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
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Apr 28 2005
STATUS
approved