%I
%S 1,1,0,1,1,0,1,2,4,0,1,3,9,20,0,1,4,15,48,113,0,1,5,22,85,282,688,0,1,
%T 6,30,132,519,1762,4404,0,1,7,39,190,837,3330,11488,29219,0,1,8,49,
%U 260,1250,5516,22135,77270,199140,0,1,9,60,343,1773,8461,37404,151089
%N Triangle, read by rows, where diagonals are successive selfconvolutions of A108447.
%C A108447 equals the central terms of pendular triangle A118340 and the diagonals of this triangle form the semidiagonals of the triangle A118340. Row sums equal A054727, the number of forests of rooted trees with n nodes on a circle without crossing edges.
%F Since g.f. G=G(x) of A108447 satisfies: G = 1  x*G + x*G^2 + x*G^3 then T(n,k) = T(n1,k)  T(n1,k1) + T(n,k1) + T(n+1,k1). Also, a recurrence involving antidiagonals is: T(n,k) = T(n1,k) + Sum_{j=1..k} [2*T(n1+j,kj)  T(n2+j,kj)] for n>k>=0.
%e Show: T(n,k) = T(n1,k)  T(n1,k1) + T(n,k1) + T(n+1,k1)
%e at n=8,k=4: T(8,4) = T(7,4)  T(7,3) + T(8,3) + T(9,3)
%e or 837 = 519  132 + 190 + 260.
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, 1, 0;
%e 1, 2, 4, 0;
%e 1, 3, 9, 20, 0;
%e 1, 4, 15, 48, 113, 0;
%e 1, 5, 22, 85, 282, 688, 0;
%e 1, 6, 30, 132, 519, 1762, 4404, 0;
%e 1, 7, 39, 190, 837, 3330, 11488, 29219, 0;
%e 1, 8, 49, 260, 1250, 5516, 22135, 77270, 199140, 0;
%e 1, 9, 60, 343, 1773, 8461, 37404, 151089, 532239, 1385904, 0; ...
%o (PARI) {T(n,k)=polcoeff((serreverse(x*(1x+sqrt((1x)*(15*x)+x*O(x^k)))/2/(1x))/x)^(nk),k)}
%Y Cf. A108447, A054727 (row sums), A118340.
%K nonn,tabl
%O 0,8
%A _Paul D. Hanna_, Apr 26 2006
