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%I #30 Jan 31 2023 08:29:48
%S 1,2,8,42,252,1636,11188,79386,579020,4314300,32697920,251284292,
%T 1953579240,15336931928,121416356108,968187827834,7769449728780,
%U 62696580696172,508451657412496,4141712433518956,33872033298518728,278014853384816184,2289376313410678312
%N Row sums of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324.
%C Self-convolution yields A100328, which equals column 1 of triangle A100326 (omitting leading zero).
%H Vincenzo Librandi, <a href="/A100327/b100327.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: (1/x)*Series_Reversion( x*(1-x + sqrt(1 - 4*x)) / (2+x) ). - _Paul D. Hanna_, Nov 22 2012
%F G.f. A(x) = (1+G(x))/(1-G(x)), also A(x)^2 = (1+G(x))*G(x)/x, where G(x) = x*(1+G(x))/(1-G(x))^2 is the g.f. of A003169.
%F a(n) = 2*A003168(n) for n>0 with a(0)=1.
%F a(n) = Sum_{k=1..n} 2*binomial(n, k)*binomial(2n+k, k-1)/n for n>0 with a(0)=1.
%F Recurrence: 20*n*(2*n+1)*a(n) = (371*n^2 - 395*n + 96)*a(n-1) - 6*(27*n^2 - 103*n + 96)*a(n-2) + 4*(n-3)*(2*n-5)*a(n-3). - _Vaclav Kotesovec_, Oct 17 2012
%F a(n) ~ sqrt(4046 + 1122*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(136*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 17 2012
%F a(n) = 2^n*binomial(3*n,2*n)*hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2). - _Peter Luschny_, Jun 10 2017
%p A100327 := n -> simplify(2^n*binomial(3*n,2*n)*hypergeom([-1-2*n,-n], [-3*n], 1/2)/ (n+1/2)): seq(A100327(n), n=0..22); # _Peter Luschny_, Jun 10 2017
%t Flatten[{1,Table[Sum[2*Binomial[n,k]*Binomial[2n+k,k-1]/n,{k,1,n}],{n,1,20}]}] (* _Vaclav Kotesovec_, Oct 17 2012 *)
%o (PARI) a(n)=if(n==0,1,sum(k=0,n,2*binomial(n,k)*binomial(2*n+k,k-1)/n))
%o (PARI) a(n)=polcoeff((1/x)*serreverse(x*(1-x+sqrt(1-4*x +x^2*O(x^n)))/(2+x)),n)
%o for(n=0,25,print1(a(n),", ")) \\ _Paul D. Hanna_, Nov 22 2012
%o (Magma)
%o A100327:= func< n | n eq 0 select 1 else (2/n)*(&+[Binomial(n, k)*Binomial(2*n+k, k-1): k in [1..n]]) >;
%o [A100327(n): n in [0..30]]; // _G. C. Greubel_, Jan 30 2023
%o (SageMath)
%o def A100327(n): return 2^n*binomial(3*n,2*n)*simplify(hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2))
%o [A100327(n) for n in range(31)] # _G. C. Greubel_, Jan 30 2023
%Y Cf. A003168, A003169, A100326, A219538.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Nov 17 2004
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