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? 
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 (696606)/26 is congruent to 13 mod 43 (920206)/26 is congruent to 13 mod 43 (5414566)/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 (54145676913*93)/1763=306 (6966076913*824)/1763=33 (9202076913*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*2^210^3 or equivalently to  (19*2^61) mod 13 69660 (even) is congruent to 307*2^21001 or equivalenly to (19*2^61) mod 13 92020 (even) the same 541456 (even) the same (19*2^6*(54145692020)/(13*43)+1)/13+10=75215 is a multiple of 307=(215*101)/7=(54145601)/17637 pg(51456) is another probable prime with 51456 congruent to 10^n mod 41 75215=(51456*19+1)/13+10=(19*2^6*(54145692020)/(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=(51456/22*13*10^243)/19 ...so summing up... 215 (odd) is congruent to 3*19*2^2 mod((41*43307)/(7*2^4)=13) where 41*43307 is congruent to 10^3+3*19*2^2 mod (3*19*2^2=228) 69660 (even) is congruent to (3*19*2^21) mod 13 ... the same for 92020 and 541456 another way is 215 (odd) is congruent to 15*81 mod ((41*43307)/(30715*13)) 69660 (even) is congruent to 15*81 mod ((41*43307)/(30715*13)) the same for 92020 and 541456 215 is congruent to (41*4330710^3)/2 mod ((41*43307)/(7*2^4)) 69660 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 92020 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 541456 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 215 is congruent to 3*19*2^2 mod ((41*43307)/(2*(19*31))) 69660 is congruent to 3*19*2^21 mod ((41*43307)/(2*(19*31))) and so 92020 and 541456 215 is congruent to 3*19*2^2 mod ((3^61)/(3*191)) 69660 is congruent to 3*19*2^21 mod((3^61)/(3*191)) 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=2131307*426^2 So 215 is congruent also to 2131307*426^2 mod 13 And 69660 is congruent to 2131307*426^2+1 mod 13 and so also 92020 and 541456 69660 is congruent to 1763307*51 mod ((1763307)/112)=13) And so 92020 and 541456 215 which is odd is congruent to  1763+307*5+1 mod 13 215+1763307*51 is divisible by 17 and by 13 And also 5414561763+307*5+1 is divisible by 13 and 17 215 and 541456 have the same residue 10 mod 41 ((5414561763+307*5+1)/(13*17)+1) *200+51456=541456 69660 92020 541456 are congruent to 7*2^6 mod 26 7*2^6(1763307*51)=221=13*17 215+7*2^6 is a multiple of 221 5414567*2^6 is a multiple of 221 
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^q1 mod 13 with q a nonnegative integer 215 is odd and pg(215) is prp 215 is congruent to 7*2^q mod 13 
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 2^696601  2^695591 is prime 2^920201  2^920191 is prime!!! 
please show us a proof that they are prime

...
Well...
actually they are only probable primes... maybe in future they will be proven primes 
[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%5E920201%29*10%5E27701%2B2%5E920191[/url] 
... I note also...
69660 I note also that
(lcm(215,344,559))^2=4999*10^5+6966060 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*181) and 69660 (multiple of 43) is 215 mod (18*181) I note that the polynomial X^2X*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 429^44*7967780460 which is a perfect square and lcm(215,344,559) divides 429^44*79677804601 Pg(331259) is prime and pg(92020) is prime. 92020+(92020/2151)*559+546=331259 Again magic numbers 559 and 546 strike! So 331259 is a number of the form 215*(13s1)+(13*b2)*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*8546)*311# 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*181) and 69660 (multiple of 43) is 215 mod (18*181) 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*181)*215 92020=69660+18*18*69+4 215 is congruent to 108 mod (18*181=323) 541456 is congruent to 108 mod (18*181) 92020 is congruent to (17*171) mod (18*181) 69660 is congruent to 215 mod (18*181) 108=6^318^2 17*171+36216=17*171+6^26^3=108 215=6^31 To make it easier 215, 69660, 541456 are congruent to plus or minus 215 mod 323 92020 is congruent to (17*171) mod (18*181) 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 6^31 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 (17^21) mod 323 and to  (6^21) 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)+6^31=(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 17^21 215, 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)^21 mod 323 288 is 17^21 288 in base 16 is 120 120=11^21 also 323=18^21 in base 16 is 143=12^21 344*((14444561763*2^9) /3441)=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 3smooth number mod 323 108 and 288 are both divisible by 36 
Q77I [QUOTE=enzocreti;541360]69660 I note also that
(lcm(215,344,559))^2=4999*10^5+6966060 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*181) and 69660 (multiple of 43) is 215 mod (18*181) I note that the polynomial X^2X*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 429^44*7967780460 which is a perfect square and lcm(215,344,559) divides 429^44*79677804601 Pg(331259) is prime and pg(92020) is prime. 92020+(92020/2151)*559+546=331259 Again magic numbers 559 and 546 strike! So 331259 is a number of the form 215*(13s1)+(13*b2)*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*8546)*311# 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*181) and 69660 (multiple of 43) is 215 mod (18*181) 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*181)*215 92020=69660+18*18*69+4 215 is congruent to 108 mod (18*181=323) 541456 is congruent to 108 mod (18*181) 92020 is congruent to (17*171) mod (18*181) 69660 is congruent to 215 mod (18*181) 108=6^318^2 17*171+36216=17*171+6^26^3=108 215=6^31 To make it easier 215, 69660, 541456 are congruent to plus or minus 215 mod 323 92020 is congruent to (17*171) mod (18*181) 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 6^31 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 (17^21) mod 323 and to  (6^21) 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)+6^31=(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 17^21 215, 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)^21 mod 323 288 is 17^21 288 in base 16 is 120 120=11^21 also 323=18^21 in base 16 is 143=12^21 344*((14444561763*2^9) /3441)=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 3smooth number mod 323 108 and 288 are both divisible by 36[/QUOTE] 215 69660 92020 541456 are congruent to plus or minus (6^k1) 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^(a3)*3^(b3)) 69660 is multiple of 3 and congruent to 0 mod (6^21) 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 (42^21)*(18^21)(6^3+1)*(2^7+1)=541456 I also note that 69660=(2^7+1)*(6^3+1)+(2^7+1)*(18^21) And by the way 215=(42^21)*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*(6^61) 27993 in base 6 is 333333 27993 has also the representation: (42^21)*4^2(6^31)=27993=2*43*(18^21)+(6^31) x/5+(42^21)*(18^21)3/5*(6^61)=20*3/5*(6^61) The solution of this equation x=92020 this identity: (42^21)*(18^21)=(10^3+18^2)*430+43*3 Maybe it is not a chance that pg(10^3+18^21=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^b1) 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 (p1) mod 323 where p is a perfect power 92020 has the factorization 2*(6^31)*(6^32) lcm(215,344,559)=(6^31)*(2*(6^32)18^2)=(2*(6^32)18^2)*(2*(6^32)18^2+111) 69660=92020lcm(215,344,559) 215=2*(6^32)+2*(6^3+1)2*18^2+1 92020=(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2 so 215=2*(6^32)+2*(6^3+1)2*18^2+1 69660==(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2((6^31)*(2*(6^32)18^2)) 92020=(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2 541456=(42^21)*(18^21)(7^36^3+2)*(6^3+1) 69660=1111115*11111+8*(42^21) 344=7^3+1=2*(1111115*11111)/(18^21)=A so lcm(215,344,559)=lcm(215, A, A+215) 541456=5*111112(42^21)*8 (12^21) divides (1111115*111116^3+1) 559=(1111115*111116^3+1)/(12^21)+(1111115*11111)/(18^21) Quite clear that 215, 69660, 92020,541456 multiple of 43 are congruent to plus or minus (18^26^k) mod (18^21) where k is 2 or 3 and 3 indeed is the maximum exponent such that 18^26^k i 1s positive (54145618^2+6^3)/323=41^25=1676 (9202018^2+6^2)/323=17^25=284 (69660+18^26^3)/323=6^3 (215+18^26^3)/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*q5)*(18^21) plus or minus (18^26^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 (18^26^3) as in the cases k=56238 and k=75894 Curio of the curios: pg(331259) is probable prime and 331259 is prime magic: 71*6^6331*(10^43*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 (11^21) mod (13^21) and for n odd Is congruent to (7^21) mod (13^21) 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*(6^31)) 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)*(429^21) 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*(2^61)1 I suspect that something in field F(239) is in action! 331259^(1) mod (215*239)=49999=(10^52)/2 ((71*(2^61))1)/2=lcm(344,559)=92026966 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*(429^21)/10541456) /3=9202 So 10* (239*10*215+3*(429^21)/10541456) /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*(429^21)/10 is a multiple of 559 239*10*215+3*(429^21)/10=(42^21)*(18^22)+43*2^5 This identity (239*10*215+(3/10)*(429^21)541456+92020)/559=92020/(215*2) One can play around with this expression containing 215 and 559 And substitute for example 92020 with (429^21)/2 Pg(331259) is prime and also 331259 is prime The inverse modulo (215*239) of 331259 is the prime 5*10^41=49999 (331259*499991)/(215*239)=7*(18^21)*(12^21)10^3+1 So 331259=((((18^21)*(10^3+1)10^3+1)*239*215)+1)/(5*10^41) 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+(30^21)*(42^21) 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+(30^21)*(42^21))/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 9202359 is a multiple of 239 Pg(92020+239239=331259) is prime I note also 331259=71*6^6331*(10^43*331)=6^2*920213=92020+239239 239239 is congruent to 13 mod 107 and mod 43 22360=(10*(239239+13))/107 92020=69660+22360 92020=331259239239 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*6^6182^23^2331*(10^43*331)10=546^2 546215=331=546(6^31) by the way 182^2+3^2+546^2 is prime 546^2 is congruent to 6^3 mod 331 541456((3*239) ^2239+(429^21)/20)=(429^21)/10 541456=(18/5)*(429^21)/2+13*((429^211)/102235) 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 323k108, or 323k288 I notice that (323108)=6^31 and (323288)=6^21 211 and 284 are also the residues of 71*6^4 and 71*6^6 mod 323 the 18th pg prime is pg(1323), the 36th is pg(360787) modulo (18^21=323), 1323 is 31 and 360787 is 319 the difference between 319 and 31 is again 288 1323 is congruent to 31 mod (6^44) 360787 is congruent to 6^2*2^3+31 mod (6^44) 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 (6^41008=288) mod 323 so multiple of 43 are either congruent to plus or minus (18+90) mod (18^21) or to 6^4(18+90+900) mod (18^21) 92020 is congruent to (11*6^2108) mod (18^21) and to  11*6^2 mod (304^2) so great fact 215, 69660, 92020, 541456 are either congruent to plus or minus (108*21) mod (108*31) or to plus/minus (108/31) mod (108*31) [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+331711[/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 (5414569202*3)/215/239=10 (13*3311)*7*11+57*11*13*239=92020 541456 is congruent to (9203+(71*6^41)/5) (mod (239*5*43)) 9203 is prime (5414569203(71*6^41)/5)/215/239=10 33125913*(9202*21)=92020 9202013*1720=69660 lcm(215,344,559)=13*1720 67*(16683/67+22360/431)=51456 where 16683=9202*211720 and 22360=lcm(215,344,559) 331259 has the representation 331259=36*(7/180+(920204)/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*6^22^2)/40=(239*6^22^2)/(6^2+2^2) 331259=(3*71*6^5+7)/5 71*6^6(3^3*71*6^57)/5=331259 (3^3*71*6^57)/5=(10^43*331)*331 so we have 92020=71*6^6(3^3*71*6^57)/513*(71*6^41)/5 (69660/3860)/13=1720 I see where 215 comes from (33125923666)/(12^21)=9*2391=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*9215) mod 71 215*9=44^21 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)/99971.107107107=541456 or 92020*(1763*10+7)/299771.107107...=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*27*71*6^4)/2 331259=(71*6^3*43)/2+307*5 331259=(2^10+1)*(18^21)+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*9991) 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*9991) is congruent to 172 mod 559 (271*9991) 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 (700^2164^2)/51456=3^2 (10*700^2164^2)/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*218 271*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*1111)*2(2439*201239)=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*2211239 51456=245*211239 Where 211 is prime and 2211 is a multiple of 67 as 51456 (54145692020) is a multiple of 559 and 67 (54145692020)/(67*3)=2236 69660+22360=92020 But ed can substitute 67*3 with 2211/11 So (54145692020)*11/2211=2236 i note also that (541456210*1001+13)=331259 92020 is so congruent to (245*2211239) mod (3*67*43*13) and (245*2211239=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 (34413s) 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)*6^213 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*(239239546)/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=(429^21)/2) (239239546)/559+2=429 so( ((239239546)/559+2)^21)/2=92020 (239239+(12^21)*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 2392111 =451 congruent to 10^m mod 41 pg(451) is prime ((71*6^4+4)344)/559=164 but we saw before that 51456=(700^2164^2)/9 541456=(10*700^2164^2)/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*6^4340)/52 +r with r being the residue 1763s+r such that is congruent to 10^m mod 41 34052=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 (92020344)/559+(92020344)/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*100181*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)+13210*1001=0 So this could be useful if you have in mind that 331259 is congruent to 71*3^3 mod (239*13) 92020=54145613*(449*771) 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 (92913) mod 1001 69660*559 is congruent to (331259559331) mod (1001) 69660*559 is so also congruent to (92020559331) mod (1001) so 69660*559 is either congruent to (x890) mod 1001 or to (x903) 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 (x890) mod (1001) with x=331259, 92020 if x=541456 not congruent to  72 mod 1001, then 69660*559 is congruent to (x903) 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*2111) or to (13*3111) 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 (14^2111) mod 1001 541456 is congruent to (14^211113) mod 1001 so 69660 is congruent to 591 mod (1001) 331259 is congruent to 559*591111 mod (1001) 541456 is congruent to 559*59111113 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)86013) 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*86041*5*13 mod (1001) 331259 and 92020 are congruent to 3*86041*5*13+13 mod (1001) 69660 is congruent to 2*41*5 mod 1001 541456 is congruent to 34413*2^513 mod 1001 331259 and 92020 are congruent to 34413*2^5 mod 1001 541456 is congruent to (344+13*2^513*2^613) mod 1001 92020 and 331259 to (344+13*2^513*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 (34413*2^5) is congruent to 72 mod 1001 (34413*2^513) 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 (920202131) is congruent to 9 mod 4280 (920202131*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 344559*3 mod (2132) pg(56238) and pg(331259) are probable primes the difference 33125956238 is congruent to  7 mod (2132*43) 331259 is 7 mod 13, 56238 is 0 mod 13 331259+756238=43*2132*3 this is equivalent to 92020+239*1001+756238=43*2132*3 92020+239239=331259 239239 is congruent to (562383447) mod (2132*43) 92020 is congruent to 344 mod (2132*43) 331259 is congruent to (562387) mod (2132*43) 92020 is congruent to 344 mod 2132 331259 is congruent to 344*2+111 mod 2132 562387 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*91)/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*1041*43*11 This means that 331259 and 92020 are congruent to (2131*10451*43) mod (239*13) and to (21310451*43+13) mod (1001) 541456 is congruent to (21310451*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*(108^21) 69660 is congruent to 204 mod 984 204=64521^2 69660 is a multiple of 645 1323=42^221^2 1323*108 mod 984 is 204 as 69660 mod 984 69660 is congruent to 21^2645 mod 984 92020 is congruent to 13^2645 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*11171) mod 2236 I notice that 92020 and 331259 are both congruent to 5 mod (239*7*11) 331259 is congruent to (239*7*1117113) mod 2236 modulo 2236 infact 239*7*11171 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+1001s287 (case 92020 and 331259) and 541456 which is of the form 215+1001s28713 so k is of the form 72+1001s or 85+1001s 215 is congruent to (92911*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 (92913) 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 (7113=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 13smooth 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 215eulerphi(324)=107 92020 is a multiple of 107, 69660 is 3 mod 107...the thing becomes too difficult multiple of 215=6^31 are 69660 and 92020 69660=324*215 92020=428*215 324 and 428 are numbers n such that neulerphi(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 (10^2881) 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*1677215 215=0*(41s+r)+215 69660=323*215215 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^n1)/9 1323 and 39699 are congruent to 645 mod (41*3*2^3) (397*10^n1) is divisible by 9209 for a certain n... 397*10^5((397*101)/9126)*10015^3 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 (78713=774) 541456(222*2021+78713)=92020 331259118*2021(78713*2)=92020 from the equation above 541456(222*2021+78713)=92020 if we work mod 787 0+715+13=728 so 92020 is congruent to (3^61) mod 787 92020 is also congruent to 43*2^n mod (78713) 9202069660=22360 is also congruent to 43*2^n=688 mod (78713) 69660 is congruent to (3^6118^2) mod 787 22360 is congruent to 18^2 mod 787 92020=69660+22360 92020 is congruent to 3^61 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 3^61 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(787428=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*10phi(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*4281=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*(11/172) is congruent to 0 mod (3^2*5*19) 92020*(11/172) is congruent to 0 mod (3^2*5*19=855) 541456*(11/172) is congruent to 513=19*3^3 mod (3^2*5*19) 855 is congruent to 1 mod 107 69660*(11/172) is congruent to 9^2 mod 107 92020*(11/172) is congruent to 0^2 mod 107 541456*(11/172) is congruent to 3^2 mod 107 2^97^3 is congruent to 6^31 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),(6^31))=22360 69660+22360=92020! curious that 559 is congruent to (12^9) 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+2^918^2) mod 787 6^3+2^9=3^61 so 3^6 congruent to 27^3 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 541456107*10*2^9 is divisible by 456 in other words: 541456=(6^3+7^3+2^91)*2^984*76 if you reduce the right side mod 107 6^3=2 7^3=22 2^91=83 2+22+83=107 69660 is congruent to (6^3+1)*(6^31) 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 (6^61) congruent to 3 mod 107 69660 is congruent to (6^61) 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(6^61) divides 92020 92020=69660+22360 69660 is a multiple of 645 22360=215*107215*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 6^61 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*2151) ((215*1071)*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) 69660/3222/3786=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) 69661138889763=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)^29^2*10^2=0 whcih has solutions: x=71 x=109 71*109+1=6^2*(6^31) from this we obtain (9^2*10^219^2+1)=6^2*(6^31)=7740, which divides 69660 69660 has also the factorization 6^2*(44+1)*(441) pg(331259) is probable prime 331259 is congruent to (71*18^21)=215*1072 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*10715*19+6^2*(6^31)90^2=22360 where 215*107 15*19 and 6^2*(6^31) 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 6966071*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 (32436=288) mod 323 pg(359) is prime 35971=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=19^22 pg(359) is prime 215, 69660, 92020, 541456 are congruent to plus or minus (19^273s) mod 323 infact 215=19^273*2 and 288=19^273 pg(51456) is prime 2^4*(73*49361)=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 (1456691)=1386 1386*239+5=331259 and pg(331259) is prime (13861001)*239+5=92020 and pg(92020) is prime 215 288 and 1456 in the field Z 73 have 18 as inverse...curious that 1456*181=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 (18^21) 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*35)*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=23005645 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=(18^21)*21 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*71)/426=1 541456 is a multiple of 787 in the ring F426 the inverse is 367 367*787=288828 288828=6^2*(8*10031) (359*891)/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*1431)/426=288 lcm(426,288)=143^21 maybe there is a link to the fact that 541456429*10^3143^2+13=92020 in the ring Z428 (215*2151)/428=108 (211*711)/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 (72b13), as Peter Scholze pointed out I note that 331259 equiv (congruent) 999 mod (33711) 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*460113 is 331259 so 3371 and 331259 are primes of the form 72*k13 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*41)*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 (337113)/46=73 which is congruent to 1 mod 72 (33125913)/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+14602+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*721=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 100112 mod 1093 working mod (1093) 3371 is 1001 331259=3371+327888k working mod (1093) 331259=100112s 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^x18^2 Also 69660 should have the form 288*3^x18^2 6^6 is 648 mod (71*324) (69660+648)/(288*3^2324)=31 69660 is divisible by 540=288*3324 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*1293371=5544661=10*(2^101)*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 33125992020 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*1071=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=10^43*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 (920202=92018, which is multiple of 331) (331*1391)/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*139215*107)+6^6=69660 331259 is congruent to 2000 mod 9007=10^43*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 (6^56^2) 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 (3312599203)/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)*(441)*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=(44^21)*36=(193*10+6)*36 (6^61) is congruent to 1935 mod 2236 1935=(44^21) 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*1091)/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=21^213 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 96^213 69660 and 92020 can be written in the form 3+215x+23004y, for some x,y 92020=3*215^26^6+1 (6^61) 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 (6^61) mod 107 this leads to 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=(22622262) 331259 is congruent to 257 congruent to (22622262)+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 920204601*435*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 (6^21)*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^n8)/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 (3594) 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 17^2359 mod 23004 is 2271522645=45360 69660 is congruent to 9 congruent to 45360 mod 71 45360 is even congruent to 9 mod (213^2) 7740 divides 69660 (71^21)=5040 is congruent to 7740 mod (71*3) from here we have 1008=10^32^3 is congruent to 43*6^2 mod (71*36) it holds: ((71^21)+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 this should lead to 204*24 is congruent to 6^3 mod 5112 69660 has the curious factorization (387/28)*(71^21) 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) this leads to (2^8+1)*5 is congruent to (2^31) mod 426 331259*56 is a semiprime congruent to 1 mod 426 and congruent to 17^2 mod 4600 (4601 divides 92020) 331259*56=1151*1439 and the difference is 14391151=288=17^21 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 6^65 mod (213^2) 6^6 infact mod 639 is 9 this leads to 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 72^221^2*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 (60^21) 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*3^21)) 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=(2^21)*(2^31)*(2^51) so is the product of (2^p1) 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(6^44) 2*6^6 is congruent to 6^4 mod 23004 this leads to 2*12^2 is congruent to 4 mod 71 i think that something is involved related to padic 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 (17^21) 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 (17^21)=2*12^2 mod 323 69660 is congruent to (17^2+18^2+(6^21))=80=9 mod 71 92020 is congruent to (18^2+(6^21))=75=4 mod 71 69660 is congruent to 2*18^2 mod 71 92020 is congruent to (2*18^217^2) 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 (17^21) mod 71 69660 is congruent to 2*17^2+2*(6^21) mod 71 92020 is congruen to 17^2+2*(6^21) mod 71 359=17^2+2*(6^21) 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)*(6^3)^2 is about 331259 ((71/10)*6^2)*(19^21)+4=92020 331259/92020 is about 3.599...about (19^21)/10^2 about 359/100 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=6^6323k 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 331259/6^6 is about 7.10003000686 (1/18)*(331259*(2^2+1)+(2^6+1))=92020=(5/18)*(331259+13) the curious 71*72=5112=512+4600 (4601 divides 92020) (331259*72331259*71)/(71*72) is about 3*6^3/10 92020/5112 is very near to 18 (331259/71)=6^6/10+7/355 (331259/6^6)=71/10+7/(2^5*3^6) (92020/(71*72)) is about 3*6=18 (331259/(71*72)) is about 3*6^3/10 104*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: (6^31) congruent to 69660*6^2 congruent to 213 mod 107 213+215=428=107*4 which divides 92020 we return to the fact that 324*71 is congruent to 1 mod 107 324*71=23005 which divides 92020 (6^61)*215 congruent to 69660 mod (43*107*5*433) 6^6 is congruent to 325=18^2+1 mod (107*433) (6^61)*(6^31)(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 (13^21) 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 (3^31) mod 71 for example (331259+27)/71=4666 331259*(6^31) is congruent to 7*43 mod (71*43*18) this leads to 1535*(6^31) 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*2020 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*23920 which is congruent to 9189 mod 23005 331259=92020+239*1001 (331259+539*239+20)/92020=5 43*107*k1 i think this is the key , for k integre 331259 for example: (43*107*61)*6*21=331259 43*107*91=(3*6^31)*2^6 331259=(3*6^31)*2^95 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 33172=259) 71*6^6 is congruent to 215+71 mod (43*107*5) From here 71*(6^61) 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*(6^61) mod (43*5*107) 4601 divides 92020 4601*7213=331259 71*(6^61) is congruent to 215 mod 23005 (6^61) 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=(6^31)*3 mod 4601 69660 is congruent to 4600*(6^61) 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*324129)*3*2^313 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*6^42 which is congruent to (3312592) whcih is congruent to 257 mod 331 so 239*1001 is congruent to 3*239*1012 which is congruent to 3312592 whcih is congruent to 257 mod 331 331259=92020+239*1001 so 0 is congruent to 239*3*1012239*1001 which is congruent to 920202 whcih is congruent to 257239*1001 mod 331 so 0 is congruent to 239*(698)2 which is congruent to 920202 mod 331 or 0 is congruent to 239*6^22 which is congruent to 920202 mod 331 so 92020 is congruent to 2 mod 331 so 0 is congruent to 92*6^22 which is congruent to 920202 mod 331 so 0 is congruent to 33122 which is congruent to 920202 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*721=21455 mod 331 21455 is congruent to 271 mod 331 it holds: 331259+85 is congruent to 72 which is congruent to 71*6^672 mod 4601 so 4673 is congruent to 72 which is congruent to 445772 mod 4601 4673+4457+72=9202 331259 is congruent to 9203 mod (4601*51=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*6^66^4+4 (215*107*91)*2^42^7=71*6^6 (215*107*91) *2 is congruent to 7 mod (331*139) so 92020 is congruent to 7*((215*107*91)^(1)) which is congruent to 2 mod (139*331) 
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*(6^61)) 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) 
[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? 
...
[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 
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 (13374^21)/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*1840331 which is a multiple of 429^2 and 71 ((184033*230051)/71429^2)/429^2=18^21 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*(429^22^2)^(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*6^624^2)/10^3)*10^m, for some nonnegative integer m (46009x1)/(y^31)+y^3=(x^21)/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 (429^21)/10 mod 230051=31*41*181 92020 Is 4 mod 11502 and 4 mod 11503 92020 Is congruent to (71*6^6/11502=288=17^21) mod 323=18^21 6^6/162=288 162 divides 69660 (2*(46009*6)^2+144*(2+46009*2))/313/49/36^3=71*6^6 46009+2 Is a multiple of 313*49*3 2*46009+2=92020 69660=3/2*(6^66^3) 9202069660=22360 which is divisible by 860 22360=860*(3^31) 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*6^21)=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*1391) is congruent to 7 mod 9203 331259 is congruent to 7^2 mod 9203 or 9203=33125946008*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*313648)/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 MathCelebrity START HERE OUR STORY VIDEOS PODCAST Upgrade to Math Mastery Enter math problem or search term (algebra, 3+3, 90 mod 8) Invia 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 Share the knowledge! Chinese Remainder Theorem Video Tags: equationmodulustheorem Add This Calculator To Your Website <!— Math Engine Widget Copyright MathCelebrity, LLC at [URL="http://www.mathcelebrity.com"]www.mathcelebrity.com[/URL]. Use is granted only if this statement and all links to [URL="http://www.mathcelebrity.com"]www.mathcelebrity.com[/URL] are maintained. ><a href="https://www.mathcelebrity.com/chinese.php" onclick="window.open('https://www.mathcelebrity.com/chinese.php?do=pop','Chinese Remainder Theorem Calculator','width=400,height=600,toolbar=no,menubar=no,scrollbars=yes,resizable=yes');return false;">Chinese Remainder Theorem Calculator</a> Run Another Calculation Email: [EMAIL="donsevcik@gmail.com"]donsevcik@gmail.com[/EMAIL] Tel: 8002342933 MembershipMath AnxietyCPC PodcastHomework CoachMath GlossarySubjectsBaseball MathPrivacy PolicyCookie PolicyFriendsContact UsMath Teacher Jobs 331259/(2*12^2=288=17^21) is about 11502/10... 71*6^6 is congruent to  12^2 mod 4601 and mod 774 331259=11502*(17^21)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^x1)) mod 1001 for some nonnegative integer x (71*6^6+72331*999)/972=331259=(71*6^6+72331*9993*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((331259243*292020)/257)=72 331259 and 92020 are 72 mod 1001 28 is congruent to (429^21) mod 257 239239 is congruent to 28 mod 257 92020+239239=331259 14 is congruent to 129*(429^21) 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=(71^21)/(6^21) mod 359 So (6^31)=(71^21)/(6^21) 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 13d7(1143f) for some dnand f So (541456+7)/13 Is 71 mod 77 (92020+7)/13 and (331259+7)/13 are 72 mod 77 ((X^21)*(46009+1/4)1)/46009x^2=0 This Is a parabola for x=+ or  429 this goes to zero... 429^21=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) semiaxis length  1/184037≈5.43369×10^6 focal parameter  2/184037≈0.0000108674 eccentricity  1 directrix  y = 33870353521/736148 This Is the parabola ((X^21)*(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 (429^29)/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 92020/2592/51/3240=71/10 324 divides 69660....i think there Is something involving 18^2 2*92020/25921/324=71 (429^21)/25921/324=71 1/69660=(1/215)*((184040/259271)) 2592=72^2/2 215/10*(20000+72^2)=541456 71*6^6/331259+1/(239*99) Is about 10... 1/(1/(71*6^6/33125910)/99+239)=277.199999... The inverse of 5 mod 46009 Is 9202 429^25 Is a multiple of 46009 (429^25) 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 (33125944)/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 (9203131)/648 mod 331259 (331259131)/648=2^91 71*6^6/6484601=2^91 4601 divides 92020 Numbers of the form 512, 5112, 511...12,... The difference 5112512, 511125112,...Is a multiple of 46 (3312599203)/5112=2^61 331259/5112 is about 64,8...=648/10 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=(13801092020) mod 666 (20*71*6^64601*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 6966069005 is a multiple of 131 pg(331259) is prime 331259=(6^21) mod 13801 370=92020X((696609155)/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*70007370*511)/511^2=10 9203131=7*6^4 5112=71*72 (331259131)/146^4+4=22360=9202069660 92020+(6^44)=0 mod 6^6 9203=5=331259 mod 14 774*(1301131)/13=69660 71*6^6=14 mod 331259 71*6^6=15 mod 43 (71*6^6+15)=4472 mod (43*107) 4472 divides (9202069660) (71*6^6+154472)/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=12915 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 (429^22^2)=(429+2)x(4292) 71*6^6=90 mod (431X7X61) curious that 69660=90X774=71X6^6*774=(7^32)x7^3 mod 431 69660+(7^32)x7^3=431X433=432^21 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=1396^2 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=(92037) mod (46009x7) 331259=(9196+7) mod 23004x7 331259=(91967) mod (4601*7) 4601 divides 92020 and 4601x5=23005 (3312599196)=7*139*331 331x7=2317, whose last digits are 317 71*6^6331259=2981317, whose last three digits are 317 71*6^6331259(331*7)=331*3^2*10^3 3312599196 is a multiple of 331*139 pg(69660) and pg(2131*3^2=19179) are primes, pg(92020) is prime 69660=19179 mod 639 (6966019179)=4472 mod 46009 4472 divides (9202069660) curious that pg(2131), pg(19179=2131*9) and pg(69660=19179 mod 213) have this property: pg(2131) is the 19th 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*13^21)x260 mod 359 331259=98 mod (359x71x13) 331259=(2601) mod 331 by the way 331259=46009x81 mod 359 331259=(260x358^(1)1) mod 359 because 358x(10^21)=260 mod 359 71x6^6=83 mod 359 358x(10^21)=1 mod 83 so 331259=98 mod (71*13*359) 71*6^6=987x2^7=994 mod (71*13*359) 71*6^6=14*71 mod 359 331259=98=+7*6^6 mod 13x359 35999=260 331259=261 mod 359 99*(3311)=1 mod (359x13) 331*99=1 mod (2^15) 331259=(12^15) mod (359x13) 7*6^6+2^151=359*1001 so 71x6^6=12^15 mod (359x13) 2^15=1 mod 331 ((2^(15+330*(1+138*k))+1)) is a multiple of 139x331 (71*6^6+2^151359359)=12^6 12^6=(71*6^6331259) mod (359x13) 12^6=7*2^7 mod (359x13) (71*6^6331259)=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*6^66^6=2^7x3^6x(6^21) where 2^7x3^6 is a 3 smooth number 3^(n+1)x2^n. 92020=2^7x3^6(6^44) (6^44)=215=6^31 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) (6966014)/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(6966014)/359=3^6 (6966014)/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 25714=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 3312593371 is divisible by 138 and by 6^3 71*6^6=(33125959)/13800=24 mod 138 6^6+14 is a multiple of 359 6966014 is a multiple of 359 6^6+15 is a multiple of 331x3 696615 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 (6966014)=(14^21) mod 331 (6966014) 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(10011/31=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=35914 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 (70984261) 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 7098469660=1324 PG(1323) Is prime 9202069660=22360 (92020=4) mod 71 (92016=71*6^4). 9201669660=22356 331259=22356 mod (197*359) 22356=18^2*69 197*359 Is 70984261 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 (23923971*6^4)/231=80^2 by the way 92020=69660+239239 22360=9202069660=4 mod 23 4 is the square root of 16 by the way (23923916)/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=(3336359) mod 331 92020=21297 mod (359*197) 21297=21296+1 21297=11*44^2+1 44^21=1935 divides 69660 mod (359x197=359x(14^2+1)) I can see: 239239=27070 mod (359x197) 27070=541456/2014^2/70 331259=92020+239239 (6^31)*(6^21)+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+223563337+331259)/10011=6^21 (107+223563337+331259)=1 mod 359 22357=223603 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 (6966019179) 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: (758943336)=4=92020 mod 71 curiously ((758943336)+4)/(72^21)=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(29xs5) 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? 9202069660=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 (71x471)+139x29s 139x29x23338/2=2131x3 pg(2131) is prime and also pg(2131x9) indeed 2131=(46x1391)/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=9202069660=1=17^2=(3x6^3359) mod 29 75894=22360=17^2=(3x6^3359) mod (29x71) I think that that is the rub because 69660=3x6^3 mod 23004 22360=9202069660 359 mod 71=4 92020=4 mod 71 because 359=1700 mod (71x29) and because the inverse of 1700 mod (17x29) is (14^21)=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 (21,220,12,3) 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=53x1011 mod (19x1009) 92020+2=6^669660 mod (19x1009) 92020+69660=6^66 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=a1 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)/13802771*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 (920201) 71x6^3x10=8 mod (19171x8) 9202019171x4=71x6^3 9202019171x4=1 mod (313x49) 9202019171x4=1 mod (3067) 92020=10 mod 3067 153367(9202019171*4+1)=138030 consider 429^219171x the maximum value of x such that 429^219171x >0 is x=9 429^219171x9=11502 429^219171x8=37x829 37x829 divides (920201) 429^219171x6=69015 69015+645=69660 645 divides 69660 92020x2=429^21 19171x6=1=429^2 mod 23005 69660=645 mod 23005 69660=6^2*71*3^3+3*6^3 46011=19171*12429^2 i observe that 19171=3^92^9 541456+13449449=92020=46010x2=331259239239 (46010x219171x22222)=51456 Pg(51456) and pg(541456) are primes 460101111+1=449x10^2 19171x2=51456+2 mod 449 51456+2+19171x2+1 is a multiple of 1009 1111=71x3 mod 449 46010=71x31 mod 449 4601=111 mod 449 so 4601x20=92020=111x20=22222 mod 449 92020+2=2222 mod 449 (92020+2) is divisible by (71x6^3+1) 92020=5^2=2220 mod 449 331259=(4601x32) mod 23004 331259+2 is a multiple of 37 whereas 92020=331259239239=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. 9223392020=71x3 
92233=427x6^3+1
92233=6^3=51456 mod 427 ((9223351456)1)/312592=10^3 12592 divides 541456 PG(51456) and PG(541456) are primes 9223351456=1=69660 mod 1699 13592=1699*8 43*(((51456+(3^61)*4)/4)10^3)=541456 69660(48751456)=69660(9223351456)=0 mod (71x1699) (69660487)=0 mod 313 (69660486)=0 mod 427 486=162x3 162 divides 69660 44944913=64=541456+13 mod 139 so 541456=51 mod 139 (449x13=1 mod 139) 92020=541456+13449449 331259=71*6^6=259 mod 331 331259=44 mod 71 33124944=331215 that Is the concatenation of 331 and 215 33125944215 Is a multiple of 331 44+215=259 71x6^6331259+92020=6^26^5 mod 311 (6^26^5)=7740 which divides 69660 (7740=1 mod 71) multyplying by 9 both sides 71x6^6x9331259x9+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 7589492020/2=215x1391 546=215+331 215x139 can be cancelled in both sides this leaves 4600946010=1 56238 can be factorized as (1396^2)*(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 6966019179=71x79x3^2 239239=9 mod 12592 541456=239239+9 mod 12592 92020=3867+9 mod 12592 23004*460091(920206)=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) 331259132132=107x1861 75894+562381=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=(100172) mod 394 i suspect that this is connected to the fact that 331259=72 mod (1001x331) 331259=1323 mod (100172=929) 331259=394=1323 mod 929=100172 3x6^33=394x(72)^(1) mod 1001 In particolar 394x431+33x6^3=169169 92020=1323394=929 mod 1001 541456+13=132339413=916 mod 1001 (920201323+1)=0 mod 449 (541456+131323+1)=0 mod 449 541456449449=9202013 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*230051)/(313x49)=2^3x3^4=3x6^3 92020=2 mod (313*49) 313x49x3x6^31=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 (920204)=71x6^4 239239=77=752 mod 259 239239=78 mod 449 54145677x5837+13=92020 curious that 449449(239239+78)=210132=13*6^2x449 which is congruent to (6^42) mod 2131 239239+78 is 449x41x... 449x41 is 1840...anything to do with 429^21??? 541456=7740 mod 18404 7740 divides 69660 18404=(429^21)/10=449x415 pg(3336) is prime 3336=24x139=1 mod 71 (24+71x9)x139=1 mod 71 (24+71x9)x139+1138=92020 (24x71x9)x139+1)/71=6^4+2 so 92020=(6^4+2)x71138 or equivalently 92020=71x6^4+142138=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)*13970)/71)=19179 71x6^6=6^444 mod (331x4=1324) 331259=44+215 mod 1324 331259=44 mod 71 (71*6^66^4+44)/(139*181)=1324 may be 44 is not random... 69660=(44^21)x6^2 331259=44 mod (311x5x71) 331259^2=1936 mod (311x5x71) 1935 divides 69660 6^5=1 mod (311x5) 6^56^2 divides 69660 136=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+215^21 pg(51456) and pg(6231) are primes with 51456 and 6231 multiple of 67 curiously (51456/676231/67)=26^21 19179=648 mod 23004 331259=(19179648)/71 mod 359 331259x71=222 mod 359 so 222=19179648 mod 359 note that 331331(331259(19179648)/71)=333 ((92020*36)*3*41)/71=6^6+1 ...3312648...i think that 3x6^3 has something to do with these numbers... 331259/71=4665+44/71=6^6/106/10+44/71=6^6/10+7/355=6^10/10+7/(3594) 331259=44 mod 4665 106x44+44/71+1=331259/71 4665=359*132 331259 =71 mod 44 331259=44 mod 71 so 331259 is a number of the form 115+(5^51)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... (429^26)=1 mod 46009=331x139 (429^26)=3 mod 639 from this 429^2=80^21=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 71x6^6541456=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+1498)/13/359=359x2^8 331259=98 mod (359x13) (541456+69660)=611116=35998=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 (696603x6^3)/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 69660642(=3x6^36) is divisible by 11503 23005x(2+46009)1(69660642) is a multiple of 23003 2*(23005*(2+46009*72)1)/460092=331272x10 33127213=331259 23005x(2+46009x721) is divisible by 449 541456+13449x1001=331259 something mysterious is boiling in Z46009 46009x7272=71x6^6 x^2/(6^62sqrt(2x+1))x*(sqrt(2x+1)1)/215=0 has solution x=92020 min (x^2/(6^62sqrt(2x+1))x*(sqrt(2x+1)1)/215)=19394.4=19179215.4 this is a parabola from 331259x71=1917917^2 mod 359 we have 222=1917917^2 mod 359 so (2^91)=19179 mod (359x13) curious that 19179511+1 is divisible by (2^21), (2^31) and (2^71) 331259=98=(512414) mod 359 19179=414 mod 139 19179=138x1393 69660=19179=6^6 mod 639 the inverse mod 639 of 2131 is 427 69660x427=9 mod 639 69660x4279=639x46549 46549 is prime =6^6107 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=(2^91) mod (359x13) from here 6^6=7x7119179 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 (3312595) is divisible by 717x6=6x(3x6^3+69) 6x(3x6^3+69)=1 mod (331x13) so (3312595) is divisible by (331x131) (3312595)/(331*131)(920205)/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 supertools (3312595)/76^6=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 92019x692020x5 is a multiple of 3539 a wagstaff prime... 331259=2132 mod 3539 6^63x6^31=3539x13=46007 239239=13 mod (9202) (239239+13)/107=2236 69660+2236x10=92020 331259=331 mod 2236 239239x2236x2=(331259331) mod 331 239239x4472=72 mod 331 2236=6^2x239239^(1) mod 331 22360=(19^21)x239239^(1) mod 331 from here 22360=148 mod 331 239239=(148/2) mod 331 22360=25774 mod 331 239239=257 mod (331x19^2) (331259=257) mod (71x111) 331259x6^2=(22360257) mod 71 I have the impression that powers of 6 and 71 are involved in the logic behind these neme primes 92020+25714 (25714=243 a power of 3) is 359x257 331259243 is a multiple of 257 1001((331259243)/257359)=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=6^6541456 mod (359x13) (696606^6+331)*214=6^6 92020=10 mod 3067 331259=22 mod 139 331259=23 mod 3067 3312592392020+10=23923913 92022x366^3=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=3539x10101 mod (3539x13) curious that (5414563539x10) is divisible by 23003=71x2^2*3^41 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 78713=774 divides 69660 787429=3591 so for example 331259=429x774787 774 divides 69660 429=sqrt(92020x2+1) 541456=344 mod 774 there is a hidden structure it is clear that 331259774 is a multiple of 4601 and 4601 divides 92020 54145613=456+331 541456+13456=359x11x137 331259=358 mod (773x429) 358=3591 773=7741 331259+773 is divisible by 1297 a prime of the form 6^s+1 331259=(259215=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 (71x6^6216)/3/3067=359+1 92020=10 mod (3067x13) (9201^21)/4601/2313=787 
Qqqq[QUOTE=enzocreti;597012]92233=427x6^3+1
92233=6^3=51456 mod 427 ((9223351456)1)/312592=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=30671 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 9202071x6^3 is a multiple of (191798=19171) 331259=(3x3067)^2 mod 359 after some steps (inverse of 3067 mod 359 is 139x2) 331259x(10^21)=9 mod 359 so it is clear that 71 and powers of 6 are involved in these primes (71*6^6216(9202010))/(71*6^31)=210 anything to do with the fact that 541456+13210*1001=331259??) 71x6^6216 is a multiple of 3067 9202010 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 mathchamp 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 (6^63*6^33)=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 54145651456=700^2 which is divisible by 490 700^2=359x490+51456 after some steps: 490x(1000359)=51456 mod 541 51456=480 mod 541 (51456+480)=51936=5x10^4+44^2 (44^21)=1935 divides 69660 1935479=1456 479394=85 pg(394) is prime 541456=(44^21)+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 54145651456=490000 curious thta 429^21=540x21^2 mod 541 (429^21)=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 (429^21)=3x394 mod 541 dividing by 2 92020=3x197 mod 541 920203x197=91*10^4+429 92020=14=331259=(69660+1=69661 prime) mod 257 curious that 696606^6+1=23005 so 92020=(696606^6+1)x4=(3*6^6(7^31)^2+1)*4 from here we come to the curious: 69660=432^2342^2 (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 23^2x9202051456+1=365^3 365^3=1 mod 1297 (920204)+6^6=107x6^6 107 is the inverse mod 71 of 215 331259=1001 mod 449 541456+13449x1001=92020 331259=92020+239239 331259x449=541456+1392020 mod (449^2) maybe manipulating this you get something 449x7401001=331259=92020+239x1001=541456+13210*1001 curious that (5414566) 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)*4331259)/359=35987 331259=3^2*29 mod 359 92020=2^2*29 mod 359 331259+6^6=541456 mod (359x13) 33125914=84 mod(359x13) 331259=98 mod (359x13) (54145684)/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)/359/131=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=(88^24)x3^2 (88^22)=0 mod 79 (6966019179)=(88^22) mod (79x541) 69660+541456=611116=17x19x44x43 69660+541456=611116 this strange palindromic number 611115 is divisible by 4665 (=359x132) 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)*131261=611116 261 is the residue of 331259 mod 359 92020(541456316)/4665=0 mod 359 69660=315 mod 4665 541456=316 mod 4665 4665=(3/5)x(6^51) ((3/15)*(6^51)+1) divides (92020216) 54145651456=700^2 (3x13x3591)x(6^21)=700^2=(3x13x3591)x(394359) i think that a key passage is this: 331259=(191793x6^3)/71=261 mod 359 92020=(696603x6^3)/284=243 mod 359 222=191793x6^3 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=(2^91) mod (359x13) 19179=(2^9+1) mod 51 69660=2^9 mod 331x2 18^2x213+3x6^3=69660 18^2x2136^6=22356 (920204=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 (331259261)/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: 213114^2=1935 which divides 69660 from this 19179x484^223004=14 mod (359x13) after some steps... 19179x482^2=13 mod (359x13) 2131x6^2=82^213 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) 277^282^2=511x137(=70007)2 19179=511 mod (359x13) by the way 19179=1 mod 137 any possible connection to the fact that (54145651456)=700^2=7^2 mod 70007??? (700^2=35 mod (359x13x35)) it could be a chance...anyway (277^21) is divisible by 23,139 and 24...if you divide by 23 that is (277^21)/23=3336 and pg(3336) is prime 3336=1 mod 23 331259=3336x23=261 mod 359 541456344 is divisible by 559x44 44x559=2236x11 9202069660=22360 69660=(44^21)x6^2 19179=2131x3^2=(3^72^7)x3^2+3x6^3 19179=1=71x3^3 mod (137) 69660=6^6=14=(71x3^3+1)/(359222) mod 359 331259x71=222 mod 359 191793x6^3 is divisible by 261 (i think it is not chance that 331259261=0 mod 359) (19179648+1) is divisible by 41 and 452 ( pg(451) is prime) so (19179648)=452*411 pg(51) is prime so 452x411=222=331259x71 mod 359 452x411222=51x359 (331259*71452*41+1)/359/71=922=1 mod 71 (331259261)/359=922 pg(1323) is prime pg(39699) is prime 39699=9 mod 1323 (396999)/2666=19179 pg(19179=2131*9) is prime pg(2131) is prime so 19179=2131x3^2=666 mod 1323 19179=5x63^2666 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 (9202069660)=22360=(191793x6^3)/71 mod 451 19179=2x344=3x79 mod 451 6966019179 is a multiple of 79 71x6^469660=261=331259 mod 359 71x6^4+469660=261 mod 451 71x6^414=261=71x6^4+6^6=331259 mod 359 71*6^469660+331259=0 mod 359 71*6^469660+331259=1 mod (139x106) I think that in 139Z there is something (71*6^469660+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 2x51x2612x2131=9202069660=22360 541456=2^11 mod (3216x13^2) 3216 divides 51456 I arrived to this from 6^6(331259x7119179+2^91222)/541=1=51457 mod 3216 so 541456=2^11 mod ((51456/16)) (541456+2^11)=2132 mod (359x13) (541456+2^112132)/359/13=116 92020=116 mod 359 (541456+2^11)=92020=2 mod 331 (541456+(2^111)2131)/(2131x3^2+2^91)=29 541456+69660=611116 611116x71 mod (359x13)=137 19179511=0 mod (359x13) 19179511137=19179648=0 mod 71 and mod 261 611116=261 mod (359x13) 331259=261 mod 359 (19179648+611116*71) is divisible by 359x13 (19179648+611116*71)+1 is divisible by 394 and pg(394) is prime (611116*71+19179648)/(331259+359261) is an integer 541456*71=6^4+1 mod 4667 2131=72 mod 2059=29x71 19179=648 mod (29x71) 541456x716^41=359x13x8237 8237=1 mod (29x71) 611116331259=6^614=6966014 mod 359 (541456+69660)331259=6^614=6966014 mod 359 71x3^3+1=1918 divides (611116+3312596^61419179) (611116+331259)/359=2625=26261 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+3312596^614)=(541456+69660+92020+2392396^614) 19179+1 is a multiple of 137 19179=511=648137 mod (359x13) 191791=30003 mod (359x137) 3x6^3261=387 92020=387 mod (2131x43) (331259257)*9/(2300419179+648)=666 ((331259257)492016)/331=722=2x19^2 2300419179+648=4473 4472 divides (9202069660) 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+666239239/7+1)=7 2629x7+1=18404 429^21=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*22629*73)*9/4473=6^4 (2629x7+3)/2=9203=331259 mod 23004 (541456+13331259=210210=4472 mod (359x13)) 4472 divides (9202069660) 541456+13449449=92020 from above 541456331259=31x44918404 mod (359x13) so (92020331259)=970x449+1318404 mod 4667 23923913=970x44918404 mod 4667 970x449=12x18404 mod 4667 or 3168=12x18404 mod 4667 multiplying by 5 3168x5=12x92020 mod 4667 210210=4472 mod 4667 1051050=(9202069660) mod 4667 16170=344 mod 4667 344 divides 541456 541456+13210210=331259 mod 359x13 infact 541456=84 84+13+44724667=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=70001400x(9202069660)=331259 mod 4667 but this is: 98=1400x(5x7x11x239+69660)=331259 mod 4667 so it seems to pop up 7x11x239 whcih divides 239239=33125992020 bringing 5 out 98=7000x(18403+13932)=331259 mod 4667 98=4471x7000=331259 mod 4667 98=2333x(18403+13932)=331259 mod 4667 35998=261 54145613x16169=331259 54145613x1616913x18403=92020 16169,13932, 18403 are not random maybe this is the reason why 541456=84 mod (359x13) 84=9814 and 541456+69660=61116=98 mod (359x13) follows that (1840313932) which is 4471 =14^2 mod (359x13) this means that 6^121=4472 mod (359x13) 1616913932=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+13210210=331259 mod (359x13) +13210210=4472+13 359*13*43196*100113=4472 this maybe is the reason why 541456=84=9814 mod (359x13) and (69660+541456)=611116=261 mod (359x13) 261=35998 541456=6^7=84 mod (359x13) 14^2=196 69660=6^2x(213114^2)=6^2x(44^21) maybe not a achance??? 4x1917984^2=69660 541456^2=4x1917969660 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+21021013 92020=331259239239 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 6^57+18403s 92020 and 331259 are of the form 92020+18403s (920206^5+6)=3370x5^2 pg(3371) is prime probably i got something 92020+6^5=2^51 mod 3371 this implies 92020=88^2+1 mod 3371 (920205)=239x7x11x5=88^24 mod 3371 I remember that 69660=6^2x(44^21) so some conclusions can be done 92015x9=69660 mod 3371 92015=71x6^41=239x7x11x5 18403=1548 mod 3371 69660=(44^21)*36 Is congruenti to 18403 mod 3371 Dividing by 45 18403 congruenti to 43*6^2 mod 3371 18403*18=3312595 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=920202 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+13449449=92020 92018=14=331259=10x1001 mod 51 (9201810x1001) is divisible by 1608 1608 divides 51456 and pg(51456) is prime (331259+72)=331331=16=92020 mod 51 92092=9202072=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) 331331239 Is also divisibile by 541 and 36 i notice that (9202016) is divisible both by 51 and by 451 pg(51) and pg(451) are primes by the way famous 23004=696606^6 is congruent to 3 mod (451x51) curious fact (920204^2) 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 (5414562131) is divisible by (239+16=255=2^81) and by 2115=46^21 so 541456=(46^21)x(16^21)+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+2^91)=5^3x(71x3^3+1) 71x3^3+1 divides 19180=2131x3^2+1 23004=3x7667+3 7667 is palindromic in base 10 and 6 (9202016) is divisible by 451*12 (54145616^2) is divisible by 451*12 4472 divides (9202069660) 4472=16=92020 mod (51x11) 215=22360/2=541456 mod 51 22360=9202069660 69660=3 mod 107 92020=0 mod 107 69660323005 is divisible by (108^21) 23004=696606^6=3 mod (451x51) so (696603) (multiple of 107) 6^6=0 mod (451x51) 69660=3x6^3 mod 23004 696603x6^3=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 64523005=22360=(9202069660) also 6^61 is a number of this form I notice the incredible fact that (696603)=651x107 where 651 is the product of the first 3 Mersenne primes. 651=3x6^3+3 6^619179=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 (108^21) and with the fact that (696609) is a multiple of 71 and 109 23008=3 mod 107 23008=4 mod 108 230083=23005 230084=23004 69660428 (428 divides 92020) is a number congruent to 3 mod 107 and to 4 mod 108 (696604283)=107x647=107x(3x6^31) 107 and 23005 are number of the form 11449+11556k 69660=6^2 mod 264^2 264 is multiple of 44 69660=(44^21)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) (541331)=210 (541456+13210210=331259) 239239=210 mod (7x79) this is equivalent to 239239=7^3 mod (7x79) 541456=22^2 mod (7^3x79) 210x10011=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 6966019179 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+41^21=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 (9202016) is divisible by (71x1081)=7667 a palindrome 7667=11x41x17 pg(451) and pg(17x3) are primes 71*108*121=92015=239x5x11... (920206) is divisible by (71*108*61) (9202010) is divisible by (71x108x21) 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 (9202069660) 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 (6966014448)*613=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 9202016 is divisible by 11 92020 is even , 67 and 181015 are odd (9202016)=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=(429^21)/2 i think that this could explains something mod (429^21)/2 for example 92020=0 181015=55^2 mod (429^21) 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+13449449=92020 541456=5^213 mod 449 inverse of 25 mod 449 is 18 92020x18=1 mod 449 92020x18*(1/5)=331272 33127213=331259 92020*(1/5)=18404 18x1840413=331259 becasue 18404=1 mod (239x7x11) 18x18404135 is a multiple of (239x7x11) 429^2x18=16 mod (449x17) 92020=(429^21)/2 92020=16 mod 17 331259=18404*(5+13)13 Mod 449 18404*5=25 mod 449 18404*1313=239239 331259239239=92020 18404*13=65=69660 mod 449 33125965=541456=33125969660 mod 449 92020x18=1 mod 449 this is the starting point 331259*592020*18=65 69660=65 mod 449 92020x1813x5=0 mod 331259 5x(18404x1813)=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 2513s mod 449 for some nonngeative s 331259+13=90 mod 18409=41x449 9065=25 69660=541456331259 mod 449 18404x18=359 mod 449 18404x18=90 mod 449 18404x1813=331259 18404=5 mod 449 331259=(449359)13 mod 449 pg(359) is prime 69660=(449359)5^2 mod 449 331259=(35913) mod 449 331259+13=0 mod 18404 18404=5 mod (449) so 92020=5x18404=5x5 mod (449) 18404x18=90=359 mod 449 331259=18404x1813 so 331259=9013=103=(35913) 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 3312599013=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=9013=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=7^21(429^21)/2 429^2=7^2 mod 449 (429^21)/2=92020 mod 449 (429^21)/2=5^2 mod (449) 541456+13=5^2 mod 449 541456=5^213 mod 449 from thsi follows that 541456+1392020=0 mod 449 92020=5^2 mod 449 18404x359=1 mod 449 92020=18404x5 359x5=1 mod 449 54145613239239=331259=103 mod 449 54145613+78=331259=103 mod 449 so 541456+65=103 mod 449 10365=38=13+25 541456=2513 mod 449 69660=65 mod 449 so 541456331259=65=69660 mod 449 210210=239239=78 mod 449 541456+13210210=331259 331259239239=92020 92020=359+65 mod 449 so 22360=9202069660=359=331259+13 mod 449 the numbers are clearly structured, but unfortunally there is no elementary method to solve the puzzle of the giant megastructure 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((429^21)/2=92020) is prime 181015=429^2155^2!!! 429 and 55 have 11 as common divisor 11x(16731275)1=181015 181015=11^2x(39^25^2)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*416) 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 (9202069660359) 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 9202067=51 mod 451 pg(181015) is prime 11^2*(39^25^2)1=181015 11^2x39^2=429^2=2x92020+1 because (a^2b^2)=(a+b)x(ab) 11^2x(39+5)x(395)1=181015 39+5=44 divides (9202016) 395=34 divides (9202016) 181015 has the curious representation 44x41141 with 4114 a palindromic number 92020x18=1 mod 449 92020x18=17^21 mod (451x51x6^3) 92020 has the curious representation 828180/9 828180... 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 (6966019179) is a multiple of 79*3^2 pg(79) is prime 79*3^k =1 mod 71 for k=2+70xs (6966019179)/79+9=3x6^3=2^nx3^(n+1) (77402131)/71=79 19179 and 69660 arebof the form 31302+50481s 6^6=2 mod 41 69660=1 mod 41 6966016^62=23001 divisibile by 51 and 451 69660=1=51456 mod 41 and mod 4551=111x41 92020=10^3 mod 41 and mod 4551=111x41 (6966051456) divides (9202010^3) 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=7^46^6 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*471=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(3^51)=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(19179667x6)=667=260x1776 mod 2131 3335=667x5 54145651456=700^2 I suspect that there Is something to do with Numbers 54656=490 7*(1+7*111...) 56 546 5446 5444...6.... 546=7*(1+7*11) 490=54656 92020 and 541456 are of the form 8643s3053. 541456+13449449=92020 449449=13 mod 8643 8643/43=201 which divides 51456 3053 mod 449=359 54145610*51456=164^2 430*(71*(108+7667*3)1)/451/17=92020 Amazing that 71*6^4*5331259=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*1001 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 92020215*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*((429^29)/71+10^4)=541456 Is there a meaning?? (92020x28)/71 Is a 3 smooth Number 2592. 2592+10000=12592 which divides 541456 43*(429^29)/71 Is divisible also by 6966=69660/10 429 mod 139=12 PG(1323) Is prime (429^212^2)=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 (9202069660) 429^2=1 mod 107 I think that here Is the Key 429 is the 8th catalan number I suspect that partitions are involved in these exponents, in a way, however, that is not easy there is this identity: ((44^21)x6^26^6+1)x4=(429^21)/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+13210210*229029=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*((44^21)*366^6+1) Manipulating this identity You get a 2 degree equation in x 4031x+3338=4x144x^24x46691 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(6^31)x(6^32) 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 139x331215x107=696606^6=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... 24329=214 (451359)x100172=9209272=92020... 69660=2 mod 29 (696602)/29=7^4+1 something in the 29Z ring? 92020=(429^21)/2 429 is a Catalan numero 56238+75894 =132132 where 132 is catalan too 920203 is a multiple of 29,19,167 (920203) is divisible by (138x231) 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=918=27 mod 71 6966019179=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 6966019179 Is a multiple of 711 79x3^2=2131 mod 71 So 331259=19179x3 mod 71 331259+19179x3+1 Is a multiple of 359 6966019179 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 9202069660=22360=215x104 696606^6=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 .(429^23x6^3) is 0 mod 71 (429^23x6^3)/41=4473 4472 divides (9202069660) 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 696609=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,... (6^63x6^3+1)*2=920202...I think that...yes...powers of 6 are involved! 69660=645 mod 23005 From here (69660/5645/5)*24=331272=331259+13 (429^21)/2=3338+4031x11x2 if I factor out 11 607x/26677/2=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) (9202023000)/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 45×4463=71x3^3=1917 19179=2131x9=9 mod 1917 331259×45=2^61 mod 71xg Where g is a 3th smooth number (331259*4563)/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=920203338 mod 3335 19179x29/3=920203=71x6^4+1 mod 3335 3058=22x139 2131*27=1 mod 29 (19178)x36^4=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*63^2666 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 (69657203+1)/29+6=7^4 i think that in 29Z there is something... (696602)/291=7^4 429^2=1323x2=(79×671)/2 mod (139x29) From here it should follow something Because 92020=(429^21)/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=71x6^4x430218 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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