A155719 Triangle t(n,m)=A039757(n,m)+A039757(n,n-m) read by rows.

2, 0, 0, 4, -8, 4, -14, 14, 14, -14, 106, -192, 172, -192, 106, -944, 1664, -720, -720, 1664, -944, 10396, -19560, 12644, -6960, 12644, -19560, 10396, -135134, 264158, -176358, 47334, 47334, -176358, 264158, -135134, 2027026, -4098304, 2925880

Offset

0,1

Comments

Row sums are zero if n>0.

Building the symmetric form A(n,m)+A(n,n-m) as here is equivalent to tabulation of the coefficients of a polynomial p_n(x) of order n associated with A(.,.) plus its reverse: t(n,m) = [x^m] ( p_n(x)+x^n*p_n(1/x)), here with p_n(x)=product(x-(2i-1)). Note that the product of the polynomial p_n(x) = sum_{m>=0} A(n,m)*x^m and the polynomial p'_n(x)= sum_{m>=0} A(n,n-m)*x^m is given in terms of (terminating) generating function by a convolution, related to the reversal of the sense of the second index in the A(n,m). So the fact that one can obtain A(n,n-m) by using the reverse polynomial p'_n(x) = x^n/p_n(x) is by no means special to this sequence here. The consequence that A(n,m)+A(n,n-m) defines a left-right symmetric row is then obvious.

Links

Table of n, a(n) for n=0..38.

Example

2;
0, 0;
4, -8, 4;
-14, 14, 14, -14;
106, -192, 172, -192, 106;
-944, 1664, -720, -720, 1664, -944;
10396, -19560, 12644, -6960, 12644, -19560, 10396;
-135134, 264158, -176358, 47334, 47334, -176358, 264158, -135134;
2027026, -4098304, 2925880, -1062656, 416108, -1062656, 2925880, -4098304, 2027026;
-34459424, 71697024, -53806368, 20516768, -3948000, -3948000, 20516768, -53806368, 71697024, -34459424;

Mathematica

Clear[p, x, n, b, a, b0];
p[x_, n_] := Product[x - (2*i + 1), {i, 0, Floor[n/2]}];
Table[Expand[ CoefficientList[ExpandAll[p[x, n]], x] + Reverse[CoefficientList[ExpandAll[p[x, n]], x]]], {n, 0, 20, 2}];
Flatten[%]

Crossrefs

Cf. A039683, A039757

Sequence in context: A098699 A021837 A236934 * A153772 A197007 A011449

Adjacent sequences: A155716 A155717 A155718 * A155720 A155721 A155722

Keyword

sign,tabl

Author

Roger L. Bagula, Jan 25 2009

Status

approved