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

%I #9 Jun 05 2018 17:43:00

%S 1,7,207,14226,1852800,409408077,142286748933,73448832515952,

%T 53835885818473473,54041298732304775000,72129250579997923194091,

%U 124900802377559946754633602,274851919918333747166200590840,755158633069275870471471631726803,2551279948230221759814139760682442500

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

%C O.g.f. equals the logarithm of the e.g.f. of A300618.

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

%H Paul D. Hanna, <a href="/A300619/b300619.txt">Table of n, a(n) for n = 1..200</a>

%e O.g.f.: A(x) = x + 7*x^2 + 207*x^3 + 14226*x^4 + 1852800*x^5 + 409408077*x^6 + 142286748933*x^7 + 73448832515952*x^8 + 53835885818473473*x^9 + ...

%e where

%e exp(A(x)) = 1 + x + 15*x^2/2! + 1285*x^3/3! + 347065*x^4/4! + 224232501*x^5/5! + 296201195791*x^6/6! + 719274160258585*x^7/7! + ... + A300618(n)*x^n/n! + ...

%e such that: [x^n] exp( n * A(x) ) = n^3 * [x^(n-1)] exp( n * A(x) ).

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

%o for(n=1, 20, print1(a(n), ", "))

%Y Cf. A300618, A296171, A300591, A300593, A300595, A300597, A300617.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Mar 10 2018