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%I #5 Mar 10 2018 10:06:26
%S 1,16,19683,142475264,3436799053125,212148041589128016,
%T 28458158819417861315152,7380230750280159370894934016,
%U 3385049575573746853297963891959753,2561548157856026756893458765378989150000,3026444829408778969259555715061437179090541565,5340113530831632053993990154143996936096662034267136
%N O.g.f. A(x) satisfies: [x^n] exp( n^5 * A(x) ) = n^5 * [x^(n-1)] exp( n^5 * A(x) ) for n>=1.
%C Compare to: [x^n] exp( n^5 * x ) = n^4 * [x^(n-1)] exp( n^5 * x ) for n>=1.
%H Paul D. Hanna, <a href="/A300615/b300615.txt">Table of n, a(n) for n = 1..150</a>
%F O.g.f. equals the logarithm of the e.g.f. of A300614.
%e O.g.f.: A(x) = x + 16*x^2 + 19683*x^3 + 142475264*x^4 + 3436799053125*x^5 + 212148041589128016*x^6 + 28458158819417861315152*x^7 + ...
%e where
%e exp(A(x)) = 1 + x + 33*x^2/2! + 118195*x^3/3! + 3419881993*x^4/4! + 412433022394701*x^5/5! + 152749066271797582081*x^6/6! + ... + A300614(n)*x^n/n! + ...
%e such that: [x^n] exp( n^5 * A(x) ) = n^5 * [x^(n-1)] exp( n^5 * A(x) ).
%o (PARI) {a(n) = my(A=[1]); for(i=1, n+1, A=concat(A, 0); V=Vec(Ser(A)^((#A-1)^5)); A[#A] = ((#A-1)^5*V[#A-1] - V[#A])/(#A-1)^5 ); polcoeff( log(Ser(A)), n)}
%o for(n=1, 20, print1(a(n), ", "))
%Y Cf. A300614, A296171, A300591, A300593, A300595, A300597.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Mar 10 2018
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