A106272 Antidiagonal sums of number triangle A106270.

1, -1, -1, -6, -15, -48, -147, -477, -1577, -5339, -18373, -64125, -226385, -807025, -2900825, -10501870, -38258495, -140146660, -515897195, -1907409850, -7080017615, -26373676870, -98562581255, -369433290520, -1388466728579, -5231379691972

Offset

0,4

Comments

To prove R. J. Mathar's conjecture, let A(x) be the g.f. of the current sequence. We note first that

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

which proves the conjecture. - Petros Hadjicostas, Jul 15 2019

Links

Table of n, a(n) for n=0..25.

Formula

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

a(n) = Sum_{k = 0..floor(n/2)} 2*0^(n-2k) - C(n-2k).

Conjecture: (n+1)*a(n) + 2*(1-2*n)*a(n-1) - (n+1)*a(n-2) + 2*(2*n-1)*a(n-3) = 0. - R. J. Mathar, Nov 09 2012

Prog

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

Crossrefs

Cf. A000108, A106270.

Sequence in context: A272449 A270455 A318414 * A056423 A056347 A271332

Adjacent sequences: A106269 A106270 A106271 * A106273 A106274 A106275

Keyword

easy,sign

Author

Paul Barry, Apr 28 2005

Status

approved