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Let $Y(k)$ be the number consisting of $1$, repeated $k$ times. We know that $Y(2) =11$ is prime. It so happens that $Y(19)$ and $Y(23)$ are also prime.

Are there any more?

Regards,

David

J. W. Tanner
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David Sycamore
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1 Answers1

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To summarize the comments, the repunit $$Y(k)=\frac{10^k-1}9,$$ the number $11...111$ consisting of $k$ copies of the digit $1,$ is known to be prime for

$k=2, 19, 23, 317,$ and $1031.$

Source: The On-Line Encyclopedia of Integer Sequences.

J. W. Tanner
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  • Note that if $k$ is composite, then $Y(k)$ is. – J. W. Tanner Mar 06 '19 at 23:15
  • Thanks to everyone who replied; much appreciated, now I know what a repunit prime is! I came across these when looking at something else, namely: – David Sycamore Mar 07 '19 at 13:01
  • Let P(n,k) be the number Y(k)prime(n)Y(k) and sequence a(n) be least k>0 such that P(n,k) is prime, and 0 if no such k exists. I find the following: 0,1,1,3,0,3,1,0,1,1,2,0,3,2,1,0,14,3,0,3,2,1,4,3,0.... The problem is with the zeroes because it is hard to prove that no such k exists. Therefore, though I am certain of the non zero terms the zeroes are best guesses at the moment. I would like to submit this to oeis when and if more is known. Any ideas for proofs most welcome, though the case for prime(5)=11 looks obvious. – David Sycamore Mar 07 '19 at 13:27
  • It so happens that Y(19) is the greatest prime divisor of P(2,18), and Y(23) is the greatest prime divisor of P(2,22), which is how I came across them in the first place. – David Sycamore Mar 07 '19 at 13:38
  • Why the down vote ? – J. W. Tanner Mar 07 '19 at 21:08
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    ???What vote? I have not (to my knowledge) voted on anything. Was very happy with responses to my first question (thanks). Have tried to ask a follow up question but no replies as yet. – David Sycamore Mar 09 '19 at 10:03
  • Somebody downvoted; I don’t know who, but am wondering why – J. W. Tanner Jun 16 '19 at 12:30