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a(n) = Sum_{k=0..n} (2*k-1)!!*binomial(2*n-k, n).
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%I #11 Oct 18 2017 05:09:39

%S 1,3,12,57,330,2436,23226,277389,3966534,65517210,1220999208,

%T 25279328958,575024187192,14247595540542,381846383109030,

%U 11004598454925405,339324532631899110,11146022446431209490,388535338484934710040,14324570939127320452350,556887682690152668745660

%N a(n) = Sum_{k=0..n} (2*k-1)!!*binomial(2*n-k, n).

%H Robert Israel, <a href="/A293469/b293469.txt">Table of n, a(n) for n = 0..403</a>

%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>

%F a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - 6*x/(1 - ...)))))))), a continued fraction.

%F a(n) = Gamma(n+1/2)*hypergeom([1/2, 1, -n], [-2*n], 2)*4^n/(n!*sqrt(Pi)). - _Robert Israel_, Oct 09 2017

%F a(n) ~ 2^(n + 1/2) * n^n / exp(n - 1/2). - _Vaclav Kotesovec_, Oct 18 2017

%p seq(add(doublefactorial(2*k-1)*binomial(2*n-k,n),k=0..n),n=0..40); # _Robert Israel_, Oct 09 2017

%t Table[Sum[(2 k - 1)!! Binomial[2 n - k, n], {k, 0, n}], {n, 0, 20}]

%t Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 + ContinuedFractionK[-k x, 1, {k, 1, n}]), {x, 0, n}], {n, 0, 20}]

%t Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) Sum[(2 k - 1)!! x^k, {k, 0, n}], {x, 0, n}], {n, 0, 20}]

%Y Cf. A001147, A076795, A084262, A092392, A270447, A293468.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Oct 09 2017