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        riddles >> hard >> Recurrence relation divisibility
        
(Message started by: NickH on May 21^st, 2003, 1:51pm)

Title: Recurrence relation divisibility
Post by NickH on May 21^st, 2003, 1:51pm

Let a₁ = 1, and a_n = (n - 1)a_n-1 + 1, for n > 1.

For what values of n is a_n divisible by n?

Title: Re: Recurrence relation divisibility
Post by visitor on May 22^nd, 2003, 6:14am

Well, my calculator could only go up to 85, but I got the numbers [hide]1,2,4,5,10,13,20,26,37,52,65,74[/hide]. There seems to be some pattern there but I don't know if it can be formalized.

Title: Re: Recurrence relation divisibility
Post by Icarus on May 23^rd, 2003, 9:42pm

I haven't gotten very far with it, but it might help to express

a_n = SUM_k=0^n-1 (n-1)!/k!

Title: Re: Recurrence relation divisibility
Post by hyperdex on May 27^th, 2003, 1:15pm

Using Icarus's formula it is easy to show that if p|n then a_p and a_n are congruent mod p. This means that if n|a_n and p|n, then p|a_p.

There appear to be very few primes p such that p|a_p. I'm running a test for p<100000 and so far I've found 2, 5, 13, 37, 463, and 83507. Also note that 4|a_4. This means that any integer that is a product of any subset of the odd primes listed here along with 1, 2, or 4 will satisfy the divisibility condition.

I wouldn't be surprised if the number of such n is finite seeing as the density of good primes is so small. I would be surprised if there were an easy way to tell whether p is a prime such that p|a_p without actually calculating a_p mod p.

Cheers,

Dave

After thinking about this a bit more, I would be surprised if the number of primes such that p|a_p were finite. Handwavingly, we would expect that the probability that a given prime p satisfies p|a_p would be roughly 1/p. This means that the expected number of primes <n with p|a_p would be \sum_{p<n} 1/p which is approximately ln(ln n). Since this goes to infinity with n (albeit slowly), we would expect the number of primes to be infinite.

I'd still be surprised if there were an easy test to check whether p|a_p.

Update: I found all of the p's less than 800000 with p|a_p and in addition to the above primes found 306759, 412021, and 466123. The number of primes seems less dense than my handwaving argument would imply, but perhaps my argument was wrong.