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

0,2

Comments

Binomial transform of A001147.

Links

Table of n, a(n) for n=0..18.

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

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 *)

Crossrefs

Cf. A001147, A053871.

Sequence in context: A303344 A355205 A196555 * A231621 A084870 A111342

Adjacent sequences: A084259 A084260 A084261 * A084263 A084264 A084265

Keyword

easy,nonn

Author

Paul Barry, May 26 2003

Status

approved