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A106271 Row sums of number triangle A106270.

%I A106271 #33 Jul 22 2019 13:07:12
%S A106271 1,0,-2,-7,-21,-63,-195,-624,-2054,-6916,-23712,-82498,-290510,
%T A106271 -1033410,-3707850,-13402695,-48760365,-178405155,-656043855,
%U A106271 -2423307045,-8987427465,-33453694485,-124936258125,-467995871775,-1757900019099,-6619846420551,-24987199492703
%N A106271 Row sums of number triangle A106270.
%C A106271 From _Petros Hadjicostas_, Jul 15 2019: (Start)
%C A106271 To prove _R. J. Mathar_'s conjecture, let A(x) be the g.f. of the current sequence. We note first that
%C A106271 Sum_{n >= 2} (n+1)*a(n)*x^n = (x*A(x))' - 1,
%C A106271 Sum_{n >= 2} (1-5*n)*a(n-1)*x^n = x*A(x) - 5*x*(x*A(x))' + 4*x, and
%C A106271 Sum_{n >= 2} 2*(2*n-1)*a(n-2)*x^n = 4*x*(x^2*A(x))' - 2*x^2*A(x).
%C A106271 Adding these equations (side by side), we get
%C A106271 Sum_{n >= 2} ((n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2))*x^n = 0,
%C A106271 which proves the conjecture. (End)
%H A106271 Michael De Vlieger, <a href="/A106271/b106271.txt">Table of n, a(n) for n = 0..1669</a>
%F A106271 G.f.: c(x)*sqrt(1 - 4x)/(1 - x), where c(x) is the g.f. of A000108.
%F A106271 a(n) = Sum_{k = 0..n} 2*0^(n-k) - C(n-k), where C(m) = A000108(m) (Catalan numbers).
%F A106271 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 A106271 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 A106271 a(n) ~ -2^(2*n + 2) / (3*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Jul 22 2019
%t A106271 Table[1 - Sum[(2n)!/n!/(n+1)!,{n,1,k}],{k,0,30}] (* _Alexander Adamchuk_, Feb 23 2007 *)
%Y A106271 Cf. A014138, A014137 (partial sums of Catalan numbers (A000108)).
%Y A106271 Cf. A106270.
%K A106271 easy,sign
%O A106271 0,3
%A A106271 _Paul Barry_, Apr 28 2005
%E A106271 More terms from _Alexander Adamchuk_, Feb 23 2007