A189176 Row sums of the Riordan matrix (1+x/sqrt(1-4*x),(1-sqrt(1-4*x))/2) (A189175).

1, 2, 5, 15, 49, 168, 594, 2145, 7865, 29172, 109174, 411502, 1560090, 5943200, 22732740, 87253605, 335897865, 1296447900, 5015206350, 19439895090, 75487384830, 293595204240, 1143532045500, 4459774977450, 17413705988874, 68067249620328, 266326619546204

Offset: 0

Author

Emanuele Munarini, Apr 18 2011

Keywords

• nonn
• easy

Examples

			a(3) = 15 since the top row of M^3 = (6, 5, 3, 1, 0, 0, 0, ...)
		

Crossrefs

Cf. A090181, A189175.

Programs

Mathematica

T[n_,k_] := If[n==k,1,Binomial[2n-k,n-k](n^2+n k-k^2-k)/((2n-k)(2n-k-1))]; Table[Sum[T[n,k], {k,0,n}], {n,0,22}]

Maxima

T(n,k):=if n=k then 1 else binomial(2*n-k,n-k)*(n^2+n*k-k^2-k)/((2*n-k)*(2*n-k-1));
makelist(sum(T(n,k),k,0,n),n,0,22);

Formula

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

G.f.: (1-5*x+4*x^2-(1-5*x)*sqrt(1-4x))/(2*x*(1-4*x))

a(n) = Sum_{k=1..n} (3-k)*binomial(2*n-k-1,n-1), n>0, a(0)=1. - Vladimir Kruchinin, Oct 18 2011

From Gary W. Adamson, Nov 14 2011: (Start)

a(n) is the sum of top row terms in M^n, M = an infinite square production matrix as follows, with the Fibonacci sequence as the left border:

1, 1, 0, 0, 0, 0, ...

1, 1, 1, 0, 0, 0, ...

2, 1, 1, 1, 0, 0, ...

3, 1, 1, 1, 1, 0, ...

5, 1, 1, 1, 1, 1, ...

which means the top row of M^n is the n-th row in A189175. (End)

Conjecture: (n+1)*a(n) + 3*(1-3*n)*a(n-1) + 10*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 15 2011

a(n) = Sum_{k=0..n} (k+1) * A090181(n,k). - Alois P. Heinz, Apr 04 2024