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A317337 O.g.f. A(x) satisfies: [x^n] exp( n^3*x*A(x) ) * (n+1 - n*A(x)) = 0 for n>=1. 4

%I #14 Aug 06 2018 05:41:22

%S 1,1,12,729,111440,31377625,14001201036,9064452341847,

%T 8027821828474816,9322437359885669613,13746212321035446900300,

%U 25094943743950232692612534,55574014665416527079564569056,146797467684802516650481763597455,456012687037844090869850529901126900,1645914373011657806464530612985244787000

%N O.g.f. A(x) satisfies: [x^n] exp( n^3*x*A(x) ) * (n+1 - n*A(x)) = 0 for n>=1.

%C Compare: the factorial series F(x) = Sum_{n>=0} n!*x^n satisfies

%C (1) [x^n] exp( n^2*x*F(x) ) * (n + 1 - n*F(x)) = 0 for n>=1,

%C (2) [x^n] exp( n^3*x*F(x) ) * (n^2 + 1 - n^2*F(x)) = 0 for n>=1.

%C It is remarkable that this sequence should consist entirely of integers.

%H Paul D. Hanna, <a href="/A317337/b317337.txt">Table of n, a(n) for n = 0..300</a>

%F a(n) ~ sqrt(1-c) * 3^(3*n) * n^(2*n - 3/2) / (sqrt(2*Pi) * c^n * (3-c)^(2*n) * exp(2*n)), where c = -LambertW(-3*exp(-3)) = 0.1785606278779211065968086697... = -A226750. - _Vaclav Kotesovec_, Aug 06 2018

%e O.g.f.: A(x) = 1 + x + 12*x^2 + 729*x^3 + 111440*x^4 + 31377625*x^5 + 14001201036*x^6 + 9064452341847*x^7 + 8027821828474816*x^8 + ...

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp( n^3*x*A(x) ) * (n+1 - n*A(x)) begins:

%e n=1: [1, 0, -23, -4376, -2674995, -3765464504, ...];

%e n=2: [1, 6, 0, -8908, -5494464, -7640806512, ...];

%e n=3: [1, 24, 549, 0, -8632395, -12056269968, ...];

%e n=4: [1, 60, 3616, 204712, 0, -17114998496, -45010750350080, ...];

%e n=5: [1, 120, 14505, 1750880, 197597325, 0, -60559334101475, ...];

%e n=6: [1, 210, 44352, 9406044, 1987128000, 391935493296, 0, ...]; ...

%e in which the coefficients of x^n in row n form a diagonal of zeros.

%o (PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp( (m-1)^3*x*(Ser(A)) ) * (m - (m-1)*Ser(A)) )[m]/(m-1) ); A[n+1]}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A305110, A305114, A305115, A305116, A317338.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Aug 01 2018

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