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

[Karol A. Penson]

  Re: A005572

I thank Max Alekseyev, Robert Israel and Paul Hanna for important remarks.
Robert's compact form can be further transformed using formula 
8.3.2.135, ch.8, p.666
  of  Yury A. Brychkov, "Handbook of Special Functions, Derivatives, 
Integrals, Series and Other Formulas",
(CRC Press, Taylor and Francis, New York, 2008),
and the following relation obtains using  the classical Gegenbauer 
polynomials, in Maple notation:

  A005572(n)=2*(12^(n/2))*(n!/(n+2)!)*GegenbauerC(n, 3/2, 2/sqrt(3)), 
n=0,1... .

Robert, would you like to enter your formula; I will then enter my 
Gegenbauer version.

Best,


Karol A. Penson






Le 03/02/2015 02:46, israel at math.ubc.ca a écrit :
> 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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