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
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..650
Programs
-
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)))))}
Formula
G.f.: (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
From Peter Bala, Sep 08 2024: (Start)
a(n) = (2/n) * Sum_{k = 0..n} binomial(n+1, n-k-1)*binomial(2*n, k)*2^(n-k) for n >= 1.
a(n) = 4*Jacobi_P(n-1, 2, n+1, 3)/n for n >= 1. Cf. A003168. (End)
Comments