%I A113310
%S A113310 1,2,1,2,1,1,2,1,0,1,2,1,1,1,1,2,1,0,2,2,1,2,1,1,2,4,3,1,2,1,0,3,6,7,4,
%T A113310 1,2,1,1,3,9,13,11,5,1,2,1,0,4,12,22,24,16,6,1,2,1,1,4,16,34,46,40,22,
               7,
%U A113310 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
%V 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,-6,
               7,-4,1,2,1,1,
%W A113310 -3,9,-13,11,-5,1,2,1,0,4,-12,22,-24,16,-6,1,2,1,1,-4,16,-34,46,-40,22,
               -7,1,2,1,0,5,
%X A113310 -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, i)}; T(n, k)=sum{j=0..n-k, (-1)^(n-k-j)C(n-j-2, 
               n-j-k); 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;
%Y A113310 Sequence in context: A003639 A061916 A076348 this_sequence A081653 A096860 
               A128185
%Y A113310 Adjacent sequences: A113307 A113308 A113309 this_sequence A113311 A113312 
               A113313
%K A113310 easy,sign,tabl
%O A113310 0,2
%A A113310 Paul Barry (pbarry(AT)wit.ie), Oct 25 2005