A304401 O.g.f. A(x) satisfies: [x^n] exp( n^3 * x*A(x) ) * (2 - A(x)) = 0 for n > 0.

1, 1, 32, 3618, 845824, 332389375, 196888240512, 164288952970296, 184344892426059776, 268830705445490506509, 496348897291481486672000, 1136486246811467501138927540, 3173564392477075053313688696832, 10660730426979559461604460186833401, 42595326050479099018430338636152049280, 200526023793925980859314834103239034380000

Offset: 0

Author

Paul D. Hanna, May 25 2018

Keywords

• nonn

Comments

Note: the factorial series, F(x) = Sum_{n>=0} n! * x^n, satisfies:

(1) [x^n] exp( n * x*F(x) ) * (2 - F(x)) = 0 for n > 0,

(2) [x^n] exp( x*F(x) ) * (n + 1 - F(x)) = 0 for n > 0.

It is remarkable that this sequence should consist entirely of integers.

Note: a(n) is divisible by n^3 for n >= 1.

Examples

			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 + ...
ILLUSTRATION OF SEFINITION.
The table of coefficients of x^k/k! in exp( n^3 * x*A(x) ) * (2 - A(x)) begins:
n=0: [1, -1, -64, -21708, -20299776, -39886725000, ...];
n=1: [1, 0, -63, -21710, -20300931, -39887501724, ...];
n=2: [1, 7, 0, -21052, -20280064, -39880261512, ...];
n=3: [1, 26, 665, 0, -19381155, -39710564418, ...];
n=4: [1, 63, 4032, 252340, 0, -37416032136, ...];
n=5: [1, 124, 15561, 1977542, 245086349, 0, ...];
n=6: [1, 215, 46592, 10194660, 2254128384, 485581472376, 0, ...]; ...
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.
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, ...].
RELATED SERIES.
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! + ...
Note that the factorial series
F(x) = 1 + x + 2!*x^2 + 3!*x^3 + 4!*x^4 + 5!*x^5 + ... + n!*x^n + ...
satisfies [x^n] exp( n*x*F(x) ) * (2 - F(x)) = 0 for n > 0.
		

Crossrefs

Cf. A304400.

Programs

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]}
for(n=0,20, print1(a(n),", "))