71x6^6=6^4-44 mod (331x4=1324)
331259=44+215 mod 1324

331259=44 mod 71

(71*6^6-6^4+44)/(139*18-1)=1324

may be 44 is not random...
69660=(44^2-1)x6^2

331259=44 mod (311x5x71)

331259^2=1936 mod (311x5x71)

1935 divides 69660

6^5=1 mod (311x5)

6^5-6^2 divides 69660

1-36=35

I suspect that there could be some link to the fact that 541456=51456+700^2

700^2 is divisible by 35

curious that pg(75894) is prime 75894 is multiple of 139

-75894=(2^16+16) mod (359x197)

2^16=-2 mod 331

2^16=1 mod (255x257)

i think that something big is happening on some field

sqrt(71x215x3+1)=214

92020=sqrt(71x215x3+1)x430

92020=(71*215*6+1)+429

92020=214^2+215^2-1

pg(51456) and pg(6231) are primes with 51456 and 6231 multiple of 67

curiously (51456/67-6231/67)=26^2-1

19179=648 mod 23004
331259=(19179-648)/71 mod 359

331259x71=222 mod 359

so

222=19179-648 mod 359

note that

331331-(331259-(19179-648)/71)=333

((92020*3-6)*3*4-1)/71=6^6+1

...3312648...i think that 3x6^3 has something to do with these numbers...

331259/71=4665+44/71=6^6/10-6/10+44/71=6^6/10+7/355=6^10/10+7/(359-4)

331259=44 mod 4665

106x44+44/71+1=331259/71

4665=359*13-2

331259 =71 mod 44
331259=44 mod 71

so 331259 is a number of the form 115+(5^5-1)k this is curious

43*(1+sqrt(9x+1))=9x

solution x=215

215*9+1=44^2

i think that you can obtain a continued fraction from that

69660=6^2x43x(1+sqrt(1+43x(1+sqrt(1+43x(1+...

curious fact:

69660=19179=3x6^3 mod 639

19179/3=(639)3 and 6393=-1 mod 139

I think that something is in action over some field...

(429^2-6)=-1 mod 46009=331x139

(429^2-6)=3 mod 639

from this

429^2=80^2-1=79x3^4=711x9 mod (71x139)

6393+((71*4+1)*139+1)=46009 i think that in Z46009 something is in action as well as in Z23004

71x6^6-541456=-1 mod 359

541456+14=-261=-331259 mod 359

69660=14=-6^6 mod 359

541456+69660=611116=-261=-331259 mod 359

611116=131x4665+1

92020-(541456+14-98)/13/359=359x2^8

-331259=98 mod (359x13)

-(541456+69660)=-611116=359-98=261 mod (359x13)

pg(1323) is prime pg(39699=13233*3) is prime

13233=18^2 mod 331

69660=215x18^2

so

13233x215=69660=150 mod 331

and 13233x3=39699=-150 mod (359x111) (359=331+18)

69660=3x6^3 mod 71

(69660-3x6^3)/71=-1 mod 139

this suggests me that something is in action over Z139 or maybe Z46009

23005*(2+331x139)-1 is divisible by 11503

69660-642(=3x6^3-6) is divisible by 11503

23005x(2+46009)-1-(69660-642) is a multiple of 23003

2*(23005*(2+46009*72)-1)/46009-2=331272x10

331272-13=331259

23005x(2+46009x72-1) is divisible by 449

541456+13-449x1001=331259

something mysterious is boiling in Z46009

46009x72-72=71x6^6

x^2/(6^6-2-sqrt(2x+1))-x*(sqrt(2x+1)-1)/215=0 has solution x=92020

min (x^2/(6^6-2-sqrt(2x+1))-x*(sqrt(2x+1)-1)/215)=-19394.4=-19179-215.4

this is a parabola

from
331259x71=19179-17^2 mod 359 we have

222=19179-17^2 mod 359

so

(2^9-1)=19179 mod (359x13)

curious that 19179-511+1 is divisible by (2^2-1), (2^3-1) and (2^7-1)

331259=-98=-(512-414) mod 359

19179=414 mod 139

19179=138x139-3

69660=19179=6^6 mod 639

the inverse mod 639 of 2131 is 427

69660x427=9 mod 639

69660x427-9=639x46549

46549 is prime =6^6-107

I think this is not random but connected to the fact that 107 divides 92020

19179=7x71+14 mod (359x13)

6^6=-14 mod (359x13)

71x270+9=7x71+14=(2^9-1) mod (359x13)

from here

6^6=7x71-19179 mod (359x13)

form here

after some steps...

331x72=7x71 mod (359x13)

curious that

19179x(7^4+1)=1 mod (359x13)

3x6^3+70 is divisible by 359

(331259-5) is divisible by 717x6=6x(3x6^3+69)

6x(3x6^3+69)=-1 mod (331x13)

so (331259-5) is divisible by (331x13-1)

(331259-5)/(331*13-1)-(92020-5)/239/77=72

331x13=-5 mod 359

359x12=-1 mod 139

331259 has the curious representation:

65^2*77+77^2+5=331259=325325*77+77^2+5

further steps toward a theory of these numbers need super-tools

(331259-5)/7-6^6=666

6^6=-666 mod (239x11)

anything to do whit 92020=5 mod (239x11)
331259=5 mod (239x11)???

curious fact:
(239239-(6^6+666)+2)/111=1729 the Ramanujan number

I will call these primes Neme primes (Neighboured Mersenne)
Neme(3)=73

Instead of Neme primes I could call them Hopeless primes, no hope to find a logic behind them

curious that 69660=-342^2 mod (432^2)

I think that starting from 23005x(2+46009x8)=1 mod 429^2 and 23005x8=-1 mod 429^2 one can develop someting useful

Using Chinese remainder theory

numbers multiple of 23005 (=0 mod 23005) and =1 mod 331x139 should form a ring or something similar...and I think that in that ring one can get something

(2+46009x8+2x92020) is divisible by 92019

92019x6-92020x5 is a multiple of 3539 a wagstaff prime...

331259=2132 mod 3539

6^6-3x6^3-1=3539x13=46007

239239=-13 mod (9202)

(239239+13)/107=2236

69660+2236x10=92020

331259=331 mod 2236

239239x2236x2=-(331259-331) mod 331

239239x4472=72 mod 331

2236=6^2x239239^(-1) mod 331

22360=(19^2-1)x239239^(-1) mod 331

from here

22360=-148 mod 331
239239=-(148/2) mod 331

22360=257-74 mod 331
239239=257 mod (331x19^2)

(331259=257) mod (71x111)

331259x6^2=(22360-257) mod 71

I have the impression that powers of 6 and 71 are involved in the logic behind these neme primes

92020+257-14 (257-14=243 a power of 3) is 359x257

331259-243 is a multiple of 257

1001-((331259-243)/257-359)=72

69660=13 mod 257
69660=14 mod 359

there is a logic but it is so complex that it is almost hopeless to find a pattern

331259+14+84=71x13x359

541456=84 mod 359

69660=-345=-(331+14) mod 359

6^6=-14 mod 359

i cannot put toghether the entire pieces of the puzzle anyway

331259=6^6-541456 mod (359x13)

(69660-6^6+331)*2-14=6^6

92020=10 mod 3067

331259=22 mod 139
331259=23 mod 3067

331259-23-92020+10=239239-13

92022x36-6^3=71x6^6
71x6^6=6^3 mod 3067

71x6^3=-1 mod (313x7^2)
71x6^3=1 mod 3067

I would call these prime Neme primes or maybe desperate primes

I stronly suspect that the exponents of these neme primes are connected among them with a logic that it is impossible to understand...only a God could find a pattern...or maybe a new Gauss...

curious that 541456=353(7)9 mod (3539x13)

with that "7" inserted

92020=6 mod (359x13)

in other words

541456=3539x10-10-1 mod (3539x13)

curious that (541456-3539x10) is divisible by 23003=71x2^2*3^4-1

71x6^6=72 mod (3539x13)

-(541456+13)=787 mod 858
-331259=787 mod 858
69660=162 mod 858

92020=214 mod 858

359x239=1 mod 429

541456=0 mod 787

787-13=774 divides 69660

787-429=359-1

so for example 331259=429x774-787

774 divides 69660

429=sqrt(92020x2+1)

-541456=344 mod 774

there is a hidden structure

it is clear that 331259-774 is a multiple of 4601 and 4601 divides 92020

-541456-13=456+331

541456+13-456=359x11x137

331259=-358 mod (773x429)

358=359-1
773=774-1

331259+773 is divisible by 1297 a prime of the form 6^s+1

331259=(259-215=44) mod 71

331259=259 mod 331
71x6^6=259 mod 331

there is something...

23005*(2+46009*k)-1=N^2

for k=8
for k=3680
,...

23005*k+1 is a square

k=8

k=3680
...

3680*23005+1=(3x3067)^2=9201

(71x6^6-216)/3/3067=359+1

92020=10 mod (3067x13)

(9201^2-1)/4601/23-13=787

 enzocreti 2022-03-06 12:54

Qqqq[QUOTE=enzocreti;597012]92233=427x6^3+1

92233=6^3=51456 mod 427

((92233-51456)-1)/3-12592=10^3
[500 lines of excessive quote]
[/QUOTE]3067 is a prime

there is a very complex hidden structure

331259/3680 is very close to 90

7775*23005+1 Is a Square

-331259=3067-1 mod 7775

(331259+3066)-10001=18^2*(10^3+1)

I think these primes taste very exotic

23005 X+1=Y^2

with X and Y integers

X=92019 is a solution

Elliptic curves???

92016( =71x6^4) x 23005+1=46009^2

3067*5+1=71x6^3

92020-71x6^3 is a multiple of (19179-8=19171)

-331259=(3x3067)^2 mod 359

after some steps (inverse of 3067 mod 359 is 139x2)

-331259x(10^2-1)=9 mod 359

so it is clear that 71 and powers of 6 are involved in these primes

(71*6^6-216-(92020-10))/(71*6^3-1)=210

anything to do with the fact that 541456+13-210*1001=331259??)

71x6^6-216 is a multiple of 3067

92020-10 is a multiple of 3067

3067x6=-1 mod (239x77)

239239=13 mod (3067x6)

-3067x3=1 mod 107

239239=-13 mod 107

331259=-13 mod 107

it seems to be a perfect complex interlocking of modules

I think that only a math-champ chould develop a theory for these numbers...I think that one should know very well Galois theory at least

331259=9203 mod (3067x5+1)
331259=5 mod (3067x6+1)
92020=5 mod (3067x6+1)

239239=0 mod (3067x6+1)

541456=7740 mod (3067x3+1)

7740 divides 69660

71x6^6=6^2 mod (3067x6+1)

19179=777 mod (3067x3)

strange at least curious

19179=3067x6+777

(6^6-3*6^3-3)=46005 is divisible by 3067,5,3

777/3=259

-331259=71x6^3x777 mod 331

359x18^2=1 mod 541

-541456=85 mod 541

-541456=359x(18^2x85) mod 541

-541456=359x490 mod 541

541456-51456=700^2 which is divisible by 490

-700^2=359x490+51456

after some steps:

490x(-1000-359)=51456 mod 541

51456=-480 mod 541

(51456+480)=51936=5x10^4+44^2

(44^2-1)=1935 divides 69660

1935-479=1456

479-394=85

pg(394) is prime

541456=-(44^2-1)+1456+394 mod 541

51456=-490x(10^3+359) mod 541
541456=-490x359 mod 541

51456=129360 mod 541
51456=129359 mod 359

541456=18309 mod 541

541456-51456=490000

curious thta

429^2-1=540x21^2 mod 541

(429^2-1)=3x394 mod 541

pg(394) is prime pg(3x21^2=1323) is prime

92020=-491=3x394x271 mod 541

curious that 69660=-129 mod (541x129)

curious that

(429^2-1)=3x394 mod 541

dividing by 2

92020=3x197 mod 541

92020-3x197=91*10^4+429

92020=14=-331259=(69660+1=69661 prime) mod 257

curious that

69660-6^6+1=23005 so

92020=(69660-6^6+1)x4=(3*6^6-(7^3-1)^2+1)*4

from here we come to the curious:

69660=432^2-342^2 (432 is just apermutation of 342)

or =432^2-(18x19)^2

92020=-67=4 mod (6^4+1=1297 prime) and mod 71

pg(67) is by the way prime

-768x92020=67x768=51456 mod (1297)

so

23^2x92020=51456 mod 1297

curiously

23^2x92020-51456+1=365^3

365^3=1 mod 1297

(92020-4)+6^6=107x6^6

107 is the inverse mod 71 of 215

-331259=1001 mod 449

541456+13-449x1001=92020

331259=92020+239239

-331259x449=541456+13-92020 mod (449^2)

maybe manipulating this you get something

449x740-1001=331259=92020+239x1001=541456+13-210*1001

curious that (541456-6) which is 0 mod 13 is congruent to (1001+13)/13=78 mod 359
the most surprising fact about these primes for me is this: why the inverse concatenation (13 instead of 31, 715 instead of 157, 1531 instead of 3115 does not give patterns???)

71x6^6+72=0 mod 46009

(71x6^6+72)=261x449 mod 359

331259=261 mod 359

((71*6^6+72)*4-331259)/359=35987

331259=3^2*29 mod 359
92020=2^2*29 mod 359

-331259+6^6=541456 mod (359x13)

-331259-14=84 mod(359x13)

-331259=98 mod (359x13)

(541456-84)/359/13=116

116 is the residue mod 359 of 92020

-23004=331 mod (359x13)

-23004x4=-92016=1324 mod (359x13)

pg(1323) is prime

92016+1323 is a palindromic number

331259=13x359 mod 6^6

i think that 13x359 is important for these numbers....and powers of 6...

69660=14 mod 359
6^6=-14 mod 359

-331259=14^2/2=98 mod (359x13)

pure chance???

-331259=2^11x3^12 mod (359x13)

(331259+2^11x3^12)/359/13-1=5x6^6

69660=19179=2131x3^2=6^6 mod 639

69660=7740x9

2131 and 7740 are numbers of the form -3478+5609s (using chinese remainder theorem 2131=-2 mod 79 2131=1 mod 71 7740=-2 mod 7 7740=1 mod 71)

69660=(88^2-4)x3^2

(88^2-2)=0 mod 79

(69660-19179)=(88^2-2) mod (79x541)

69660+541456=611116=17x19x44x43

69660+541456=611116 this strange palindromic number

611115 is divisible by 4665 (=359x13-2) 4665 again shares the same digits with 6^6=46656

i think these numbers are more mysterious than pyramids of Giza

331259=4665x71+44

(359*13)*131-261=611116

261 is the residue of 331259 mod 359

92020-(541456-316)/4665=0 mod 359

69660=-315 mod 4665
541456=316 mod 4665

4665=(3/5)x(6^5-1)

((3/15)*(6^5-1)+1) divides (92020-216)

541456-51456=700^2

(3x13x359-1)x(6^2-1)=700^2=(3x13x359-1)x(394-359)

i think that a key passage is this:

331259=(19179-3x6^3)/71=261 mod 359

-92020=(69660-3x6^3)/284=243 mod 359

222=19179-3x6^3 mod 359

19179=51x359+870

i dont know if this has something to do with the fact that pg(51) and pg(359) are orimes

19179=(2^9-1) mod (359x13)
19179=(2^9+1) mod 51

-69660=2^9 mod 331x2

18^2x213+3x6^3=69660

18^2x213-6^6=22356

(92020-4=71x6^4)-22356=69660

i started the hunt for a new neme prime Ne(k) with k of the form 648+213s. Probably I will never find it

(331259-261)/359=922=-1 mod 71

I think that everything is inset in these numbers, modular congruences,fields,...very very difficult...

no other prime Neme(23005k) found after 92020 (up to 1300000)

I realized that I forgot the most stupid thing: 2131-14^2=1935 which divides 69660

from this

19179x4-84^2-23004=-14 mod (359x13)

after some steps...

19179x4-82^2=-13 mod (359x13)

2131x6^2=82^2-13 mod (359x13)

2131x6^2(=19179x4)+13 is a perfect sqaure!!! (277^2)=1 mod 139

so

277^2=82^2 mod (359x13)

277^2=1 mod (138x139)

277^2-82^2=511x137(=70007)-2

19179=511 mod (359x13)

by the way 19179=-1 mod 137

any possible connection to the fact that (541456-51456)=700^2=-7^2 mod 70007??? (700^2=-35 mod (359x13x35))

it could be a chance...anyway

(277^2-1) is divisible by 23,139 and 24...if you divide by 23 that is (277^2-1)/23=3336 and pg(3336) is prime

3336=1 mod 23

331259=3336x23=261 mod 359

541456-344 is divisible by 559x44

44x559=2236x11

92020-69660=22360

69660=(44^2-1)x6^2

19179=2131x3^2=(3^7-2^7)x3^2+3x6^3

19179=-1=71x3^3 mod (137)

69660=-6^6=14=(71x3^3+1)/(359-222) mod 359

331259x71=222 mod 359

19179-3x6^3 is divisible by 261 (i think it is not chance that 331259-261=0 mod 359)

(19179-648+1) is divisible by 41 and 452 ( pg(451) is prime)

so

(19179-648)=452*41-1

pg(51) is prime

so

452x41-1=222=331259x71 mod 359

452x41-1-222=51x359

(331259*71-452*41+1)/359/71=922=-1 mod 71

(331259-261)/359=922

pg(1323) is prime pg(39699) is prime

39699=9 mod 1323

(39699-9)/2-666=19179

pg(19179=2131*9) is prime pg(2131) is prime

so 19179=2131x3^2=-666 mod 1323

19179=5x63^2-666

19179=-666 mod 1323
19179=-665 mod 451

pg(1323) and pg(451) are primes

pg(451) and pg(1323) are two consecutive pg primes!

this is equivalent to

19179=-666 mod 1323
19179=-214 mod 451

-19179x430=16=92020 mod 451

19179x430+16 is a multiple of 41^2
19179x430+92020 is a multiple of 43^2

(92020-69660)=22360=(19179-3x6^3)/71 mod 451

19179=2x344=3x79 mod 451

69660-19179 is a multiple of 79

71x6^4-69660=-261=-331259 mod 359

71x6^4+4-69660=261 mod 451

71x6^4-14=-261=71x6^4+6^6=-331259 mod 359

71*6^4-69660+331259=0 mod 359
71*6^4-69660+331259=-1 mod (139x106)

I think that in 139Z there is something

(71*6^4-69660+331259+1)/106=3336 and pg(3336) is prime

106=-1 mod 107 (anything to do with the fact that 92020 is a multiple of 107???)

curio:

pg(4) is prime pg(51) is prime pg(451) is prime pg(92020) is prime

92020=4x51x451+16

2x51x261-2x2131=92020-69660=22360

-541456=2^11 mod (3216x13^2)

3216 divides 51456

I arrived to this from

6^6-(331259x71-19179+2^9-1-222)/541=1=51457 mod 3216

so -541456=2^11 mod ((51456/16))

(541456+2^11)=2132 mod (359x13)

(541456+2^11-2132)/359/13=116

92020=116 mod 359

(541456+2^11)=92020=2 mod 331

(541456+(2^11-1)-2131)/(2131x3^2+2^9-1)=29

541456+69660=611116

611116x71 mod (359x13)=137

19179-511=0 mod (359x13)
19179-511-137=19179-648=0 mod 71 and mod 261

611116=-261 mod (359x13)

331259=261 mod 359

(19179-648+611116*71) is divisible by 359x13
(19179-648+611116*71)+1 is divisible by 394 and pg(394) is prime

(611116*71+19179-648)/(331259+359-261) is an integer

541456*71=6^4+1 mod 4667

2131=72 mod 2059=29x71

19179=648 mod (29x71)

541456x71-6^4-1=359x13x8237

8237=1 mod (29x71)

-611116-331259=-6^6-14=69660-14 mod 359

-(541456+69660)-331259=-6^6-14=69660-14 mod 359

71x3^3+1=1918 divides (611116+331259-6^6-14-19179)

(611116+331259)/359=2625=2626-1

239239+92020=331259

239239=-214 mod 359

430x239239=-92020 mod 359

the inverse of 430 mod 359 is 268

(359x499) divides both (92020x268+239239) and (611116+331259-6^6-14)=(541456+69660+92020+239239-6^6-14)

19179+1 is a multiple of 137
19179=511=648-137 mod (359x13)

-19179-1=30003 mod (359x137)

3x6^3-261=387

92020=387 mod (2131x43)

(331259-257)*9/(23004-19179+648)=666

((331259-257)-4-92016)/331=722=2x19^2

23004-19179+648=4473

4472 divides (92020-69660)

331259=92020=5 mod 2629

(2629*5+1) divides (92020+2)

it turns out that 331259 has this curious representation:

(6^6+666)x7+5=331259

(92022=92015+7)/(6^6+666-239239/7+1)=7

2629x7+1=18404

429^2-1=184040

239239 is a multiple of 7 and 2629

331259=92020+239239

429^2=10^2+1 mod (2629x35)

92015 is divisible by 2629x35

(331259*2-2629*7-3)*9/4473=6^4

(2629x7+3)/2=9203=331259 mod 23004

(541456+13-331259=210210=-4472 mod (359x13))
4472 divides (92020-69660)

541456+13-449449=92020

from above

541456-331259=31x449-18404 mod (359x13)

so

(92020-331259)=-970x449+13-18404 mod 4667

-239239-13=-970x449-18404 mod 4667

970x449=12x18404 mod 4667

or

3168=-12x18404 mod 4667

multiplying by 5

3168x5=-12x92020 mod 4667

210210=-4472 mod 4667

1051050=-(92020-69660) mod 4667

16170=-344 mod 4667

344 divides 541456

541456+13-210210=331259

mod 359x13 infact

541456=84
84+13+4472-4667=98

331259=-98=261 mod 359

i think there is something more subtle but it's too hard to understand for me.

manipulating a bit the above equations i obtained

-98=7000-1400x(92020-69660)=331259 mod 4667

but this is:

-98=1400x(-5x7x11x239+69660)=331259 mod 4667

so it seems to pop up 7x11x239 whcih divides 239239=331259-92020

bringing 5 out

-98=7000x(-18403+13932)=331259 mod 4667

-98=-4471x7000=331259 mod 4667

-98=2333x(-18403+13932)=331259 mod 4667

359-98=261

541456-13x16169=331259
541456-13x16169-13x18403=92020

16169,13932, 18403 are not random

maybe this is the reason why 541456=84 mod (359x13)

84=98-14

and 541456+69660=61116=98 mod (359x13)

follows that

(18403-13932) which is 4471 =-14^2 mod (359x13)

this means that
6^12-1=-4472 mod (359x13)

16169-13932=2237=4472/2+1

14^2+14=210

-210210=4472 mod (359x13)

-(14^2+14)*1001=4472 mod (359x13)

-14^2*1001=4472+13 mod (359x13)

from here I think you can derive why 541456+13-210210=331259

mod (359x13) +13-210210=4472+13

359*13*43-196*1001-13=4472

this maybe is the reason why 541456=84=98-14 mod (359x13)

and -(69660+541456)=-611116=261 mod (359x13)

261=359-98

541456=-6^7=84 mod (359x13)

14^2=196

69660=6^2x(2131-14^2)=6^2x(44^2-1)

maybe not a achance???

4x19179-84^2=69660

541456^2=4x19179-69660 mod (359x13)

541456^2+1=239x10 mod (359x13)

6^5=-1 mod 77

541456=-8 mod 77

331259=5=92020 mod 77

5+8=13

i guess that there is something to do with the fact that 541456=331259+210210-13
92020=331259-239239

so

331259=5=92020=(541456+13)=-5x6^5 mod 77

5x6^5=-77 mod 239

i think that here is the key

541456=11^2 mod 239

541456=-8 mod 77

541456 is of the form 6^5-7+18403s

92020 and 331259 are of the form 92020+18403s

(92020-6^5+6)=3370x5^2

pg(3371) is prime

probably i got something

-92020+6^5=2^5-1 mod 3371

this implies

92020=88^2+1 mod 3371

(92020-5)=239x7x11x5=88^2-4 mod 3371

I remember that 69660=6^2x(44^2-1)

so some conclusions can be done

92015x9=69660 mod 3371

92015=71x6^4-1=239x7x11x5

18403=1548 mod 3371

69660=(44^2-1)*36 Is congruenti to 18403 mod 3371

Dividing by 45

18403 congruenti to 43*6^2 mod 3371

18403*18=331259-5 Is congruenti to 43*6^2*18 mod (3371*18)

It utns out that

331259=29x31^2 mod (3371*18)

2x139x331=1001 mod 3371

2x139x331=92020-2

92018=14=331259 mod 51 pg(51) is prime

the inverse mod 51 of 14 is 11

-541456=11 mod 51

92018=14=331259=-449449 mod 51

541456+13-449449=92020

92018=14=331259=10x1001 mod 51

(92018-10x1001) is divisible by 1608

1608 divides 51456 and pg(51456) is prime

-(331259+72)=-331331=16=92020 mod 51

-92092=-92020-72=14=331259=-449x1001 mod 51

449=92=41 mod 51

239=-16=-92020=331331

92020+239 Is also divisibile by 67 PG(67) Is prime After PG(51)

331331-239 Is also divisibile by 541 and 36

i notice that (92020-16) is divisible both by 51 and by 451

pg(51) and pg(451) are primes

by the way famous 23004=69660-6^6 is congruent to 3 mod (451x51)

curious fact

(92020-4^2) is divisible by 4, 51, 451

pg(4), pg(51), pg(451) are primes

pg(215) and pg(541456) pg(2131) are primes

-541456=215= +2131 mod 51

(541456-2131) is divisible by (239+16=255=2^8-1) and by 2115=46^2-1

so 541456=(46^2-1)x(16^2-1)+2131

19179=2131x3^2=-51456 mod (51x5)

541456=2131x3^2x28+4444

239x1001=239239=-2^9=-2 mod 51

-92020x1001=-92020x32=2^9=2 mod 51

so 92020=16 mod 51

(239239+2^9-1)=5^3x(71x3^3+1)

71x3^3+1 divides 19180=2131x3^2+1

23004=3x7667+3

7667 is palindromic in base 10 and 6

(92020-16) is divisible by 451*12
(541456-16^2) is divisible by 451*12

4472 divides (92020-69660)

-4472=16=92020 mod (51x11)

215=22360/2=-541456 mod 51

22360=92020-69660

69660=3 mod 107

92020=0 mod 107

69660-3-23005 is divisible by (108^2-1)

23004=69660-6^6=3 mod (451x51)

so (69660-3) (multiple of 107) -6^6=0 mod (451x51)

69660=3x6^3 mod 23004

69660-3x6^3=9 mod (451x51x3=23001)

69660=3 mod 107
69660=0 mod 215

using chinese remainder theorem

69660 is a number of the form 645+23005k

if k=-1

645-23005=22360=(92020-69660)

also 6^6-1 is a number of this form

I notice the incredible fact that (69660-3)=651x107 where 651 is the product of the first 3 Mersenne primes. 651=3x6^3+3

6^6-19179=0 mod (387x71)

387 divides 69660

I think that something in some field is at work...surely 23004Z has something to do

i don't know if it is even possible for a human beeing to conceive a theory for these numbers

I think that a possible clue could be

18^2=69660=18^2x215=3 mod 107

so for example I notice that

22360=-18^2 mod (106x107)

22360=3 mod 107
22360=4 mod 108

I could think that this has something to do with the fact that 6^6=4 mod (108^2-1) and with the fact that (69660-9) is a multiple of 71 and 109

23008=3 mod 107
23008=4 mod 108

23008-3=23005
23008-4=23004

69660-428 (428 divides 92020) is a number congruent to 3 mod 107 and to 4 mod 108
(69660-428-3)=107x647=107x(3x6^3-1)

107 and 23005 are number of the form 11449+11556k

69660=-6^2 mod 264^2

264 is multiple of 44

69660=(44^2-1)x36

i think that using Lagrange or some primitive root concept one can get something

((139*(47+71*5)-1))=71x787

47 is the order mod 71 that is the least integer such that 139xn=1 mod 71

I suspect that this has something to do with the fact that 787 divides 541456

curious fact

92020=71x6^4+4

331259=92020+239*7*11*13

7,11 and 13 are primitive roots mod 71

-92020=331 mod (7x79)
-69660=18 mod (7x79)
-331259=541 mod (7x79)

(541-331)=210

(541456+13-210210=331259)

-239239=210 mod (7x79) this is equivalent to 239239=7^3 mod (7x79)

-541456=22^2 mod (7^3x79)

210x1001-1=-22^2=541456 mod (7x79)

playing around with this modulus (7x79) which is not random I got

7^3+12+210x1001=-2^7=541456+7^3+13

541456+11+210x1001=-2^7 mod (7x79)

from here

because 541456=(7^3+1)x1574

-140=(7^3+1)x1575

dividing both sides by 7 and by 5

-2^2=(7^3+1)x45 mod (79x7)

from here I got

69660x(7^3+1)=444 mod (79x7)

this reduces to:

-1=86x45 mod (79x7) where 86x45=3870 which divides 69660

i think that this has something to do with the fact that 69660-19179 is a multiple of 79

curious that 71 239 359 have 7 as smallest primitive root

another curio about these crazy numbers:

pg(451=11x41) is prime

pg(2131) is prime

451+41^2-1=2131

this could be connected with the fact that it seems to exist infinitely many pg(k) primes with k multiple of 41. pg(181015) for exampe is prime and 181015 is a multiple of 41

it seems that there are infinitely many pg(k) primes with k multiple of 41 and infinitely many with k multiple of 43.

When k is multiple of 43, then k is of the form 41x43xk+r

manipulating a bit the previous things I got

(92020-16) is divisible by (71x108-1)=7667 a palindrome
7667=11x41x17

pg(451) and pg(17x3) are primes

71*108*12-1=92015=239x5x11...

(92020-6) is divisible by (71*108*6-1)

(92020-10) is divisible by (71x108x2-1)

pg(19179=2131x9) is prime

19180 is divisible by (71x27x2+2)

181015 and 92020 are numbers of the form 51s+16

92020 is 0 mod 43

using crt

92020 has the form 2107+2193s

allowing negative s

-4472 is a number of the form 2107+2193s

4472 divides (92020-69660)

4472=-16=-92020 mod (51x11x4)

i wonder if there are infinitely many primes pg(k) with k of the form 16+51s

pg(67), pg(92020), pg(181015) are primes with k of the form 16+51s

are there infinitely many pg(k) primes with k of the form 14+51s?

pg(79) and pg(331259) are primes and k is of the form 14+51s

331259=-13 mod 18404
239239=-13 mod 18404
541456=7740 mod 18404

7740 divides 69660

541456x9=69660 mod (18404x261)

69660=14448 mod 18404

(69660-14448)*6-13=331259

an extension of the conjecture could be:

there are infinitely many primes pg(k) with k of the form +/- 14+51s.
394 for example is of the form -14+51s

pg(181015=16+51s) is prime

181015=-1 mod 22^3 curious

curious that also 67=51+16 is congruent to 1 mod 11 pg(67) is prime

11 is one og the factors of 451

92020=16+51s

92020-16 is divisible by 11

92020 is even , 67 and 181015 are odd

(92020-16)=-11 mod 239x5x7 so 92020=5 mod 239x5x7

331259=5 mod 11

92020=5 mod 11

92020/5=18404

18404=1 mod (239x11...)

-181015=(5x11)^2+1 mod (429^2)

92020=(429^2-1)/2

i think that this could explains something

mod (429^2-1)/2 for example 92020=0

-181015=55^2 mod (429^2-1)

this is equivalent to

-181015=(5x11)^2 mod 92020

181015 and 92020 have the same form 16+51s

181015, 67, 92020 are of the form 16+51s

-181015=(5x11)^2 mod 92020
-92020=0 mod 92020
-67=71x(6^4+1) mod 92020

curious that

181015, 67 and 92020 (with the form 16+51s) are congruent to +/- j^2 mod 71

67 and 92020 are +/- 4 mod 71

181015=6^2 mod 71

-71x(6^4+1)=5=92020=331259 mod 11

-181016=(5x11)^2 mod (429^2)

i think that here is the rub....

-181016, 5x11 and 429 have 11^2 as divisor

so you can divide by 11

it turns out thst

-1496=5^2 mod 39^2

181015=16 mod 51 and mod 39^2

39^2*7+1=22^2

67 92020 and 181015 are of the form 16+51s

residues mod 11 and mod 17 and mod 13 are not random I think

as you can see

-67=-1 mod 17 and -67=1 mod 11 67=2 mod 13

92020=4^2 mod 41

-92020=5^2 mod 41

-92020=5^2 mod 449

541456+13-449449=92020

541456=-5^2-13 mod 449

inverse of 25 mod 449 is 18

92020x18=-1 mod 449

92020x18*(1/5)=331272

331272-13=331259

92020*(1/5)=18404

18x18404-13=331259

becasue 18404=1 mod (239x7x11)

18x18404-13-5 is a multiple of (239x7x11)

429^2x18=16 mod (449x17)

92020=(429^2-1)/2

92020=16 mod 17

331259=18404*(5+13)-13

Mod 449

18404*5=-25 mod 449
18404*13-13=239239

331259-239239=92020

-18404*13=65=69660 mod 449

-331259-65=-541456=--331259-69660 mod 449

92020x18=-1 mod 449

this is the starting point
331259*5-92020*18=65

69660=65 mod 449

92020x18-13x5=0 mod 331259

5x(18404x18-13)=0 mod 331259

(331259+13+65)x18=-1 mod 449

331259+13=0 mod 18404x18

-(331259+13+65)=5^2=-92020 mod 449

92020, 331259 and 541456 are congruent to -25-13s mod 449 for some nonngeative s

331259+13=-90 mod 18409=41x449

90-65=25

69660=541456-331259 mod 449

18404x18=359 mod 449
-18404x18=90 mod 449

18404x18-13=331259

-18404=5 mod 449

331259=-(449-359)-13 mod 449
pg(359) is prime

69660=(449-359)-5^2 mod 449

331259=(359-13) mod 449

331259+13=0 mod 18404

-18404=5 mod (449)

so 92020=5x18404=-5x5 mod (449)

-18404x18=90=-359 mod 449

331259=18404x18-13

so 331259=-90-13=-103=(359-13) mod 449

5x359=-1 mod 449

92020=-(1/359^2) mod 449

so 92020=-25 mod 449

92020=-1/18=-(1/359^2)=-25 mod 449

359^2x18404=-90 mod 449
18x18404=-90 mod 449

90^2x18404=-90 mod 449
(1/25)x18404=-90 mod 449
90 is the inverse of 5 mof 449

331259-90-13=0 mod 449

-5x18x18=(331259+13)/5^2=359/5^2 mod 449

-5x18=331259+13=359 mod 449

5x18x18=-(331259+13)/5^2=90/5^2 mod 449

331259=-90-13=-103 mod 449

92020=-(1/359^2)=(-1/90^2)=-25 mod 449

18404x90=-1 mod 449

the invers mod 449 of 5 is 90

18404x90x90=-90 mod 449
this means

331272=-90=359 mod 449

331259=-103 mod 449

331272x444=1 mod 449
331272x(-5)=1 mod 449

331272x(-5)=-92020x18

from here follows necause inverse of 18 mod 449=25

92020=-25 mod 449

but the question I think is more subtle than I think

429^2=-7^2 mod 449

541456+13=-7^2-1-(429^2-1)/2

429^2=-7^2 mod 449

(429^2-1)/2=92020

mod 449

(429^2-1)/2=-5^2 mod (449)

541456+13=-5^2 mod 449

541456=-5^2-13 mod 449

from thsi follows that

541456+13-92020=0 mod 449

92020=-5^2 mod 449

18404x359=1 mod 449

92020=18404x5

359x5=-1 mod 449

-541456-13-239239=-331259=103 mod 449

-541456-13+78=-331259=103 mod 449

so

-541456+65=103 mod 449

103-65=38=13+25
541456=-25-13 mod 449
69660=65 mod 449

so

541456-331259=65=69660 mod 449

210210=-239239=78 mod 449

541456+13-210210=331259

331259-239239=92020

92020=359+65 mod 449

so 22360=92020-69660=359=331259+13 mod 449

the numbers are clearly structured, but unfortunally there is no elementary method to solve the puzzle of the giant mega-structure that generetes these primes.
Beeing structured, no surprise we do not find any prime of this type congruent to 6 mod 7.

exponets leading to such type of primes appear to assume only certain particular forms. This maybe obstrues the possibility of a 6 mod 7 prime of this form

look at this crazy curio:

427x428x429x430=1 mod 449
33712999320=427x428x429x430 is the concatenation of 3371 (pg(3371) is prime) and 2999320 which is divisible by 449

i think that with new tecnologies just for recreational purposes it would be worth to find other exponents leading to a prime of this type

429^2x394=1 mod 449

pg(394) is prime

becasue 429^2=92020x2+1 (pg(92020) is prime)

92020x2x394=-3x131 mod 449

92020=-3x131x300^2 mod (449x359)

i think that these exponents leading to a prime are connected to each other in a very deep and mysterious way

exist pg(K) primes with k multiple of 215 (3 found)

exist pg(k) primes with k multiple of 43 (4 found three of which are multiple of 215)

exist pg(k) primes with k multiple of 139 (2 found)

exist pg(k) primes with k of the form 16+51s (3 found)...

it seems clear that the exponents leading to a prime are not random at all.

Incredible:

pg(181015) is prime pg((429^2-1)/2=92020) is prime

181015=429^2-1-55^2!!!

429 and 55 have 11 as common divisor

11x(16731-275)-1=181015

181015=11^2x(39^2-5^2)-1

181015=92020=67=55^2=4^2 mod 51

Neme(k) this is the name of these numbers

pg(1323) is prime and pg(39699) is prime

1323=11=39699 mod 41

this is another case in which exponents leading to a prime seem to have a certain form, in this case 41s+11

39699=11 also mod 11x41=451

pg(6231) and pg(2131) are primes

6231 and 2131 have the form 41s+40

I notice that 1323 and 39699 have also the same residue 10 mod 13

so 1323 and 39699 have the form 257+533s

incredibly 1323 and 39699 are of the form 257+41x(t^2+1) for some t.

1323=257+41x(5^2+1)
39699=257+41x(31^2+1)

and remarkably 5^2 and 31^2=-1 mod 13

These numbers contain a lotq of surprises because they are structured...but the problem Is that only an alien of type 5 civilisation could solve this kind of problems I think that when Riemann hypotesis Will be solved this kind of problems Will be still open and for many other centuries

(449*41-6) divides both 92015 and 331254

A Little pompously I could call these numbers numbers for the end of the world or at least for the next geologic era

It's simpler to say that 1323 and 39699 are of the form 298+41xp^2

neme(176006) (or pg(176006) is prime)

(176006+2) is divisible by (92020-69660-359)

176006=-2=-451 mod 449

pg(451) is prime

also pg(2) or neme(2) is prime

pg(92020), pg(67) pg(51) and pg(451) are primes

92020-67=-51 mod 451

pg(181015) is prime

11^2*(39^2-5^2)-1=181015

11^2x39^2=429^2=2x92020+1

because (a^2-b^2)=(a+b)x(a-b)

11^2x(39+5)x(39-5)-1=181015

39+5=44 divides (92020-16)
39-5=34 divides (92020-16)

181015 has the curious representation 44x4114-1 with 4114 a palindromic number

92020x18=-1 mod 449

92020x18=17^2-1 mod (451x51x6^3)

92020 has the curious representation 828180/9

82-81-80...

mod 449

the inverse of 359 is 444

444=-5 mod 449

22360=-90 mod 449

90x5=1 mod 449

90x5+1=451

I think that probably there is a connection to the fact that:

69660=14 mod 359
6^6=-14 mod 359

331259=-14^2/2 mod 359

but I am not sure

a curious thing I noticed is this:

6231, 19179, 39699, 51456, 56238, 69660, 75894. Seven multiples of 3 in a row.

pg(19179=2131x3^2) is prime

69660=9 mod 71
19179=9 mod 71

(69660-19179) is a multiple of 79*3^2

pg(79) is prime

79*3^k =1 mod 71 for k=2+70xs

(69660-19179)/79+9=3x6^3=2^nx3^(n+1)

(7740-2131)/71=79

19179 and 69660 arebof the form -31302+50481s

6^6=-2 mod 41

69660=1 mod 41

69660-1-6^6-2=23001 divisibile by 51 and 451

69660=1=51456 mod 41 and mod 4551=111x41

92020=10^3 mod 41 and mod 4551=111x41

(69660-51456) divides (92020-10^3)

10^3 mod 41=16

6^6=-14=-69660 mod 359

77x6^6=-98x11=-69660x77 mod 359

77x98x6^6=-98=-69660x77x98=331259 mod 359

77x6^6+98*11=3593590

7x6^6=-98=-7x69660=331259 mod 359

7x6^6=-2x7^2=7^4+14=331259=7^4+69660=7^4-6^6 mod 359

I think that 6^6 69660 are the way to beat...
the role of 71 and 359 are mysterious.

I think that it needs an extremely complex tool for understanding these numbers.

PG(3336) and PG(75894) are primes wirh 3336 and 75894 multiple of 139

I suspect that somerhing Is happening in Z139

3336=1=75894 mod 29

75893/29=2617 which Is a wagstaff prime exponent.

I conjecture that there are infinitely many PG(3336+29*139s) primes.

It seems that exponents leasing to a PG prime can assume only certain forms

92020=2 mod 139
92020=3 mod 29
I think that there Is something in 139Z and 29Z a mysteriius force in these fields generaring the exponents

I think that there Is a connection wirh the fact that the modular multiplicative inverse of 71 mod 139 Is 47

71*47-1=3336

71x6^4+2=0 mod 139
71x6^4+2=1 mod 29

75894=-19X3 mod 87

3336=-19x3 mod 87

i think that gigantic groups are at work in these numbers, but it's very very hard to find our rosebud

139 is the inverse of 546 mod 139

i think that there should be a connection with the other number 56238 which is divisible by 546 but I don't understand how

3335=(4X10^3+2)X(5/6)

just curious

I conjecture that there are infinitely many pg(k) primes with k of the form (2x10^n+16)/6

the two found are pg(36) and pg(3336)

the next 333...6 congruent to 0 mod 139 and 1 mod 29 is

(2x10^648+16)/6 ....648=3x6^3

I wonder if pg(333...6) is prime ...

i think that something periodic is at work...maybe applying Dirichlet characters???

75894 and 56238 are multiples of 546

the sum is 132132=546x(3^5-1)=546x242
546x242=1 mod 71

92020=2+92018

662x139=2 mod 71

23x139=2 mod 71

48x139=2x3336=-2 mod 71

3336=-1 mod 71

75895/5=15179=-4000 mod 2131

1776x(19179-667x6)=-667=260x1776 mod 2131

3335=667x5

541456-51456=700^2

I suspect that there Is something to do with

Numbers 546-56=490

7*(1+7*111...) 56 546 5446 5444...6....

546=7*(1+7*11)
490=546-56

92020 and 541456 are of the form 8643s-3053.

541456+13-449449=92020

449449=13 mod 8643

8643/43=201 which divides 51456

3053 mod 449=359

541456-10*51456=164^2

430*(71*(108+7667*3)-1)/451/17=92020

Amazing that 71*6^4*5-331259=128821 a palindromico Number

71*6^4=1 mod 239
331259=5 mod 239

Amazing that

PG(39699) and PG(79798) are primes

39699=401*99
79798=401*99+401*100-1

92020 is congruent to 0 mod 215

92020=3338 mod 4031

Using chinese remainder theorem

92020+866...5s are numbers =0 mod 215 and 3338 mod
4031

Allowimg negative s

92020-215*139*29=774645

774645 Is the concatenation of 774 and 645 which both divide 69660

29Zx139Z something here Is in action!

331259=717=239x3 mod (29*139)

331259=5=92020 mod 239

43*((429^2-9)/71+10^4)=541456

Is there a meaning??

(92020x2-8)/71 Is a 3 smooth Number 2592. 2592+10000=12592 which divides 541456

43*(429^2-9)/71 Is divisible also by 6966=69660/10

429 mod 139=12

PG(1323) Is prime

(429^2-12^2)=139x1323

I think that a super tool Is needed for these primes...something much more difficult than permutarions in 3x+1 problems or dirichelet characters

331259=92020+239239

I think that 429 Is involved but i canot clearly see how

239239/13=18403

429^2=11 mod 239

Dividing by 11

16731=1 mod 239

239x70=-1 mod 429

92020=331259=5 mod 1673

239239=-13 mod (2236x107)

2236 divides (92020-69660)

429^2=1 mod 107

I think that here Is the Key

429 is the 8-th catalan number

I suspect that partitions are involved in these exponents, in a way, however, that is not easy

there is this identity:

((44^2-1)x6^2-6^6+1)x4=(429^2-1)/2

i think that manipulating this identity and knowing that gcd(44,429)=11 one can get something but I have no idea how

Maybe an intuition but the idea has to be devoped:

541456+13-210210*2-29029=92020

210x2+29=449

29Z in action?

i think that ring 4031Z is in some mysterioius and hard way involved

92020=-77x3^2 mod 4031

331259=239239+92020

3338+4031*22=4*((44^2-1)*36-6^6+1)

Manipulating this identity

You get a 2 degree equation in x

4031x+3338=4x144x^2-4x46691

One solution Is x=22

The discriminant DELTA=(21313)^2 where 21313 looks like 2131 so I suspect that there Is some connection with these exponents

in the ring 215Z the zero divisors have the form q*172

69660 92020 541456 are all divisble by 172

the starting point is

215x107=1 mod 71

this is equivalent to

215x36=1 mod 71

215x36=7740 which divides 69660

pg(36) is prime pg(215) is prime pg(69660) pg(215x107x2=92020) are primes

92020=2x(6^3-1)x(6^3-2)

215=2 mod 71

69660=3 mod (6^3+1)

6^6=1 mod (6^3+1)

69660=3x6^6 mod (6^3+1)

92020=16 mod (451x51=23001)

23001 is a Poulet number Fermat pseudoprime in base 2

215x107=1 mod 71
139x331=1 mod 71

92020=2 mod 139x331

139x331-215x107=69660-6^6=23004

215x107=1 mod 71 139x331=1 mod 71

pg(1323) and PG(39699) are primes

1323 and 39699 have the form 98+y^2

For y=35 and y=199

1323=98+35^2
39699=98+199^2

35^2 and 199^2=6^2 mod 41

35=199=-6 mod 41

(239239+243)/29=1 mod 359

so

239239=-214 mod 359

92020=430x214

430 mod 359=71 or =-288

so

288x239239=-243 mod 359
288x239239=-242 mod 451

92020=-243 mod 359
these are only simple modular considerations, I think that something more complex is needed to fully undertand what is going on...

243-29=214

(451-359)x1001-72=92092-72=92020...

69660=2 mod 29

(69660-2)/29=7^4+1
something in the 29Z ring?

92020=(429^2-1)/2

429 is a Catalan numero

56238+75894 =132132 where 132 is catalan too

92020-3 is a multiple of 29,19,167

(92020-3) is divisible by (138x23-1)

667x138=1 mod 449

92020=-25 mod 449
92020=-26 mod 667

92046 is divisible by 46
pg(46) is prime

probably something involving groups is in action

331259=-(69660+18)/(79x3^2) mod (71x359)

mod 71

331259=-9-18=-27 mod 71
69660-19179=711x71

pg(79) is prime

79x3^2=1 mod 71

(69660+18)/79=-21^2 mod 1323
Pg(1323) is prime

2x21^2=-21^2x79 mod 189^2

92020x4=3337 mod (79x3^5)
92020x4=3336 mod 359

71x6^4+4=92020

Si

15232x4=71x47=3337 mod (79*3^6)

79x3^5=19197=2131x9+18

79x3^5=44 mod 71

331259=44 mod 71

79x3^2=1 mod 71

69660-19179 Is a multiple of 711

79x3^2=2131 mod 71

So

331259=-19179x3 mod 71

331259+19179x3+1 Is a multiple of 359

69660-19179 is multiple of 71x9

71x9=1 mod 29x22

92020=3338+139x29x22

331259=-79 mod (7^4x138)

(331259+79)/7=142002/3

142000=71x2x10^3=1 mod (77x331)

331259=-72 mod (331x77)

Pg(67) is prime

92020=-67 mod (71x 1297)

331259=-67 mod 79
I guess that there is some periodic modular character but it is completely out of reach

92020=99999 mod 7979

92020-69660=22360=215x104

69660-6^6=71x18^2

71x18^2=-1 mod 107

Inverse of 71 mod 107 is 104

18^2+104=428 which divides 92020

22360=104x215

18^2=3 mod 107

.(429^2-3x6^3) is 0 mod 71

(429^2-3x6^3)/41=4473

4472 divides (92020-69660)

I think that something very complex is in action on 71Z

I think these exponents are full of treasures, connections with unimmaginable secrets...the problem is they are too hard to study

69660=6^6 mod 23004

23001=451x51 pg(51) and pg (451) are primes

69660-9=3x6^3 mod 23001

69660=9 mod 71

i think that playing around with 69660=3 mod 107x651, something could be found ...651 appears in many oeis sequences i think that it has a central role, it is the product of three Mersenne primes, it is a pentagonal number,...

(6^6-3x6^3+1)*2=92020-2...I think that...yes...powers of 6 are involved!

69660=645 mod 23005

From here

(69660/5-645/5)*24=331272=331259+13

(429^2-1)/2=3338+4031x11x2

if I factor out 11

607x/2-6677/2=0

I think that that odd 6677 has some importance, it appears in many oeis sequences, but his sense is obscure

6677=11x607

607 is an exponent leading to a Mersenne prime

a curio:
pg(394) is prime

394 is the sum of the first two Honacker primes 131+263

I suspect that there Is a connection between 23001=451x51=1 mod 46

PG(46) Is prime

PG(51=17*3) Is prime

-17=29 mod 46

51x9=-1 mod 46

23001=451x51=1 mod 46

29x3=87=1 mod 46

23000=-460 mod 17x46
23000=-461 mod 29

23000=92020=-19x3^2 mod (17x29)

(92020-23000)/17=1 mod 451

92020=23002=11501x2 mod 11503

The inverse of 11501 mod 11503 is 71x3^4

92020=16 mod 23001
92020=0 mod 23005

using CRT

92020+529138005k

for k=1

92020+529138005=23005^2

I wonder if pg(529138005) is prime

331259=44 mod 71

The inverse of 44 mod 71 is 21

21 divides 1323 pg(1323) is prime

1323=45 mod 71

331259=44 mod 71

1323=63×21

45×44-63=71x3^3=1917

19179=2131x9=9 mod 1917

331259×45=2^6-1 mod 71xg

Where g is a 3th smooth number

(331259*45-63)/92016=162
92016=71x6^4
162 divides 69660

331259x26=-63 mod (71x29)

331259=21 mod 29

26=-3 mod 29

pg(19179) is prime
pg(3336) is prime

3336=1 mod 29

19179/3=3058 mod 3335

19179x29/3=92020-3338 mod 3335

19179x29/3=92020-3=71x6^4+1 mod 3335

3058=22x139

2131*27=1 mod 29

(19178)x3-6^4=56238 pg(56238) is prime

2131=-74 mod 147

2131*9=19179=-74*9 mod 1323

19179=-666 mod 1323

I think it is not random that 19179=-666 (the beast number) mod 1323

19179=5*63^2-666

2131=-73 mod 29
2131=-74 mod 147

19179=-666 mod 63^2=1323x3 pg(1323) is prime

19179=-665 mod 451

pg(451) is prime

I think there is a connection with

67x79=1 mod 1323

pg(67) and pg(79) are two consecutive ec primes

because 19179=-3 mod 139

19179=3303 mod 126^2

because

19179=-3 mod 138x139

138x139=6x19x29 mod 126^2 (and so also mod 1323)

I am prestti sure that something is boiling in 29x139Z

69660=3 mod 107

651x107=6^6 mod 451

651x107=203 mod 451

(69657-203+1)/29+6=7^4

i think that in 29Z there is something...

(69660-2)/29-1=7^4

429^2=1323x2=(79×67-1)/2 mod (139x29)

From here it should follow something

Because 92020=(429^2-1)/2

Forcexample it follows that

864864x429=1323 mod 4031

2230*429=1323 mod (79×139×29)

I think that 4031Z has something special for these primes, but no simple modular considerations are useful for sheding light on it.

71x6^4x430=92020 mod (466x394x215)

92020=218 mod 394

71x6^4x430=218 mod (466x394)

39 566 662=71x6^4x430-218

pg(394) is prime

466 are the first three digits of 6^6

because 430=6^2 mod 394

71x6^6=92020 mod 394

331259=-95 mod (29x29x394)

i really think that there is a spectacular hidden structure, but incredible complex tools are needed

It's like a perfect fit...29Z or 139Z or both are surely involved...exponent like 394, 359 leading to a prime are like pieces of a puzzle they perfectly fit on the entire structure

71x6^6=218 mod 394 ...from here

71x164=218 mod (394x29)

92020x2+1=35^2 mod (394x29)

429^2=-101^2=35^2 mod (394x29)

394x29=11426 is a number that appears in sequence oeis A255684 which has to do with Bernoulli numbers

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