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

[israel at math.ubc.ca]

And, according to Maple, these sums can be written as a hypergeometric 
function:

A005572(n) = 4^n*hypergeom([-n/2, (1-n)/2], [2], 1/4)

Cheers,
Robert

On Feb 2 2015, Max Alekseyev wrote:

>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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>>
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