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%I #13 Jul 24 2019 03:37:31
%S 1,3,120,14490,3472200,1368656352,799089568704,646368125624208,
%T 691455367099357920,945212125833646957440,1607847054961982310495360,
%U 3331923827097057308557968384,8266530005101303527012249628800,24199480411463293209043241824200960,82559430544523811331164210671638394880,324767777154180733152754873992738391947264,1459453773372794878685570187904547878429951488
%N O.g.f. satisfies: [x^n/n!] exp( n^2*(n+1)*x - n*(n+1)*A(x) ) = 0, for n >= 1.
%C It is remarkable that this sequence should consist entirely of integers.
%e O.g.f.: A(x) = x + 3*x^2 + 120*x^3 + 14490*x^4 + 3472200*x^5 + 1368656352*x^6 + 799089568704*x^7 + 646368125624208*x^8 + 691455367099357920*x^9 + ...
%e The table of coefficients of x^k/k! in exp( n^2*(n+1)*x - n*(n+1)*A(x) ) begins:
%e n=1: [1, 0, -12, -1440, -695088, -833155200, -1970719243200, ...];
%e n=2: [1, 6, 0, -4752, -2192832, -2562534144, -6002370169344, ...];
%e n=3: [1, 24, 504, 0, -4904064, -5544412416, -12577520001024, ...];
%e n=4: [1, 60, 3480, 180000, 0, -10389600000, -23099972428800, ...];
%e n=5: [1, 120, 14220, 1641600, 171104400, 0, -39055052923200, ...];
%e n=6: [1, 210, 43848, 9072000, 1838313792, 339176678400, 0, ...];
%e n=7: [1, 336, 112560, 37554048, 12444582528, 4054169581056, 1209847750465536, 0, ...]; ...
%e in which the coefficients of x^n in row n form a diagonal of zeros.
%o (PARI) {a(n) = my(A=[1], m); for(i=1, n+1, m=#A; A=concat(A, 0); A[m+1] = Vec( exp(m^2*(m+1)*x +x*O(x^#A)) / Ser(A)^(m*(m+1)) )[m+1]/m/(m+1) ); polcoeff( log(Ser(A)), n)}
%o for(n=1, 30, print1(a(n), ", "))
%K nonn
%O 1,2
%A _Paul D. Hanna_, Dec 06 2018
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