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A304313 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^3 * x ) / F(x) = 0 for n>0.

%I A304313 #18 Aug 13 2025 05:53:18
%S A304313 1,49,6877,1854545,807478656,514798204147,451182323794896,
%T A304313 519961864703259753,762210147961330421167,1384945048774500147047194,
%U A304313 3055115321627096660341307614,8043516699726480852467167758419,24915939138210507189761922944830006,89709850983809128394441772076036629240,371523831948166269091257380175120352465872
%N A304313 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^3 * x ) / F(x) = 0 for n>0.
%C A304313 Is this sequence essentially the same as A006692?
%C A304313 Conjecture: a(n) is the number of connected n-state finite automata with 3 inputs (A006692).
%C A304313 Equals row 3 of table A304321.
%H A304313 Paul D. Hanna, <a href="/A304313/b304313.txt">Table of n, a(n) for n = 0..300</a>
%F A304313 Logarithmic derivative of the o.g.f. of A304323.
%F A304313 For n>=1, a(n) = B_{n+1}((n+1)^3-0!*a(0),-1!*a(1),...,-(n-1)!*a(n-1),0) / n!, where B_{n+1}(...) is the (n+1)-st complete exponential Bell polynomial. - _Max Alekseyev_, Jun 18 2018
%F A304313 a(n) ~ sqrt(1-c) * 3^(3*(n+1)) * n^(2*n + 5/2) / (sqrt(2*Pi) * c^(n+1) * (3-c)^(2*(n+1)) * exp(2*n)), where c = -LambertW(-3*exp(-3)). - _Vaclav Kotesovec_, Aug 31 2020
%e A304313 O.g.f.: L(x) = 1 + 49*x + 6877*x^2 + 1854545*x^3 + 807478656*x^4 + 514798204147*x^5 + 451182323794896*x^6 + 519961864703259753*x^7 + ...
%e A304313 such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304323 :
%e A304313 F(x) = 1 + x + 25*x^2 + 2317*x^3 + 466241*x^4 + 162016980*x^5 + 85975473871*x^6 + 64545532370208*x^7 + 65062315637060121*x^8 + ... + A304323(n)*x^n + ...
%e A304313 which satisfies [x^n] exp( n^3 * x ) / F(x) = 0 for n>0.
%t A304313 m = 25;
%t A304313 F = 1 + Sum[c[k] x^k, {k, m}];
%t A304313 s[n_] := Solve[SeriesCoefficient[Exp[n^3*x]/F, {x, 0, n}] == 0][[1]];
%t A304313 Do[F = F /. s[n], {n, m}];
%t A304313 CoefficientList[D[F, x]/F + O[x]^m, x] (* _Jean-François Alcover_, May 21 2018 *)
%o A304313 (PARI) {a(n) = my(A=[1],L); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^3 +x^2*O(x^m)) / Ser(A) )[m] ); L = Vec(Ser(A)'/Ser(A)); L[n+1]}
%o A304313 for(n=0,25, print1( a(n),", "))
%Y A304313 Cf. A304323, A006692, A304321, A304312, A304314, A304315.
%K A304313 nonn
%O A304313 0,2
%A A304313 _Paul D. Hanna_, May 11 2018