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A128195 a(n) = (2*n + 1)*(a(n - 1) + 2^n) for n >= 1, a(0) = 1.

%I A128195 #8 Nov 16 2022 04:55:21
%S A128195 1,9,65,511,4743,52525,683657,10256775,174369527,3313030741,
%T A128195 69573667065,1600194389599,40004859842375,1080131215965309,
%U A128195 31323805263469097,971037963168557815,32044252784564570583,1121548847459764557925,41497307356011298342553,1618394986884440655806799
%N A128195 a(n) = (2*n + 1)*(a(n - 1) + 2^n) for n >= 1, a(0) = 1.
%H A128195 P. Luschny, <a href="http://www.luschny.de/math/seq/variations.html">Variants of Variations</a>.
%F A128195 a(n) = A126062(2, n), double variations.
%F A128195 a(n) = (2n+1)!/(n! 2^n) Sum(k=0..n, 4^k*k!/(2k)!) [Gottfried Helms]
%F A128195 a(n) = 2^n (2n+1) Sum(k=0..n, Gamma(n+1/2)/Gamma(k+1/2))
%F A128195 a(n) = 2^(n+1) Gamma(n+3/2) Sum(k=0..n, 1/Gamma(k+1/2))
%F A128195 a(n) = A128196(n)*A005408(n)
%F A128195 a(n) = A128196(n+1)-A000079(n+1)
%F A128195 Recursive form:
%F A128195 a(n) = 2^(n+1)*v(n+1/2) with v(x) = if x <= 1 then x else x(v(x-1)+1).
%F A128195 a(n) = (2n+1)*(a(n-1)+2^n), a(0) = 1 [Wolfgang Thumser]
%F A128195 Note: The following constants will be used in the next formulas.
%F A128195 K = (1-exp(1)*Gamma(1/2,1))/Gamma(1/2)
%F A128195 M = sqrt(2)(1+exp(1)(Gamma(1/2)-Gamma(1/2,1)))
%F A128195 Generalized form: For x>0
%F A128195 a(x) = 2^(x+1)(x+1/2)(exp(1) Gamma(x+1/2,1) + K Gamma(x+1/2))
%F A128195 Asymptotic formula:
%F A128195 a(n) ~ 2^(n+5/2)*Gamma(n+3/2)
%F A128195 a(n) ~ (exp(1)+K)*2^(n+1)*(n+1/2)!
%F A128195 a(n) ~ M(2n+1)(2exp(-1)(n-1/(24*n+19/10*1/n)))^n
%p A128195 a := n -> `if`(n=0,1,(2*n+1)*(a(n-1)+2^n));
%t A128195 a[0] = 1; a[n_] := a[n] = (2*n+1)*(a[n-1] + 2^n); Table[a[n], {n, 0, 14}] (* _Jean-François Alcover_, Jul 29 2013 *)
%Y A128195 Cf. A007526 (The number of variations), A128196 (A weighted sum of double factorials), A126062.
%K A128195 easy,nonn
%O A128195 0,2
%A A128195 _Peter Luschny_, Feb 26 2007