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For $n \ge 1$ an integer, let's denote

$u_n = \sum_{k = 0}^{n-1} 10^k$

That is $u_1 = 1$, $u_2 = 11$, $u_3 = 111$, $u_4 = 1111$, ...

My question is the following : Which of them are prime numbers ?

What I know so far :

  • If $u_n$ is prime, then $n$ is prime (meaning there's an obvious factorization when $n$ is not prime).

  • When $p$ is prime, $u_p$ can either be prime (2, 19 and 23 being the only examples I found so far) or not prime (all primes up to 67 with the exception of 2, 19 and 23). But I haven't been able to see any pattern.

Any thought is welcome. Maybe a sub-question would be to know whether there's a finite or infinite number of such primes. Thanks for your help.

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You mean k starts with 0? see this form: $(10^(n+1)-1)/9$ which is equivallent to yours, if it helps... –  tomerg May 8 '11 at 13:35
7  
Note: these are called repunit primes en.wikipedia.org/wiki/Repunit –  Douglas S. Stones May 8 '11 at 13:36
    
@tomerg : Yes that's what I meant. Thanks for pointing it out. –  Joel Cohen May 8 '11 at 13:47
    
@ Douglas S. Stones : Thanks for the reference. Also I see I had forgotten $u_{19}$ somehow. So it seems it's an open problem. This question was asked to me on an exam, so I thought there might be an easy answer I had overlooked. –  Joel Cohen May 8 '11 at 13:52

1 Answer 1

up vote 10 down vote accepted

OEIS has a list of the number of 1's where these are prime. It only has eight of them, the next is 317. So you could have looked for a while.

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Thank you. Factoring $u_{71}$ already took a considerable amount of time. I surrendered at that point. –  Joel Cohen May 8 '11 at 18:40
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@Joel Cohen: Primality tests are usually much faster than factoring. A primality test on that number revealed that it is composite in less than a millisecond. –  Charles May 12 '11 at 15:08
    
@ Charles : You're right. At first, I was also looking for a pattern in the factorization (any obvious factor). I should have had switched to just testing primality after a while. –  Joel Cohen May 12 '11 at 15:36

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