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A177780 E.g.f. satisfies: L(x) = x*Sum_{n>=0} (2^n/n!)*Product_{k=0..n-1} L(3^k*x). 3

%I #10 Nov 03 2019 09:07:11

%S 1,4,60,2496,276240,83893248,72508524480,182341191057408,

%T 1348995112077074688,29528107099434111467520,

%U 1918583757808453356238126080,370812729559366641806998574727168

%N E.g.f. satisfies: L(x) = x*Sum_{n>=0} (2^n/n!)*Product_{k=0..n-1} L(3^k*x).

%C More generally, we have the following conjecture.

%C Define the series E(,) and L(,) by:

%C E(x,q) = Sum_{n>=0} q^(n(n-1)/2)*x^n/n!,

%C L(x,q) = x*d/dx log(E(x,q)) = x*E'(x,q)/E(x,q),

%C then L(x,q) satisfies:

%C L(x,q) = x*Sum_{n>=0} (q-1)^n/n! * Product_{k=0..n-1} L(q^k*x,q),

%C 1/E(x,q) = Sum_{n>=0} (-1)^n/n! * Product_{k=0..n-1} L(q^k*x,q).

%F a(n) = n*A054941(n), where A054941(n) is the number of connected oriented graphs on n nodes.

%F Define the series E(x) and L(x) by:

%F E(x) = Sum_{n>=0} 3^(n(n-1)/2)*x^n/n!,

%F L(x) = x*d/dx log(E(x)) = x*E'(x)/E(x),

%F then L(x) satisfies:

%F L(x) = x*Sum_{n>=0} 2^n/n! * Product_{k=0..n-1} L(3^k*x),

%F 1/E(x) = Sum_{n>=0} (-1)^n/n! * Product_{k=0..n-1} L(3^k*x).

%e E.g.f.: L(x) = x + 4*x^2/2! + 60*x^3/3! + 2496*x^4/4! + 276240*x^5/5! + ... + n*A054941(n)*x^n/n! + ...

%e Given the related expansions:

%e E(x) = 1 + x + 3*x^2/2! + 27*x^3/3! + 729*x^4/4! + 59049*x^5/5! + ...

%e log(E(x)) = x + 2*x^2/2! + 20*x^3/3! + 624*x^4/4! + 55248*x^5/5! + ... + A054941(n)*x^n/n! + ...

%e then L(x) satisfies:

%e L(x)/x = 1 + 2*L(x) + 2^2*L(x)L(3x)/2! + 2^3*L(x)L(3x)L(9x)/3! + 2^4*L(x)L(3x)L(9x)L(27x)/4! + ...

%e 1/E(x) = 1 - L(x) + L(x)L(3x)/2! - L(x)L(3x)L(9x)/3! + L(x)L(3x)L(9x)L(27x)/4! -+ ...

%t m = 13; A[_] = 0; Do[A[x_] = x Sum[2^n/n! Product[A[3^k x], {k, 0, n-1}], {n, 0, m}] + O[x]^m // Normal, {m}]; CoefficientList[A[x]/x, x] * Range[1, m-1]! (* _Jean-François Alcover_, Nov 03 2019 *)

%o (PARI) {a(n,q=3)=local(Lq=x+x^2);for(i=1,n,Lq=x*sum(m=0,n,(q-1)^m/m!*prod(k=0,m-1,subst(Lq,x,q^k*x+x*O(x^n)))));n!*polcoeff(Lq,n)}

%Y Cf. A054941, A177779, A177777, A177781.

%K nonn

%O 1,2

%A _Paul D. Hanna_, May 20 2010

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