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%I #11 Jun 06 2018 21:32:45
%S 1,1,32,3618,845824,332389375,196888240512,164288952970296,
%T 184344892426059776,268830705445490506509,496348897291481486672000,
%U 1136486246811467501138927540,3173564392477075053313688696832,10660730426979559461604460186833401,42595326050479099018430338636152049280,200526023793925980859314834103239034380000
%N O.g.f. A(x) satisfies: [x^n] exp( n^3 * x*A(x) ) * (2 - A(x)) = 0 for n > 0.
%C Note: the factorial series, F(x) = Sum_{n>=0} n! * x^n, satisfies:
%C (1) [x^n] exp( n * x*F(x) ) * (2 - F(x)) = 0 for n > 0,
%C (2) [x^n] exp( x*F(x) ) * (n + 1 - F(x)) = 0 for n > 0.
%C It is remarkable that this sequence should consist entirely of integers.
%C Note: a(n) is divisible by n^3 for n >= 1.
%e O.g.f.: A(x) = 1 + x + 32*x^2 + 3618*x^3 + 845824*x^4 + 332389375*x^5 + 196888240512*x^6 + 164288952970296*x^7 + 184344892426059776*x^8 + ...
%e ILLUSTRATION OF SEFINITION.
%e The table of coefficients of x^k/k! in exp( n^3 * x*A(x) ) * (2 - A(x)) begins:
%e n=0: [1, -1, -64, -21708, -20299776, -39886725000, ...];
%e n=1: [1, 0, -63, -21710, -20300931, -39887501724, ...];
%e n=2: [1, 7, 0, -21052, -20280064, -39880261512, ...];
%e n=3: [1, 26, 665, 0, -19381155, -39710564418, ...];
%e n=4: [1, 63, 4032, 252340, 0, -37416032136, ...];
%e n=5: [1, 124, 15561, 1977542, 245086349, 0, ...];
%e n=6: [1, 215, 46592, 10194660, 2254128384, 485581472376, 0, ...]; ...
%e in which the main diagonal is all zeros after the initial term, illustrating that [x^n] exp( n^3 * x*A(x) ) * (2 - A(x)) = 0 for n > 0.
%e Terms along the secondary diagonal in the above table are divisible by the differences of cubes: [1, 7/7, 665/19, 252340/37, 245086349/61, 485581472376/91, ...] = [1, 1, 35, 6820, 4017809, 5336060136, ...].
%e RELATED SERIES.
%e exp( x*A(x) ) = 1 + x + 3*x^2/2! + 199*x^3/3! + 87625*x^4/4! + 101938881*x^5/5! + 239933646571*x^6/6! + 993998976594583*x^7/7! + 6632090620377452049*x^8/8! + ...
%e Note that the factorial series
%e F(x) = 1 + x + 2!*x^2 + 3!*x^3 + 4!*x^4 + 5!*x^5 + ... + n!*x^n + ...
%e satisfies [x^n] exp( n*x*F(x) ) * (2 - F(x)) = 0 for n > 0.
%o (PARI) {a(n) = my(A=[1],m); for(i=1,n, A=concat(A,0); m=#A; A[m] = Vec( exp( (m-1)^3 * x * Ser(A) ) * (2 - Ser(A)) )[m] );A[n+1]}
%o for(n=0,20, print1(a(n),", "))
%Y Cf. A304400.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 25 2018
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