Sequence Of Primes, a(n)+a(n-1) is divisible by n

[Max Alekseyev]

On 10/14/07, Leroy Quet <qq-quet at mindspring.com> wrote:

> >%N A134204 a(0)=2. a(n) = the smallest prime not occurring earlier in the
> >sequence such that a(n-1)+a(n) is a multiple of n.
> >%C A134204 Is this sequence infinite, and, if so, is it a permutation of
> >the primes?

Infiniteness of this sequence is equivalent to non-existence of such n
that a(n-1) divides n. If this is the case then for any n, the
arithmetic progression n*k - a(n-1) contains infinitely many primes
(due to Dirichlet theorem) and a(n) is well-defined as the smallest
prime of the form n*k - a(n-1), not occurring earlier in the sequence
a().
On the other hand, if a(n-1) divides n then it also divides n*k -
a(n-1) for every k and there is no primes, except maybe a(n-1) itself,
in this arithmetic progression, meaning that a(n) is not defined.

I've confirmed that at least 80000 first terms of this sequence are
well-defined.

Max



Max Alekseyev has given a simple proof that the infiniteness/finiteness 
of the sequence depends upon whether a(n-1) ever divides n for any n. My 
reasoning why this is the case follows his reasoning exactly.

I must point out that cases where a(n-1) is <= n seem to be relatively 
rare, as is suggested Robert Israel's post.
So I would think that it is quite possible, sans a proof to the contrary, 
that the sequence could contain MANY terms and still terminate at some 
point.
I would suspect that a proof showing the sequence to be infinite would 
terms with index greater than some large integer.

Thanks,
Leroy Quet