A106271 - OEIS

%I #33 Jul 22 2019 13:07:12

%S 1,0,-2,-7,-21,-63,-195,-624,-2054,-6916,-23712,-82498,-290510,

%T -1033410,-3707850,-13402695,-48760365,-178405155,-656043855,

%U -2423307045,-8987427465,-33453694485,-124936258125,-467995871775,-1757900019099,-6619846420551,-24987199492703

%N Row sums of number triangle A106270.

%C From _Petros Hadjicostas_, Jul 15 2019: (Start)

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

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

%C Sum_{n >= 2} (1-5*n)*a(n-1)*x^n = x*A(x) - 5*x*(x*A(x))' + 4*x, and

%C Sum_{n >= 2} 2*(2*n-1)*a(n-2)*x^n = 4*x*(x^2*A(x))' - 2*x^2*A(x).

%C Adding these equations (side by side), we get

%C Sum_{n >= 2} ((n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2))*x^n = 0,

%C which proves the conjecture. (End)

%H Michael De Vlieger, <a href="/A106271/b106271.txt">Table of n, a(n) for n = 0..1669</a>

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

%F a(n) = Sum_{k = 0..n} 2*0^(n-k) - C(n-k), where C(m) = A000108(m) (Catalan numbers).

%F 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

%F 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

%F a(n) ~ -2^(2*n + 2) / (3*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Jul 22 2019

%t Table[1 - Sum[(2n)!/n!/(n+1)!,{n,1,k}],{k,0,30}] (* _Alexander Adamchuk_, Feb 23 2007 *)

%Y Cf. A014138, A014137 (partial sums of Catalan numbers (A000108)).

%Y Cf. A106270.

%K easy,sign

%O 0,3

%A _Paul Barry_, Apr 28 2005

%E More terms from _Alexander Adamchuk_, Feb 23 2007