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A320669 O.g.f. A(x) satisfies: [x^n] exp(-n^3*A(x)) / (1 - n^2*x)^n = 0, for n > 0.

%I A320669 #7 Oct 19 2018 17:50:21
%S A320669 1,2,27,2176,316125,92433420,38689900249,24036220587520,
%T A320669 19705732103751309,21228545767337495500,28631298365231328948940,
%U A320669 47701162183511368703635200,95797470923250302955913961043,228907109818475997814838969598324,641132565508623116202107427900402750,2082400670957118326405938988144017645568
%N A320669 O.g.f. A(x) satisfies: [x^n] exp(-n^3*A(x)) / (1 - n^2*x)^n = 0, for n > 0.
%C A320669 It is remarkable that this sequence should consist entirely of integers.
%e A320669 O.g.f.: A(x) = x + 2*x^2 + 27*x^3 + 2176*x^4 + 316125*x^5 + 92433420*x^6 + 38689900249*x^7 + 24036220587520*x^8 + 19705732103751309*x^9 + ...
%e A320669 ILLUSTRATION OF DEFINITION.
%e A320669 The table of coefficients of x^k/k! in exp(-n^3*A(x)) / (1 - n^2*x)^n begins:
%e A320669 n=1: [1, 0, -3, -160, -52191, -37930176, -66549456875, ...];
%e A320669 n=2: [1, 0, 0, -1040, -414720, -303430848, -532404700160, ...];
%e A320669 n=3: [1, 0, 135, 0, -1237275, -1019993472, -1799293659165, ...];
%e A320669 n=4: [1, 0, 768, 22400, 0, -2155144704, -4259850874880, ...];
%e A320669 n=5: [1, 0, 2625, 136000, 25862625, 0, -7511859284375, ...];
%e A320669 n=6: [1, 0, 6912, 524880, 192513024, 36792874944, 0, ...];
%e A320669 n=7: [1, 0, 15435, 1591520, 938926485, 280095248832, 121196964253015, 0, ...]; ...
%e A320669 in which the coefficient of x^n in row n forms a diagonal of zeros.
%o A320669 (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^2*x +x^2*O(x^m))^m)[m+1]/m^3 ); A[n]}
%o A320669 for(n=1, 20, print1(a(n), ", "))
%Y A320669 Cf. A320418, A320668, A319939.
%K A320669 nonn
%O A320669 1,2
%A A320669 _Paul D. Hanna_, Oct 19 2018