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The sequence $\frac{4\cdot10^{L-1}-7}{3}$ generates numbers of the form 11, 131, 1331, 13331, etc. When $L$ is even then the term is divisible by 11. When $L$ is $15+18k$ then the term is divisible by 19. It would appear that when the length is 27+28k they are divisible by 29 and when length is 21+30k they are divisible by 31.

Is there some general way to determine arithmetic progressions of $L$ for which terms of the sequence $\frac{4\cdot10^{L-1}-7}{3}$ is guaranteed to have a given factor beyond just guessing and trying to prove each individually?

Goldbug
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    When $L$ is $6+16k$, the term is divisible by $17$ – J. W. Tanner Dec 05 '19 at 15:19
  • We don't even know all prime repunit numbers – J. W. Tanner Dec 05 '19 at 15:54
  • @J.W.Tanner I guess asking for all values is not a simple question. I guess I am just looking for ways to determine a way to find arithmetic progressions which are composite other than looking at factorization of terms, guessing, and trying to prove individual cases. – Goldbug Dec 05 '19 at 16:14
  • @J.W.Tanner This sequence already has more known primes than repunits... This type of sequence is one way to generalize repunits/repdigits. I am wondering if anyone has looked at sequences of the form $\frac{a\cdot10^n-b}{c}$ in general. It is also an interesting problem to determine when this form is guaranteed to return an integer for all values of n. – Goldbug Dec 05 '19 at 16:31
  • I see you posted this problem – J. W. Tanner Dec 05 '19 at 18:08
  • the difficulty is predicting $10^{L-1}\equiv {7\over 4}\bmod p$ about the only help is that $L$ can never be more than twice $p$, before we have it as a factor if it can factor it at all. –  Dec 06 '19 at 14:02

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