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%I #4 Mar 30 2012 18:36:44
%S 1,1,1,4,5,1,24,32,9,1,172,236,76,13,1,1360,1896,656,136,17,1,11444,
%T 16116,5828,1348,212,21,1,100520,142544,53112,13184,2376,304,25,1,
%U 911068,1298524,494364,128924,25436,3804,412,29,1,8457504,12100952
%N Triangle, read by rows, where each column equals the convolution of A032349 with the prior column, starting with column 0 equal to A032349 shift right.
%C Row sums equal A027307; the self-convolution of the row sums form A032349. Column 0 equals A032349 shift right. Column 1 is A102231. This triangle is a variant of A100326.
%F G.f.: A(x, y) = (1+x*F(x))/(1-x*y*F(x)) where F(x) is the g.f. of A032349 and satisfies F(x) = (1+x*F(x))^2/(1-x*F(x))^2.
%e This triangle is generated by the recurrence:
%e T(n,k) = Sum_{i=0..n-k} T(i+1,0)*T(n-i-1,k-1) for n>k>0,
%e T(n,0) = Sum_{i=0..n-1} (2*i+1)*T(n-1,i) for n>0, with T(0,0)=1.
%e Rows begin:
%e [1],
%e [1,1],
%e [4,5,1],
%e [24,32,9,1],
%e [172,236,76,13,1],
%e [1360,1896,656,136,17,1],
%e [11444,16116,5828,1348,212,21,1],
%e [100520,142544,53112,13184,2376,304,25,1],...
%e Column 0 is formed from the partial sums of the prior row
%e after a term-by-term product with the odd numbers:
%e T(2,0) = 1*T(1,0) + 3*T(1,1) = 1*1 + 3*1 = 4.
%e T(3,0) = 1*T(2,0) + 3*T(2,1) + 5*T(2,2) = 1*4 + 3*5 + 5*1 = 24.
%o (PARI) {T(n,k)=if(n<k|k<0,0,if(n==0,1,if(k==0, sum(i=0,n-1,(2*i+1)*T(n-1,i)), sum(i=0,n-k,T(i+1,0)*T(n-i-1,k-1)));))}
%Y Cf. A032349, A027307, A102231, A100326.
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Jan 01 2005
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