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 A118343 Triangle, read by rows, where diagonals are successive self-convolutions of A108447. 3

%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 self-convolutions of A108447.

%C A108447 equals the central terms of pendular triangle A118340 and the diagonals of this triangle form the semi-diagonals 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(n-1,k) - T(n-1,k-1) + T(n,k-1) + T(n+1,k-1). Also, a recurrence involving antidiagonals is: T(n,k) = T(n-1,k) + Sum_{j=1..k} [2*T(n-1+j,k-j) - T(n-2+j,k-j)] for n>k>=0.

%e Show: T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n,k-1) + T(n+1,k-1)

%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*(1-x+sqrt((1-x)*(1-5*x)+x*O(x^k)))/2/(1-x))/x)^(n-k),k)}

%Y Cf. A108447, A054727 (row sums), A118340.

%K nonn,tabl

%O 0,8

%A _Paul D. Hanna_, Apr 26 2006

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Last modified January 22 14:59 EST 2021. Contains 340362 sequences. (Running on oeis4.)