A100328 Column 1 of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324, with leading zero omitted.

1, 4, 20, 116, 736, 4952, 34716, 250868, 1855520, 13979192, 106901032, 827644424, 6474611984, 51100656544, 406400018092, 3253636464756, 26201323746880, 212093247874904, 1724793778005528, 14084738953391768, 115447965121881856

Offset

0,2

Comments

Self-convolution of A100327, which equals the row sums of triangle A100326.

Links

Vaclav Kotesovec, Table of n, a(n) for n = 0..650

Formula

G.f. A(x) = (1+G003169(x))*G003169(x)/x, where G003169(x) is the g.f. of A003169.

Recurrence: 4*(n+1)*(2*n+1)*(17*n^2 - 28*n + 12)*a(n) = (1207*n^4 - 1988*n^3 + 1013*n^2 - 124*n - 12)*a(n-1) - 2*(n-2)*(2*n-3)*(17*n^2 + 6*n + 1)*a(n-2). - Vaclav Kotesovec, Jul 05 2014

a(n) ~ sqrt(95+393/sqrt(17)) * ((71+17*sqrt(17))/16)^n / (4*sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Jul 05 2014

Prog

(PARI) {a(n)=if(n==0, 1, sum(j=0, n, if(j==0, 1, sum(k=0, j, 2*binomial(j, k)*binomial(2*j+k, k-1)/j))* if(n-j==0, 1, sum(k=0, n-j, 2*binomial(n-j, k)*binomial(2*n-2*j+k, k-1)/(n-j)))))}

Crossrefs

Cf. A003169, A100324, A100326, A100327.

Sequence in context: A165311 A367283 A363556 * A082298 A129378 A078944

Adjacent sequences: A100325 A100326 A100327 * A100329 A100330 A100331

Keyword

nonn

Author

Paul D. Hanna, Nov 17 2004

Status

approved