# [seqfan] Re: Guess the formula http://oeis.org/A076264

Max Alekseyev maxale at gmail.com
Tue Nov 22 20:27:32 CET 2011

Ron,
Did you mean http://oeis.org/A119851 ?
If so, an explicit formula is derivable from the given g.f. there.

In particular, T(n,k) is the coefficient of z^n in

z^(3k) / (1-3z+z^3)^(k-1)

that is

\sum_{j=0}^{n-3k}  binomial(-(k-1),j) * binomial(j,(n-3k-j)/2) *
(-3)^((3j+3k-n)/2)

where j has the same oddness as n-3k.

Here is a sample PARI/GP code implementing this formula:

{ T(n,k) = sum(j=0,n-3*k, if((n-3*k-j)%2,0, binomial(-(k-1),j) *
binomial(j,(n-3*k-j)/2) * (-3)^((3*j+3*k-n)/2) )) }

and a couple of columns:

? vector(20,n,T(n,2))
%1 = [0, 0, 0, 0, 0, 1, 3, 9, 26, 75, 216, 622, 1791, 5157, 14849,
42756, 123111, 354484, 1020696, 2938977]
? vector(20,n,T(n,3))
%2 = [0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 27, 106, 387, 1350, 4566, 15102,
49113, 157622, 500520, 1575558]

Regards,
Max

On Tue, Nov 22, 2011 at 10:59 PM, Ron Hardin <rhhardin at att.net> wrote:
> A formula for T(n.k) http://oeis.org/A076264 might be guessable, after the
> formula for
>
> column 2 given in http://oeis.org/A076264
>
> a(n)=sum{k=0..floor(n/3), binomial(n-2k, k)(-1)^k*3^(n-3k)}
>
> with the right generalization (I didn't stumble into it though).
>
>  rhhardin at mindspring.com
> rhhardin at att.net (either)
>
>
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