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A084262
Binomial transform of double factorials.
10
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.
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
Sequence in context: A303344 A355205 A196555 * A231621 A084870 A372349
KEYWORD
easy,nonn
AUTHOR
Paul Barry, May 26 2003
STATUS
approved