[seqfan] Re: Tabl formula?
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
> 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 .
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