A304315 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^5 * x ) / F(x) = 0 for n>0.

1, 961, 6737401, 172342090401, 11657788116175751, 1722786509653595220757, 489506033977061086758261063, 243968979437942649897623460813009, 199025593654123221838381793032781035510, 251774439716905627952289102887999425054599511, 472942802381336010263584088374665504251010554412128, 1273071332950625956697135571575613091625334028239417955701

Offset: 0

Author

Paul D. Hanna, May 11 2018

Keywords

• nonn

Comments

Conjecture: a(n) is the number of connected n-state finite automata with 5 inputs.

Equals row 5 of table A304321.

Examples

			O.g.f.: L(x) = 1 + 961*x + 6737401*x^2 + 172342090401*x^3 + 11657788116175751*x^4 + 1722786509653595220757*x^5 + 489506033977061086758261063*x^6 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304325 :
F(x) = 1 + x + 481*x^2 + 2246281*x^3 + 43087884081*x^4 + 2331601789103231*x^5 + 287133439746933073357*x^6 + 69929721774643572422651223*x^7 + ... + A304325(n)*x^n + ...
which satisfies [x^n] exp( n^5 * x ) / F(x) = 0 for n>0.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A304325, A304321, A304312, A304313, A304314.

Programs

Mathematica

m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n^5*x]/F, {x, 0, n}] == 0][[1]];
Do[F = F /. s[n], {n, m}];
CoefficientList[D[F, x]/F + O[x]^m, x] (* Jean-François Alcover, May 21 2018 *)

PARI

{a(n) = my(A=[1],L); for(i=0, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^4 +x^2*O(x^m)) / Ser(A) )[m] ); L = Vec(Ser(A)'/Ser(A)); L[n+1]}
for(n=0,25, print1( a(n),", "))

Formula

Logarithmic derivative of the o.g.f. of A304325.

For n>=1, a(n) = B_{n+1}((n+1)^5-0!*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

a(n) ~ sqrt(1-c) * 5^(5*(n+1)) * n^(4*n + 9/2) / (sqrt(2*Pi) * c^(n+1) * (5-c)^(4*(n+1)) * exp(4*n)), where c = -LambertW(-5*exp(-5)). - Vaclav Kotesovec, Aug 31 2020