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Binomial transform of double factorials.
10

%I #37 Apr 29 2023 23:02:35

%S 1,2,6,28,188,1656,17992,232016,3460368,58574368,1109200736,

%T 23230928832,533139875776,13304094478208,358653008619648,

%U 10387075613199616,321626829363798272,10602925778746753536,370770015836513986048

%N Binomial transform of double factorials.

%C Binomial transform of A001147.

%F a(n) = Sum_{k=0..n} C(n, k)*(2*k)!/(k!*2^k).

%F E.g.f.: exp(x)/(1-2*x)^(1/2).

%F 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

%F 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

%F 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

%F a(n) -2*n*a(n-1) +2*(n-1)*a(n-2)=0. - _R. J. Mathar_, Nov 09 2012

%F 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

%F 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

%F 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

%F a(n) ~ 2^(n+1/2) * n^n / exp(n-1/2). - _Vaclav Kotesovec_, Feb 04 2014

%F G.f.: Sum_{k>=0} (2*k - 1)!!*x^k/(1 - x)^(k+1). - _Ilya Gutkovskiy_, Apr 12 2019

%t a[n_] := HypergeometricPFQ[{1/2, -n}, {}, -2]; Table[a[n], {n, 0, 18}] (* _Jean-François Alcover_, Aug 08 2012 *)

%t CoefficientList[Series[E^x/(1-2*x)^(1/2), {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, Feb 04 2014 *)

%Y Cf. A001147, A053871.

%K easy,nonn

%O 0,2

%A _Paul Barry_, May 26 2003