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The first 5 primes of this form are of lengths 3, 5, 7, 95, and 161 (131, 13331, 1333331, etc.). Can it be shown there are an infinite number of primes with form 133...331? Can it be shown there are only a finite number of them?

Goldbug
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    I doubt it highly. – Matt Samuel Dec 04 '19 at 15:47
  • I have only been able to determine a pattern for factors 11 and 19. – Goldbug Dec 04 '19 at 15:54
  • It was only recently shown that there are infinitely many primes without a 7, or any other single digit (within the last few years, in a paper by Maynard) and as far as I know there wouldn’t yet be a proof that you have infinitely many that use only 1 and 3. For these guys in particular maybe there is a proof of finitely many, but if not then the problem is open. – guy Dec 04 '19 at 15:57
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    The closed formula for this is $\frac{4\cdot 10^n-7}{3}.$ – Thomas Andrews Dec 04 '19 at 16:05
  • @guy This is already remarkable , but still far away from solving questions like this. – Peter Dec 04 '19 at 16:07
  • Surely, someone has checked those numbers upto a very large $\ n\ $ – Peter Dec 04 '19 at 16:07
  • Heuristically, infinite many such prime exist. A quite large has $\ 2899\ $ digits. Incredibly , factordb claims it to be proven prime by the $p+1$ - method. – Peter Dec 04 '19 at 16:14
  • First $n$ , for which Thomas Andrews's expression is (probably) prime : 1 2 4 6 94 160 360 1470 2898 3094 3112 – Peter Dec 04 '19 at 16:36
  • Just found another PRP ( $\ n=15699\ $ ) – Peter Dec 04 '19 at 16:46
  • Here is a related post. I wonder if these types of generalized Mersenne numbers have been looked at more closely? – Goldbug Dec 04 '19 at 16:54
  • @Peter Thanks for extending the search! Could you elaborate on what the p+1 method is? – Goldbug Dec 05 '19 at 02:46
  • @Peter See OEIS sequence A082697 – Goldbug Dec 05 '19 at 20:36
  • @Goldbug A primality proving method based on the factorization of $\ p+1\ $ , if $\ p\ $ has to be proven prime. I am astonished however that this was possible for the huge prime. – Peter Dec 06 '19 at 09:28
  • @Peter I was curious if perhaps there was some reason why this p+1 method was easy to use on a number of this form... It seems there might be if you were astonished? – Goldbug Dec 06 '19 at 14:02
  • @Goldbug Usually, we need the prime factorization of $p+1$ or at least a sufficient partial factorization. – Peter Dec 06 '19 at 14:12
  • @Peter Could the p+1 method be easier because $133...331+1=133...332=3411...111$? I know repunits are easier to factor than the average number. – Goldbug Dec 06 '19 at 14:25
  • Let us continue this discussion in chat. – Peter Dec 06 '19 at 14:25
  • 2,3,5,7 can't divide ( okay 7 maybe but not more than 1 in 6 terms). So that makes smaller values more likely to produce primes. –  Dec 08 '19 at 16:30
  • length can't be 4 mod 16 for divisibility by 17 it looks like. –  Dec 08 '19 at 16:36

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