A106269 Expansion of 1/((1 - x^2)*(2 - c(x))), where c(x) is the g.f. of A000108.

1, 1, 4, 11, 39, 137, 501, 1853, 6936, 26163, 99314, 378879, 1451392, 5579179, 21509692, 83137939, 322049887, 1249941049, 4859617537, 18922572949, 73782881947, 288051510169, 1125832363807, 4404766873969, 17249634205357

Offset: 0

Author

Paul Barry, Apr 28 2005

Keywords

• easy
• nonn

Comments

Row sums of number triangle A106268.

From Petros Hadjicostas, Jul 19 2019: (Start)

Let A(x) be the g.f. of the current sequence. We note first that

Sum_{n >= 3} n*a(n)*x^n = x*A'(x) - (x + 8*x^2),

Sum_{n >= 3} 2*(1-2*n)*a(n-1)*x^n = 2*x*A(x) - 4*x*(x*A(x))' + (2*x + 6*x^2),

Sum_{n >= 3} (-n)*a(n-2)*x^n = -x*(x^2*A(x))' + 2*x^2, and

Sum_{n >= 3} 2*(2*n-1)*a(n-3)*x^n = 4*x*(x^3*A(x))' - 2*x^3*A(x).

Adding these equations (side by side), we get

Sum_{n >= 3} (n*a(n) + 2*(1-2*n)*a(n-1) - n*a(n-2) + 2*(2*n-1)*a(n-3))*x^n = 0,

which proves R. J. Mathar's formula.

(End)

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1664

Crossrefs

Cf. A000108, A106268.

Programs

Magma

A106269:= func< n | (-1)^n*(&+[Binomial(2*k-n, n-2*k): k in [0..Floor(n/2)]]) >;
[A106269(n): n in [0..40]]; // G. C. Greubel, Jan 10 2023

Mathematica

Array[(-1)^#*Sum[Binomial[2 k - #, # - 2 k], {k, 0, Floor[#/2]}] &, 25, 0] (* Michael De Vlieger, Jul 18 2019 *)

PARI

c(x) = (1-sqrt(1-4*x))/(2*x);
my(x='x+O('x^35)); Vec(1/((1 - x^2)*(2 - c(x)))) \\ Michel Marcus, Jul 16 2019

SageMath

def A106269(n): return (-1)^n*sum(binomial(2*k-n, n-2*k) for k in range(n//2+1))
[A106269(n) for n in range(41)] # G. C. Greubel, Jan 10 2023

Formula

a(n) = (-1)^n*Sum{k = 0..floor(n/2)} binomial(2*k - n, n - 2*k).

n*a(n) = 2*(2*n-1)*a(n-1) + n*a(n-2) - 2*(2*n-1)*a(n-3). - R. J. Mathar, Dec 10 2011

G.f.: 1/(sqrt(1-4*x)*(1-x^2)*c(x)), where c(x) is the g.f. of A000108. - G. C. Greubel, Jan 10 2023