[seqfan] Re: Question on A005572 from K. A. Penson

[Max Alekseyev]

Hi Karol,

There is a formula

A005572(n) = \sum_{k=0}^n A097610(n,k)*4^k,

which expands (with substitution k -> n-2k) into:

A005572(n) = \sum_{k=0}^{[n/2]} binomial(n,2*k) * binomial(2k,k) /
(k+1) * 4^(n-2k)

PARI/GP code:

{ A005572(n) = sum(k=0,n\2, binomial(n,2*k) * binomial(2*k,k) / (k+1)
* 4^(n-2*k) ) }

Regards,
Max



On Mon, Feb 2, 2015 at 6:24 PM, Karol <penson at lptl.jussieu.fr> wrote:
> Does anybody know how to obtain the close form of A005572(n) ?
>
> Thanking in advance,
>
>     Karol A. Penson
>
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