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A113310 Riordan array ((1+x)/(1-x),x/(1+x)).

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%I A113310 #12 Sep 07 2024 21:09:40
%S A113310 1,2,1,2,1,1,2,1,0,1,2,1,1,-1,1,2,1,0,2,-2,1,2,1,1,-2,4,-3,1,2,1,0,3,
%T A113310 -6,7,-4,1,2,1,1,-3,9,-13,11,-5,1,2,1,0,4,-12,22,-24,16,-6,1,2,1,1,-4,
%U A113310 16,-34,46,-40,22,-7,1,2,1,0,5,-20,50,-80,86,-62,29,-8,1,2,1,1,-5,25,-70,130,-166,148,-91,37,-9,1
%N A113310 Riordan array ((1+x)/(1-x),x/(1+x)).
%C A113310 Row sums are A113311. Diagonal sums are A113312. Inverse is A113313. The family of Riordan arrays ((1+x)/(1-(q-1)x),x/(1+x)) allow one to calculate the weight distribution of MDS codes.
%D A113310 F.J. MacWilliams, N. J. A. Sloane, The Theory of Error-Correcting Codes, North-Holland, 2003, p. 321.
%F A113310 T(n, k) = Sum_{j=0..n-k} (-1)^j*C(j+k-2, j).
%F A113310 T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(n-j-2, n-j-k).
%F A113310 T(n, k) = Sum_{j=k..n} (-1)^(n-j)*C(n, j)*(2^(j-k+1)-1).
%e A113310 Triangle begins
%e A113310   1;
%e A113310   2,1;
%e A113310   2,1,1;
%e A113310   2,1,0,1;
%e A113310   2,1,1,-1,1;
%e A113310   2,1,0,2,-2,1;
%t A113310 T[n_, k_] := Sum[(-1)^(n-j) Binomial[n, j] (2^(j-k+1) - 1), {j, k, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 27 2017 *)
%K A113310 easy,sign,tabl
%O A113310 0,2
%A A113310 _Paul Barry_, Oct 25 2005