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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. 2
1, 4, 20, 116, 736, 4952, 34716, 250868, 1855520, 13979192, 106901032, 827644424, 6474611984, 51100656544, 406400018092, 3253636464756, 26201323746880, 212093247874904, 1724793778005528, 14084738953391768, 115447965121881856 (list; graph; refs; listen; history; text; internal format)
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: A085456 A120915 A165311 * A082298 A129378 A078944

Adjacent sequences:  A100325 A100326 A100327 * A100329 A100330 A100331

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Nov 17 2004

STATUS

approved

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Last modified July 29 17:41 EDT 2021. Contains 346346 sequences. (Running on oeis4.)