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

[Paul D Hanna]

Also, 
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1). 
 
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A000108(k+1). 
 
This can be derived from the relation 
a(n) = [x^n] (1+4*x+x^2)^(n+1) / (n+1) 
 
which is due to 
G.f.: (1/x) * Series_Reversion( x/(1+4*x+x^2) ).  
 
However, the formula from Max seems to be more efficient. 
      Paul 
 
---------- Original Message ----------
From: Max Alekseyev <maxale at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Cc: Karol <penson at lptl.jussieu.fr>
Subject: [seqfan] Re: Question on A005572 from K. A. Penson
Date: Mon, 2 Feb 2015 18:52:29 -0500

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
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/

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