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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.

%I A100328 #15 Sep 09 2024 09:35:15
%S A100328 1,4,20,116,736,4952,34716,250868,1855520,13979192,106901032,
%T A100328 827644424,6474611984,51100656544,406400018092,3253636464756,
%U A100328 26201323746880,212093247874904,1724793778005528,14084738953391768,115447965121881856
%N 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.
%C A100328 Self-convolution of A100327, which equals the row sums of triangle A100326.
%H A100328 Vaclav Kotesovec, <a href="/A100328/b100328.txt">Table of n, a(n) for n = 0..650</a>
%F A100328 G.f.: (1+G003169(x))*G003169(x)/x, where G003169(x) is the g.f. of A003169.
%F A100328 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
%F A100328 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
%F A100328 From _Peter Bala_, Sep 08 2024: (Start)
%F A100328 a(n) = (2/n) * Sum_{k = 0..n} binomial(n+1, n-k-1)*binomial(2*n, k)*2^(n-k) for n >= 1.
%F A100328 a(n) = 4*Jacobi_P(n-1, 2, n+1, 3)/n for n >= 1. Cf. A003168. (End)
%o A100328 (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)))))}
%Y A100328 Cf. A003168, A003169, A100324, A100326, A100327.
%K A100328 nonn,easy
%O A100328 0,2
%A A100328 _Paul D. Hanna_, Nov 17 2004