OFFSET
0,3
COMMENTS
From Petros Hadjicostas, Jul 15 2019: (Start)
To prove R. J. Mathar's conjecture, let A(x) be the g.f. of the current sequence. We note first that
Sum_{n >= 2} (n+1)*a(n)*x^n = (x*A(x))' - 1,
Sum_{n >= 2} (1-5*n)*a(n-1)*x^n = x*A(x) - 5*x*(x*A(x))' + 4*x, and
Sum_{n >= 2} 2*(2*n-1)*a(n-2)*x^n = 4*x*(x^2*A(x))' - 2*x^2*A(x).
Adding these equations (side by side), we get
Sum_{n >= 2} ((n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2))*x^n = 0,
which proves the conjecture. (End)
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..1669
FORMULA
G.f.: c(x)*sqrt(1 - 4x)/(1 - x), where c(x) is the g.f. of A000108.
a(n) = Sum_{k = 0..n} 2*0^(n-k) - C(n-k), where C(m) = A000108(m) (Catalan numbers).
a(n) = 2 - A014137(n) for n >= 0 and a(n) = 1 - A014138(n) for n >= 0. - Alexander Adamchuk, Feb 23 2007, corrected by Vaclav Kotesovec, Jul 22 2019
Conjecture: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ -2^(2*n + 2) / (3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 22 2019
MATHEMATICA
Table[1 - Sum[(2n)!/n!/(n+1)!, {n, 1, k}], {k, 0, 30}] (* Alexander Adamchuk, Feb 23 2007 *)
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, Apr 28 2005
EXTENSIONS
More terms from Alexander Adamchuk, Feb 23 2007
STATUS
approved