# [seqfan] Re: Tabl formula?

Ron Hardin rhhardin at att.net
Fri Apr 1 18:56:32 CEST 2011

```I was thinking perhaps use T(n,k) to compute the first n+1 terms of a row, and
plug them into a lagrange polynomial to get the kth term of the row (or perhaps
just expand a zero n+1th difference table out to it), and perhaps some symbolic
package could rearrange the sums into something simpler.

Maybe it can only be a program, and not a formula, though.

rhhardin at mindspring.com
rhhardin at att.net (either)

> rh>  Empirical: T(n,k) = (n+1)*2^(k-1) + (1-k)*2^(k-2) for k<n+3, and then the
>entire
>
> rh> row n is a polynomial of degree n in k.
>
> No pattern so  far. An analytical formula would split the polynomial into
> two sums  associated with the 2 terms:
> 1+2^(k-1) = (B_k(3)-B_k)/k with Bell polynomials  and Bell numbers B.
> 1+2^(k-2) =  (B_(k-1)(3)-B_(k-1))/(k-1).
> 1-k+(1-k)2^(k-2) = -(B_(k-1)(3)-B_(k-1)).
> then  convert to polynomials with eq (13) in
> http://mathworld.wolfram.com/BellPolynomial.html .

```