Gottfried Helms helms at uni-kassel.de
Sat Aug 27 22:45:25 CEST 2011

Am 27.08.2011 14:30 schrieb Richard Mathar:
>
> I guess, up to signs, with offset 0 in both indices, the second column is
> T(n,1) = A000225(..)
> the third
> T(n,2) = 2^(2n+2)-6*2^n+3 = 3-6*2^n+4*4^n.
> the fourth
> T(n,3) = 2^(3n+3)-28*4^n+42*2^n-21 = -21+42*2^n-28*4^n+8*8^n
> the fifth
> T(n,4) = 420*4^n-630*2^n-120*8^n+2^(4*n+4)+315
> etc. So each column is a sum over powers of 2.
> Apparently T(n,k)=T(k,n) (mirror symmetry along the diagonal).
>
> Recurrences down each column appear to take the coefficients in A158474:
> T(n,2) =  7*T(n-1,2)-14*T(n-2,2)+  8*T(n-3,2)
> T(n,3) = 15*T(n-1,3)-70*T(n-2,3)+120*T(n-3,3)-64*T(n-4,3)
> T(n,4) = 31*T(n-1,4)-310*T(n-2,4)+1240*T(n-3,4)-1984*T(n-4,4)+1024 *T(n-5,4)
>

many thanks for the suggestions. The typical occurence of 2^n- 1 suggests
a best possible smooth expression in q-brackets/q-factorials/q-binomials
to base 2, I hoped there were an even more concise general expression
possible...

Also I'll see, whether the reference to A158474 gives some more food.

Sincerely -

Gottfried Helms