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A112293 Row sums of number triangle A112292.

%I A112293 #44 Jan 06 2024 09:22:25
%S A112293 1,2,7,36,253,2278,25059,325768,4886521,83070858,1578346303,
%T A112293 33145272364,762341264373,19058531609326,514580353451803,
%U A112293 14922830250102288,462607737753170929,15266055345854640658,534311937104912423031
%N A112293 Row sums of number triangle A112292.
%H A112293 Seiichi Manyama, <a href="/A112293/b112293.txt">Table of n, a(n) for n = 0..404</a>
%F A112293 a(n) = Sum_{k=0..n} (2n-1)!!/(2k-1)!!.
%F A112293 a(n) = Sum_{k=0..n} 2^(n-k)(n-1/2)!/(k-1/2)!.
%F A112293 a(n) = Sum_{k=0..n} n!C(2n, n)2^(k-n)/(k!C(2k, k)).
%F A112293 a(n) = Sum_{k=0..n} (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)).
%F A112293 Conjecture: a(n) - 2*n*a(n-1) + (2*n-3)*a(n-2) = 0. - _R. J. Mathar_, Nov 28 2014
%F A112293 a(n) = floor((2*n-1)!! * C) for n>0, where C = 1 + sqrt(e*Pi/2)*erf(1/sqrt(2)). - _Don Knuth_, Mar 25 2018
%F A112293 a(n) = 2^n*(C*Gamma(n + 1/2) + Gamma(n + 1/2, 1/2))*sqrt(e/2) for n >= 0, where C = sqrt(2/(e*Pi)) - erfc(1/sqrt(2)). - _Peter Luschny_, Mar 25 2018
%F A112293 a(n) = a(n-1) * (2*n-1) + 1 for n > 0 and a(0) = 1; that proves the conjecture of R. J. Mathar from Nov 28 2014; G.f. A(x) satisfies the equation: A(x) = 1/(1-x)^2 + A'(x) * 2*x^2/(1-x), where A' is the first derivative of A. - _Werner Schulte_, Oct 18 2023
%t A112293 T[n_, k_] := If[k <= n, (2 n - 1)!! / (2 k - 1)!!, 0];
%t A112293 a[n_] := Sum[T[n, k], {k, 0, n}];
%t A112293 Table[a[n], {n, 0, 18}] (* _Jean-François Alcover_, Jun 13 2019 *)
%Y A112293 Cf. A001147, A060196, A286286.
%K A112293 easy,nonn
%O A112293 0,2
%A A112293 _Paul Barry_, Sep 01 2005