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A106272
Antidiagonal sums of number triangle A106270.
1
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
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
Sequence in context: A272449 A270455 A318414 * A056423 A056347 A271332
KEYWORD
easy,sign
AUTHOR
Paul Barry, Apr 28 2005
STATUS
approved