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

[israel at math.ubc.ca]

Done.  I note that Peter Luschny has also entered a different
hypergeometric form.  I don't know how that one was derived.

Cheers,
Robert

On Feb 4 2015, Karol A. Penson wrote:

>  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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