# [seqfan] Re: Peculiar Integer Recurrences ... Proof?

Max Alekseyev maxale at gmail.com
Mon Feb 9 19:54:46 CET 2009

```>From the recurrence relations, it follows that

a(n) = B_n( 0!*2^(1^2), 1!*2^(2^2), 2!*2^(2^3), ..., (n-1)!*2^(2^n) ) / n!

and

a(n) = B_n( 0!*(2^1+1)^1*0!, 1!*(2^2+1)^2, 2!*(2^3+1)^3, ...,
(n-1)!*(2^n+1)^n ) / n!

where B_n() are complete Bell polynomials:
http://en.wikipedia.org/wiki/Bell_polynomial

Regards,
Max

On Sun, Feb 8, 2009 at 6:24 AM, Paul D Hanna <pauldhanna at juno.com> wrote:
> Seqfans,
>        Prove that the following recurrences generate only integers.
>
> (1) a(n) = (1/n)*Sum_{k=1..n} 2^(k^2) * a(n-k) for n>0, with a(0)=1.
>
> (2) a(n) = (1/n)*Sum_{k=1..n} (2^k + 1)^k * a(n-k) for n>0, with a(0)=1.
>
> Emperical evidence: a(0) thru a(400) are all integers - quite convincing
> (a(400) has 48163 digits in both recurrences).
>
> A proof would be nice!
> Anyone up for the challenge?
>       Paul
>
> P.s.: recurrence (1) was derived by Vladeta Jovovic from the g.f. for A155200.
> I think (1) may lend itself to a proof better than the g.f. given there:
> G.f.: A(x) = exp( Sum_{n>=1} 2^(n^2) * x^n/n ).
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

```