0,8

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.

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.

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

at n=8,k=4: T(8,4) = T(7,4) - T(7,3) + T(8,3) + T(9,3)

or 837 = 519 - 132 + 190 + 260.

Triangle begins:

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, 6, 30, 132, 519, 1762, 4404, 0;

1, 7, 39, 190, 837, 3330, 11488, 29219, 0;

1, 8, 49, 260, 1250, 5516, 22135, 77270, 199140, 0;

1, 9, 60, 343, 1773, 8461, 37404, 151089, 532239, 1385904, 0; ...

(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)}

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

Sequence in context: A074078 A130659 A083741 this_sequence A005657 A009332 A045872

Adjacent sequences: A118340 A118341 A118342 this_sequence A118344 A118345 A118346

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 26 2006