A304312 Logarithmic derivative of F(x) that satisfies: [x^n] exp( n^2 * x ) / F(x) = 0 for n>0.

1, 9, 148, 3493, 106431, 3950832, 172325014, 8617033285, 485267003023, 30363691715629, 2088698040637242, 156612539215405732, 12709745319947141220, 1109746209390479579732, 103724343230007402591558, 10332348604630683943445797, 1092720669631704348689818959, 122274820828415241343176467043, 14433472319311799728710020346232

Offset: 0

Author

Paul D. Hanna, May 11 2018

Keywords

• nonn

Comments

Is this sequence essentially the same as A006691?

Conjecture: a(n) is the number of connected n-state finite automata with 2 inputs (A006691). [I believe the name of A006691 should be changed to read "(n+1)-state". See my comments in A006691. - Petros Hadjicostas, Feb 26 2021]

Equals row 2 of table A304321.

Examples

			O.g.f.: L(x) = 1 + 9*x + 148*x^2 + 3493*x^3 + 106431*x^4 + 3950832*x^5 + 172325014*x^6 + 8617033285*x^7 + 485267003023*x^8 + 30363691715629*x^9 + ...
such that L(x) = F'(x)/F(x) where F(x) is the o.g.f. of A304322 :
F(x) = 1 + x + 5*x^2 + 54*x^3 + 935*x^4 + 22417*x^5 + 685592*x^6 + 25431764*x^7 + 1106630687*x^8 + 55174867339*x^9 + 3097872254493*x^10 + ... + A304322(n)*x^n + ...
which satisfies [x^n] exp( n^2 * x ) / F(x) = 0 for n>0.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A304322, A006691, A304321, A304313, A304314, A304315.

Programs

Mathematica

m = 25;
F = 1 + Sum[c[k] x^k, {k, m}];
s[n_] := Solve[SeriesCoefficient[Exp[n^2 * 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 20 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)^2 +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 A304322.

For n>=1, a(n) = B_{n+1}((n+1)^2-0!*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) * 2^(2*n + 3/2) * n^(n + 3/2) / (sqrt(Pi) * c^(n+1) * (2-c)^(n+1) * exp(n)), where c = -A226775 = -LambertW(-2*exp(-2)). - Vaclav Kotesovec, Aug 31 2020