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G.f.: A(x) = Sum_{n>=0} (1 + x)^(n^2) / 2^(n+1).
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%I #30 May 03 2018 09:48:05

%S 1,3,36,744,21606,807912,36948912,1997801520,124666314300,

%T 8817945612300,697162848757056,60925366551278592,5831682410241684192,

%U 606763511537812563648,68184018356901256320192,8229830886505821175612416,1061871008421711265790015880,145851902823090076435152800208,21247730059665104564252809209792

%N G.f.: A(x) = Sum_{n>=0} (1 + x)^(n^2) / 2^(n+1).

%C Variant of A104209, which enumerates labeled directed multigraphs.

%C Number of labeled digraphs with n edges and no vertices of degree zero, in which loops are permitted but not duplicate edges. - _David Bevan_, Apr 22 2013

%H Paul D. Hanna, <a href="/A173217/b173217.txt">Table of n, a(n) for n = 0..100</a>

%H StackExchange, <a href="http://math.stackexchange.com/questions/366280">Combinatorial puzzle</a> (2013)

%F G.f.: 1/(2 - q/(1 - q*(q^2-1)/(2 - q^5/(1 - q^3*(q^4-1)/(2 - q^9/(1 - q^5*(q^6-1)/(2 - q^13/(1 - q^7*(q^8-1)/(2 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity. - _Paul D. Hanna_, Mar 18 2018

%F G.f.: Sum_{n>=0} 1/2^(n+1) * (1+x)^n * Product_{k=1..n} (2 - (1+x)^(4*k-3)) / (2 - (1+x)^(4*k-1)), due to a q-series identity. - _Paul D. Hanna_, Mar 18 2018

%F a(n) ~ 2^(2*n - 1/2 - log(2)/8) * n^n / (exp(n) * log(2)^(2*n + 1)). - _Vaclav Kotesovec_, Mar 21 2018

%t Table[Sum[Binomial[k^2, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 21 2018 *)

%t Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -2*j, 0]/2, {j, 0, n}] / n!, {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 21 2018 *)

%o (PARI) {a(n)=local(A=sum(m=0,n^2+100,(1+x +O(x^(n+2)))^(m^2)/2^(m+1)));round(polcoeff(A,n))}

%o (PARI) /* Continued fraction expression: */

%o {a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(2 - q^(4*n-4*k+1)/(1 - q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}

%o for(n=0, 30, print1(a(n), ", ")) \\ _Paul D. Hanna_, Mar 18 2018

%Y Cf. A104209, A173218, A301466, A301468, A265936.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Mar 05 2010