# [seqfan] Re: G.F. for A245925 Sought

Paul D Hanna pauldhanna at juno.com
Sun Aug 17 20:21:52 CEST 2014

```In other words, since

gamma(1/2-n) = (-4)^n * sqrt(Pi) * n! / (2*n)!
then (in PARI code)

a(n) = (-1)^n * sum(k=0,n, binomial(2*n-k, k)^2 * (2*n-2*k)! / (n-k)!^2 )
or, perhaps better still:

a(n) = (-1)^n * sum(k=0,n, binomial(2*n-k, k)^2 * binomial(2*n-2*k, n-k) )

Thanks, Robert!
Paul

---------- Original Message ----------
From: israel at math.ubc.ca
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: G.F. for A245925 Sought
Date: 16 Aug 2014 21:44:58 -0700

Maple can simplify a(n) a bit:

a(n) = sum((-1)^k*binomial(2*n-k,
k)^2*sqrt(Pi)/(2^(-2*n+2*k)*GAMMA(n-k+1)*GAMMA(1/2-n+k)), k = 0 .. n)

Cheers,
Robert

On Aug 16 2014, Paul D Hanna wrote:

> Seqfans, Here is an interesting sequence that needs a g.f. with a closed
> form:
>
>http://oeis.org/A245925
>1, -3, 25, -243, 2601, -29403, 344569, -4141875, 50737129, ...
>
>The generating function is given by the binomial series identity:
>
> A(x) = Sum_{n>=0} x^n*Sum_{k=0..n} (-1)^k * C(n,k)^2 * Sum_{j=0..k}
> C(k,j)^2 * x^j
>
>= Sum_{n>=0} x^n / (1+x)^(2*n+1) * [ Sum_{k=0..n} C(n,k)^2*(-x)^k ]^2
>
>and the term-by-term formula is:
>
> a(n) = Sum_{k=0..n} Sum_{j=0..2*n-2*k} (-1)^(j+k) * C(2*n-k,j+k)^2 *
> C(j+k,k)^2.
>
>
> What is surprising about the terms in A245925 is that they involve
> perfect squares:
>
>a(2*n) = A245926(n)^2,
>
>a(2*n+1) = (-3)*A245927(n)^2,
>
>
>where the g.f.s for A245926 and A245927 have a closed form:
>
>A245926: sqrt( (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)) ),
>
>A245927: sqrt( (1-x - sqrt(1-14*x+x^2)) / (6*x*(1-14*x+x^2)) ).
>
>
> Perhaps someone could derive a g.f. with a ~similar closed form for
> A245925.
>
>Thanks,
>    Paul
>
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>
>Seqfan Mailing list - http://list.seqfan.eu/
>
>

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