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%I #18 Feb 24 2023 11:25:21
%S 1,1,5,19,145,981,10141,98575,1289569,16314121,258568021,4023553931,
%T 74961787825,1383475135069,29636315118957,632414472704071,
%U 15316605861040321,370875832116841105,10021723060544059429,271409166367070755843
%N E.g.f.: A(x) = 1/(1 - 2*x^2)^(1 + 1/(2*x)).
%C More generally, we have the identity:
%C Sum_{n>=0} (x^n/n!)*Product_{k=1..n} (1+k*y) = 1/(1 - x*y)^(1 + 1/y); here y=2*x.
%H Seiichi Manyama, <a href="/A193287/b193287.txt">Table of n, a(n) for n = 0..425</a>
%F E.g.f.: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (1 + 2*k*x).
%F a(n) ~ n! * 2^(n/2-1/2-1/sqrt(2))*n^(1/sqrt(2))/Gamma(1/sqrt(2)). - _Vaclav Kotesovec_, Jun 25 2013
%e E.g.f.: A(x) = 1 + x + 5*x^2/2! + 19*x^3/3! + 145*x^4/4! + 981*x^5/5! +...
%e where A(x) satisfies:
%e A(x)^(2*x/(1+2*x)) = 1 + 2*x^2 + 4*x^4 + 8*x^6 + 16*x^8 + 32*x^10 +...
%e Also,
%e A(x) = 1 + x*(1+2*x) + x^2*(1+2*x)*(1+4*x)/2! + x^3*(1+2*x)*(1+4*x)*(1+6*x)/3! + x^4*(1+2*x)*(1+4*x)*(1+6*x)*(1+8*x)/4! +...
%e The logarithm begins:
%e log(A(x)) = x + 2*x^2 + 2*x^3/2 + 4*x^4/2 + 4*x^5/3 + 8*x^6/3 + 8*x^7/4 +...
%e a(0) = 1; a(n) = (n-1)! * Sum_{k=1..n} k * 2^floor(k/2)/floor((k+1)/2) * a(n-k)/(n-k)!. - _Seiichi Manyama_, Apr 30 2022
%t CoefficientList[Series[1/(1-2*x^2)^(1+1/(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Jun 25 2013 *)
%o (PARI) {a(n)=n!*polcoeff(1/(1 - 2*x^2 +x^2*O(x^n))^((1+2*x)/(2*x)),n)}
%o (PARI) {a(n)=n!*polcoeff(sum(m=0,n,x^m/m!*prod(k=1,m,1+2*k*x+x*O(x^n))),n)}
%o (PARI) a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, i, j*2^(j\2)/((j+1)\2)*v[i-j+1]/(i-j)!)); v; \\ _Seiichi Manyama_, Apr 30 2022
%Y Cf. A193281, A193288, A193289, A193290.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 21 2011
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