Always composite numbers?

[sweety439]

Can a dozenal (duodecimal, base 12) number {111...111}21{111...111} (with the same number of 1s in the two brackets) (start with 21, 1211, 112111, 11121111, ...) be prime? I cannot find such prime with <=1000 (decimal 1728) digits, but find neither covering set nor algebra factors. (I have proved that dozenal (duodecimal, base 12) numbers 1{555...555}1 (start with 151, 1551, 15551, 155551, ...) cannot be primes, because of covering sets and algebra factors)

Also, can a dozenal (duodecimal, base 12) 414141...4141411 (start with 411, 41411, 4141411, 414141411, ...) be prime? I want to find the dozenal (duodecimal, base 12) analog of A086766, and I proved that there are no primes for n = 10 (decimal 12) and n = 33 (decimal 39), because of algebra factors, thus the conjecture in A086766 is not true in dozenal (duodecimal, base 12).

Besides, I want to find the dozenal (duodecimal, base 12) analog of many other OEIS sequences, such as A088782, A069568, A200065, A272232, A089776 (n%12 = 1, 5, 7, 11 instead of n%10 = 1, 3, 7, 9), A267720, A244424, A262300, A046035, A047777, A060421, A064118.

For A069568 case, I found the prime (12^1676*298-1)/11 for n = 23 (decimal 27) and proved that there are no primes for n = 34, 89, and 99 (decimal 40, 105, and 117), because of covering sets and algebra factors, but for the A089776 case, I cannot find prime for n = 65 (decimal 77) and n = EE (decimal 143).

[LaurV]

Are you trolling, or what?

gp > doze(n)=((12^n-1)/11)*(12^(n+2)+1)+25*12^n
%1 = (n)->((12^n-1)/11)*(12^(n+2)+1)+25*12^n
gp > doze(0)
%2 = 25
gp > doze(1)
%3 = 2029
gp > 12^3+13+288
%4 = 2029
gp > doze(2)
%5 = 273181
gp > n=0;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
...0...
 %6 = 1

HUH???

(this fast stop was really unexpected, haha, and it totally took me by surprise.
I should have known that 2029 is prime!
But somewhere in the corner of my mind I was trusting you that there are no primes...

Big mistake...

Anyhow, continuing...

 gp > isprime(2029)
%7 = 1
gp > n=2;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
%8 = 2

HUH??? Again?


continuing...

 
gp > isprime(273181)
%9 = 1
gp > 12^5+12^4+2*12^3+12^2+12^1+1
%10 = 273181
gp > n=3;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
%11 = 3
gp > doze(3)
%12 = 39109981
gp > isprime(%)
%13 = 1

HUH??? Again?


continuing...

 
gp > n=4;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
...26...
time = 11 ms.
%14 = 27
gp > doze(27)
%15 = 247034206444509265352361785372634788125166592897428447392861
gp > isprime(%)
%16 = 1
gp >

GRRRRRR....

Edit, modify the oneliner to be easier to call (just pressing up arrow, without changing the start value)

gp > 27
%18 = 27
gp > n=%+1;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
...34...
%19 = 35
gp > n=%+1;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
...41...
%20 = 42
gp > n=%+1;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
...65...
%21 = 66
gp > n=%+1;while(!isprime(doze(n)),printf("...%d...%c",n,13);n++);n
...84...
%22 = 85
gp > doze(85)
%23 = 3782523501323864273640948080633865577531395800266304515211581987031320387343953524514660357926542631809629293255791000553
6420609584690895592018538198985334828642857876896842543151543389
gp > ?digits
digits(x,{b=10}): gives the vector formed by the digits of x in base b (x and b integers).
gp > digits(doze(85),12)
%24 = [
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Be my guest to count if the form is right... (to make it easyer for you, I aligned and eliminated the spaces using editor's facilities.

Eight primes or so in the first 100... you either were dreaming during testing, or you are trolling.

[axn]

Simpler form (note: off by 1 from LaurV):

doze(n)=(12^(2*n)-1)/11 + 12^n
for(i=1, 2000, n=doze(i); if(ispseudoprime(n), print(i)))
2
3
4
28
36
43
67
86
218
427
1284

[LaurV]

yeah, well.. the second form, which I rendered like below (probably a simpler form still exists for this one too) has no primes in sight, but there is no reason why a prime should not be there, the factoring looks like our crus lists, haha, and we would need a sieving program for it, because, with 19 dividing every third, 11 and 17 dividing every 8th, 10th, etc, the chances of finding a prime is very slim, but they should still come in an infinite amount, unless some aurifeuillian factorization unknown to us, or auriferous or platinodiamantine or whatever is called. By construction, these numbers are not divisible by 2, 3 (obviously, they are 1 (mod 12)), and 5, 7 (not so obvious, but see below) and that makes it harder to find a prime. But the prime is there, well hidden..

gp > ?doze41
doze41 =  (n)->49*12*(12^(2*n)-1)/143+1
gp > for(i=1,10,print(digits(doze41(i),12)))
[4, 1, 1]
[4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1]
[4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1]
gp > for(n=1,100,print(n": "factorint(doze41(n))~[1,]))
1: [19, 31]
2: [11, 23, 337]
3: [17, 722237]
4: [13, 19, 1979, 3617]
5: [571, 991, 449929]
6: [177823, 206171347]
7: [19, 277859424790111]
8: [37, 131, 13159, 105173, 113329]
9: [2837, 38587299124605977]
10: [19, 17189, 48268299300277811]
11: [17, 133530286335052494259517]
12: [367, 37560247, 23713556318104789]
13: [11, 19, 23, 35153, 1993859951, 139708400149]
14: [41, 21070110346361, 7846310875960621]
15: [1109, 880130606633990466566760020441]
16: [19, 31, 53, 21023, 78234938983, 2737502719302277]
17: [13, 37, 998861, 279663973627, 150631914918952787]
18: [101, 797, 10891, 1328911, 4006823, 624342125794134911]
19: [17, 19, 5527, 231898913, 2642684909, 383612326598340557]
20: [189767, 2958811, 96442165805129, 1116056700437206537]
21: [8273, 88379, 35125823, 338855864639826302126776175129]
22: [19, 823, 12541, 6390436308030046891653188795484059219533]
23: [3725522325449, 48438583726617962727111389675540810021]
24: [11, 23, 67, 12563393, 505124491, 9544230624930767, 25310427402129151]
25: [19, 83, 643, 1789, 139939, 161312028629, 91378718923313704966951677181]
26: [37, 463, 877, 2387291, 121589407, 123561358204325220849241943959288199]
27: [17, 4564357509721325038561737361681120186114742434887025935677]
28: [19, 1871, 100696837, 3667879961, 851006222623598233419959001107582844037]
29: [36955415772205820717, 1524981620206945422383, 28550311910196094306159]
30: [13, 14718871, 40827524265371788247316943, 29658224926068953911305277722409]
31: [19, 31, 71, 2339, 1469022047, 12220606596293613370899217920874878772972292555153]
32: [242276653, 415099039, 47772475059814551788252488123598816293657760084470783]
33: [50591, 55762433, 121017041968289563417, 2026478573464457245639635066584383583179]
  *** factorint: user interrupt after 15,460 ms
  ***   Break loop: <Return> to continue; 'break' to go back to GP prompt
break>

[LaurV]

back to it, here is your (pseudo)prime, which took about 29 minutes to find (I let a "while is not pseudoprime" running, but busy around, didn't check for a while):

gp > #digits(doze41(4383),10)
 = 9461
gp > ispseudoprime(doze41(4383))
time = 13,595 ms.
 = 1
gp >