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A320668 O.g.f. A(x) satisfies: [x^n] exp(-n^3*A(x)) / (1 - n*x)^(n^2) = 0, for n > 0.

%I A320668 #8 Oct 19 2018 17:50:33
%S A320668 1,1,3,48,1125,74844,4538576,571979264,61768818081,11756208796500,
%T A320668 1930305045364047,501690433505046336,114627985830970025544,
%U A320668 38401761759325497631504,11530876917646339177773375,4792821920208552461683208192,1816651428077402993910096849969,911361374568809242258003199407404
%N A320668 O.g.f. A(x) satisfies: [x^n] exp(-n^3*A(x)) / (1 - n*x)^(n^2) = 0, for n > 0.
%C A320668 It is remarkable that this sequence should consist entirely of integers.
%e A320668 O.g.f.: A(x) = x + x^2 + 3*x^3 + 48*x^4 + 1125*x^5 + 74844*x^6 + 4538576*x^7 + 571979264*x^8 + 61768818081*x^9 + 11756208796500*x^10 + ...
%e A320668 ILLUSTRATION OF DEFINITION.
%e A320668 The table of coefficients of x^k/k! in exp(-n^3*A(x)) / (1 - n*x)^(n^2) begins:
%e A320668 n=1: [1, 0, -1, -16, -1143, -134816, -53867825, ...];
%e A320668 n=2: [1, 0, 0, -80, -8832, -1076928, -431006720, ...];
%e A320668 n=3: [1, 0, 27, 0, -24543, -3592512, -1464710445, ...];
%e A320668 n=4: [1, 0, 128, 896, 0, -7099904, -3495833600, ...];
%e A320668 n=5: [1, 0, 375, 4000, 371625, 0, -6020725625, ...];
%e A320668 n=6: [1, 0, 864, 11664, 2270592, 78335424, 0, ...];
%e A320668 n=7: [1, 0, 1715, 27440, 9134433, 444056032, 73395100555, 0, ...]; ...
%e A320668 in which the coefficient of x^n in row n forms a diagonal of zeros.
%o A320668 (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(-m^3*x*Ser(A))/(1-m*x +x^2*O(x^m))^(m^2))[m+1]/m^3 ); A[n]}
%o A320668 for(n=1, 20, print1(a(n), ", "))
%Y A320668 Cf. A320418, A320669, A319939.
%K A320668 nonn
%O A320668 1,3
%A A320668 _Paul D. Hanna_, Oct 19 2018