|
%I #25 Nov 26 2022 13:43:44
%S 1,1,4,33,436,8185,206046,6622945,263313688,12627149265,716160702970,
%T 47284266221401,3587061106583604,309251317536586633,
%U 30017652739792964806,3254137305364883664945,391238883136463492841136,51846176797206158144925985
%N E.g.f.: Sum_{n>=0} x^n * exp(n^2*x).
%H Seiichi Manyama, <a href="/A193421/b193421.txt">Table of n, a(n) for n = 0..290</a>
%F E.g.f.: A(x) = Sum_{n>=0} x^n*exp(n*x)*Product_{k=1..n} (1 - x*exp((4*k-3)*x)) / (1 - x*exp((4*k-1)*x)), due to a q-series identity.
%F Let q = exp(x), then the e.g.f. equals the continued fraction:
%F A(x) = 1/(1- q*x/(1- q*(q^2-1)*x/(1- q^5*x/(1- q^3*(q^4-1)*x/(1- q^9*x/(1- q^5*(q^6-1)*x/(1- q^13*x/(1- q^7*(q^8-1)*x/(1- ...))))))))), due to a partial elliptic theta function identity.
%F a(n) = n! * Sum_{k=0..n} (n-k)^(2*k)/k!. - _Paul D. Hanna_, Jan 19 2013
%F O.g.f.: Sum_{k>=0} k! * x^k / (1 - k^2*x)^(k+1). - _Ilya Gutkovskiy_, Jul 02 2019
%F log(a(n)) ~ n*(2*(log(n) - 1) + LambertW(sqrt(n))*(3*log(n) - 2*log(1 + LambertW(sqrt(n))) + 2*LambertW(sqrt(n)))) / (2*(1 + LambertW(sqrt(n)))). - _Vaclav Kotesovec_, Nov 26 2022
%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 33*x^3/3! + 436*x^4/4! + 8185*x^5/5! + 206046*x^6/6! +...
%e where
%e A(x) = 1 + x*exp(x) + x^2*exp(4*x) + x^3*exp(9*x) + x^4*exp(16*x) +...
%e By a q-series identity:
%e A(x) = 1 + x*exp(x)*(1-x*exp(x))/(1-x*exp(3*x)) + x^2*exp(2*x)*(1-x*exp(x))*(1-x*exp(5*x))/((1-x*exp(3*x))*(1-x*exp(7*x))) + x^3*exp(3*x)*(1-x*exp(x))*(1-x*exp(5*x))*(1-x*exp(9*x))/((1-x*exp(3*x))*(1-x*exp(7*x))*(1-x*exp(11*x))) +...
%t Flatten[{1,Table[n! * Sum[(n-k)^(2*k)/k!,{k,0,n}],{n,1,20}]}] (* _Vaclav Kotesovec_, Oct 21 2014 *)
%o (PARI) {a(n)=local(Egf); Egf=sum(m=0, n, x^m*exp(m^2*x+x*O(x^n))); n!*polcoeff(Egf, n)}
%o (PARI) /* q-series identity: */
%o {a(n)=local(A=1+x);for(i=1, n, A=sum(m=0, n, x^m*exp(m*x+x*O(x^n))*prod(k=1, m, (1-x*exp((4*k-3)*x+x*O(x^n)))/(1-x*exp((4*k-1)*x+x*O(x^n)))))); n!*polcoeff(A, n)}
%o (PARI) {a(n) = n!*sum(k=0,n, (n-k)^(2*k)/k!)}
%o for(n=0,20,print1(a(n),", "))
%Y Cf. A006153, A193467.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 27 2011
| 