A084262 Binomial transform of double factorials.
1, 2, 6, 28, 188, 1656, 17992, 232016, 3460368, 58574368, 1109200736, 23230928832, 533139875776, 13304094478208, 358653008619648, 10387075613199616, 321626829363798272, 10602925778746753536, 370770015836513986048
Offset: 0
Programs
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Mathematica
a[n_] := HypergeometricPFQ[{1/2, -n}, {}, -2]; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Aug 08 2012 *) CoefficientList[Series[E^x/(1-2*x)^(1/2), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Feb 04 2014 *)
Formula
a(n) = Sum_{k=0..n} C(n, k)*(2*k)!/(k!*2^k).
E.g.f.: exp(x)/(1-2*x)^(1/2).
a(n) = (1/sqrt(2*Pi))*Integral_{x=1..oo} x^n*exp((1-x)/2)/sqrt(x-1) dx. - Paul Barry, Jan 28 2008
G.f.: 1/(1-x-x/(1-2x/(1-x-3x/(1-4x/(1-x-5x/(1-6x/(1-x-7x/(1-... (continued fraction). - Paul Barry, Dec 02 2009
Let M be the infinite bidiagonal matrix with M(r,r)=1 in the main diagonal, M(r,r+1) = 2r-1, r >= 1, odd integers in the superdiagonal, and with the rest zeros. a(n) is the sum of first row terms of M^n. Example: a(4) = 188 = 1 + 4 + 18 + 60 + 105. - Gary W. Adamson, Jun 24 2011
a(n) -2*n*a(n-1) +2*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 09 2012
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1) - 2*x^2*(k+1)*(2*k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 12 2013
G.f.: 1/(1-x)/Q(0), where Q(k) = 1 - x/(1-x)*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
G.f.: conjecture: T(0)/(1-x), where T(k) = 1 - x*(k+1)/(x*(k+1) - (1-x)/T(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 13 2013
a(n) ~ 2^(n+1/2) * n^n / exp(n-1/2). - Vaclav Kotesovec, Feb 04 2014
G.f.: Sum_{k>=0} (2*k - 1)!!*x^k/(1 - x)^(k+1). - Ilya Gutkovskiy, Apr 12 2019
Comments