A112293 Row sums of number triangle A112292.
1, 2, 7, 36, 253, 2278, 25059, 325768, 4886521, 83070858, 1578346303, 33145272364, 762341264373, 19058531609326, 514580353451803, 14922830250102288, 462607737753170929, 15266055345854640658, 534311937104912423031
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..404
Programs
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Mathematica
T[n_, k_] := If[k <= n, (2 n - 1)!! / (2 k - 1)!!, 0]; a[n_] := Sum[T[n, k], {k, 0, n}]; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Jun 13 2019 *)
Formula
a(n) = Sum_{k=0..n} (2n-1)!!/(2k-1)!!.
a(n) = Sum_{k=0..n} 2^(n-k)(n-1/2)!/(k-1/2)!.
a(n) = Sum_{k=0..n} n!C(2n, n)2^(k-n)/(k!C(2k, k)).
a(n) = Sum_{k=0..n} (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)).
Conjecture: a(n) - 2*n*a(n-1) + (2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 28 2014
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
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
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