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

0,3

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

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

Mathematica

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

Prog

(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
(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
(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

Crossrefs

Cf. A000108, A106268.

Sequence in context: A203161 A050987 A137191 * A126758 A149258 A326423

Adjacent sequences: A106266 A106267 A106268 * A106270 A106271 A106272

Keyword

easy,nonn

Author

Paul Barry, Apr 28 2005

Status

approved