A100327 Row sums of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324.

1, 2, 8, 42, 252, 1636, 11188, 79386, 579020, 4314300, 32697920, 251284292, 1953579240, 15336931928, 121416356108, 968187827834, 7769449728780, 62696580696172, 508451657412496, 4141712433518956, 33872033298518728, 278014853384816184, 2289376313410678312

Offset: 0

Author

Paul D. Hanna, Nov 17 2004

Keywords

• nonn

Comments

Self-convolution yields A100328, which equals column 1 of triangle A100326 (omitting leading zero).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A003168, A003169, A100326, A219538.

Programs

Magma

A100327:= func< n | n eq 0 select 1 else (2/n)*(&+[Binomial(n, k)*Binomial(2*n+k, k-1): k in [1..n]]) >;
[A100327(n): n in [0..30]]; // G. C. Greubel, Jan 30 2023

Maple

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

Mathematica

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 *)

PARI

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

PARI

a(n)=polcoeff((1/x)*serreverse(x*(1-x+sqrt(1-4*x +x^2*O(x^n)))/(2+x)),n)
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Nov 22 2012

SageMath

def A100327(n): return 2^n*binomial(3*n,2*n)*simplify(hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2))
[A100327(n) for n in range(31)] # G. C. Greubel, Jan 30 2023

Formula

G.f.: (1/x)*Series_Reversion( x*(1-x + sqrt(1 - 4*x)) / (2+x) ). - Paul D. Hanna, Nov 22 2012

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.

a(n) = 2*A003168(n) for n>0 with a(0)=1.

a(n) = Sum_{k=1..n} 2*binomial(n, k)*binomial(2n+k, k-1)/n for n>0 with a(0)=1.

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

a(n) ~ sqrt(4046 + 1122*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(136*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012

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