Sat Aug 27 16:31:36 CEST 2011

```I decode over rows (without the signs):  a_1(n)=1; a_2(n)=2^n-1; a_3(n)=(2^n-1)^2 - 2*(2^(n-1)-1);

a_4(n)=(2^n-1)^3 - 16*(2^(n-1)-1)^2 + 4*(2^(n-1)-1);
a_5(n)=(2^n-1)^4 - 88*(2^(n-1)-1)^3 + 132*(2^(n-1)-1)^2 - 94*(2^(n-1)-1),...

Thus it is natural to expect that a_k(n)=(2^n-1)^(k-1)-P_(k-2)((2^(n-1)-1), k>=2,
where P_(k-2)(x) is a polynomial of degree k-2 with integer and alternating coefficients.

Regards.

----- Original Message -----
From: Richard Mathar <mathar at strw.leidenuniv.nl>
Date: Saturday, August 27, 2011 15:30
To: seqfan at seqfan.eu

>
> http://list.seqfan.eu/pipermail/seqfan/2011-August/015257.html says
>
> gh> From seqfan-bounces at list.seqfan.eu Sat Aug 27 09:32:33 2011
> gh> Date: Sat, 27 Aug 2011 00:22:57 +0200
> gh> From: Gottfried Helms
> gh>
> gh>  I came across this number array (should be extended to
> infinity) by q-brackets/q-binomials
> gh>  to base 2 and cannot find a good description. Does
> someone know this array or has an
> gh>  idea how to decode?
> gh>
> gh>    1    -
> 1       1        -1            1            -1                 1                    -1
> gh>   -1
> 3      -
> 7        15          -31            63              -127                   255
> gh>    1    -
> 7      43
> -211
> 931         -
> 3907             16003                -64771
> gh>   -1    15    -
> 211      2619       -26251        234795          -1985131              16323819
> gh>    1   -31
> 931    -26251
> 654811     -
> 13255291         238658491           -4050110011
> gh>   -1    63   -
> 3907    234795    -
> 13255291     662827803      -26961325147          973958217435
> gh>    1  -127   16003  -
> 1985131    238658491  -
> 26961325147     2699483026843      -220115609012251
> gh>   -1   255  -64771
> 16323819  -4050110011  973958217435  -220115609012251
>
> 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)
>
> Richard J. Mathar
>
>
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