Formula for A120014(n) -- Simplified

[Gottfried Helms]

Paul -
 just for couriosity. I did a gauss-elimination
 at the matrix of your values at the end of your
 posting (where a(n) is then the principal diagonal only).

 With A containing that coefficients

   gausselimination(A) -> Binom'

I got the transposed binomial-matrix binom, (binom
is assumed to be the lower triangular matrix containing
the binomial-coeficients / Pascal triangle) so that
actually with interpreting the Gauss-elimination
as left-multiplication with the matrix Elim

  Elim * A = Binom '

and Elim seems to be composed of

  Elim = inv(Fac) *St1^

where Fac is the diagonal matrix of factorials
and St1^ is a modified matrix of Stirlingnumbers
of first kind, possibly something like

  St1^ = St1 + X

or the like.

So

  A = inv(Fac * St1^) * Binom'

and
  A = inv(St1^) * ( Inv(Fac)*Binom' )

 where the most right parenthese provides
 a matrix, in which the binomialcoefficients
 and the inverse of the factorials cancel
 nicely to define a simple summation-scheme
 with inv(St1^) for the computation of the
 entries of A.

 But St1^ is not really transparent to me.

 To have a deeper look at this I would like to
 have more coefficients for A. Would you mind
 to provide more of these coefficients, day 8 x 8
 or 12 by 12?

Regards -

Gottfried Helms


Am 09.06.2006 07:27 schrieb Paul D. Hanna:
> Max, and Seqfans, 
>     Using the lovely formula for the n-th term of the m-th power of
> Catalan function: 
> [x^n] Catalan(x)^m = m*C(2*n+m-1,n)/(n+m), 
> I have found the simplest form yet for A120014: 
>