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%I #9 Sep 30 2018 19:10:44
%S 1,11,615,93042,26367840,11896260357,7790323334052,6971417293247088,
%T 8168310247001057784,12135241152923019699370,
%U 22293293737202063158881716,49628657379474422562364524042,131669704562621977069760097583577,410540174242581740798413641858906388
%N O.g.f. A(x) satisfies: [x^n] 1/(1-x)^(n^3) / exp( n*A(x) ) = 0 for n >= 1.
%C It is remarkable that this sequence should consist entirely of integers.
%H Paul D. Hanna, <a href="/A319835/b319835.txt">Table of n, a(n) for n = 1..200</a>
%e G.f.: A(x) = x + 11*x^2 + 615*x^3 + 93042*x^4 + 26367840*x^5 + 11896260357*x^6 + 7790323334052*x^7 + 6971417293247088*x^8 + ...
%e The table of coefficients of x^k/k! in 1/(1-x)^(n^3)/exp(n*A(x)) begins
%e n=1: [1, 0, -21, -3688, -2231679, -3163366296, -8564468186765, ...];
%e n=2: [1, 6, 0, -7796, -4645296, -6462213792, -17357826387392, ...];
%e n=3: [1, 24, 537, 0, -7554843, -10352322288, -27117968407587, ...];
%e n=4: [1, 60, 3576, 197048, 0, -15133017984, -39314407194560, ...];
%e n=5: [1, 120, 14415, 1715200, 188756385, 0, -54226299944825, ...];
%e n=6: [1, 210, 44184, 9292212, 1935426096, 373568674464, 0, ...]; ...
%e in which the coefficient of x^n in the n-th row forms a diagonal of zeros.
%e RELATED SERIES.
%e exp(A(x)) = 1 + x + 23*x^2/2! + 3757*x^3/3! + 2249353*x^4/4! + 3176162021*x^5/5! + 8585203977031*x^6/6! + 39325009824213793*x^7/7! + ...
%o (PARI) {a(n) = my(A=[1], m); for(i=1, n+1, A=concat(A, 0); m=#A; A[m] = Vec( 1/(1-x +x^2*O(x^m))^(m^3) * exp(-m*x*Ser(A)) )[m+1]/m ); polcoeff( x*Ser(A), n)}
%o for(n=1, 20, print1(a(n), ", "))
%Y Cf. A317342, A319836, A319831.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Sep 30 2018
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