A000442 a(n) = (n!)^3.
1, 1, 8, 216, 13824, 1728000, 373248000, 128024064000, 65548320768000, 47784725839872000, 47784725839872000000, 63601470092869632000000, 109903340320478724096000000, 241457638684091756838912000000, 662559760549147780765974528000000, 2236139191853373760085164032000000000
Offset: 0
References
- F. Smarandache, "Properties of the Numbers", University of Craiova Archives, 1975; Arizona State University Special Collections, Tempe, AZ
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..100
- G. S. Kazandzidis, On a Conjecture of Moessner and a General Problem, Bull. Soc. Math. Grèce, Nouvelle Série - vol. 2, fasc. 1-2, pp. 23-30, 1961.
- Index to divisibility sequences
Programs
-
Magma
[Factorial(n)^3: n in [0..15]]; // Vincenzo Librandi, Jan 13 2012
-
Maple
seq((n!)^3, n=0..14), # Karol A. Penson, Jul 28 2013
-
Mathematica
Table[(n!)^3, {n, 0, 20}] (* Stefan Steinerberger, Apr 14 2006 *) -
PARI
a(n)=n!^3 \\ Charles R Greathouse IV, Jan 12 2012
Formula
a(n) = det(S(i+3, j), 1 <= i, j <= n), where S(n, k) are Stirling numbers of the second kind. - Mircea Merca, Apr 04 2013
From Karol A. Penson, Jul 28 2013: (Start)
G.f. of hypergeometric type: sum(a(n)*z^n/(n!)^3, n = 0..infinity) = 1/(1-z);
Integral representation as n-th moment of a positive function w(x) on a positive halfaxis (solution of the Stieltjes moment problem), in Maple notation:
a(n) = int(x^n*w(x), x = 0..infinity), n >= 0, where w(x) = MeijerG([[], []], [[0, 0, 0]], []], x), w(0) = infinity, limit(w(x), x = infinity) = 0.
w(x) is monotonically decreasing over (0, infinity). The Meijer G function above cannot be represented by any other known special function. This solution of the Stieltjes moment problem is not unique.
Asymptotics: a(n) -> (1/16)*sqrt(2)*Pi^(3/2)*(32*n^2 + 8*n + 1)*(n)^(-1/2+3*n)*exp(-3*n), for n -> infinity. (End)
D-finite with recurrence: a(n) -n^3*a(n-1)=0. - R. J. Mathar, Feb 16 2020
From Amiram Eldar, Nov 09 2020: (Start)
a(n) = A000142(n)^3.
Sum_{n>=0} 1/a(n) = A271574. (End)
a(n) = [x^n] Product_{k=1..n} (1 + k^3*x). - Vaclav Kotesovec, Feb 19 2022
Comments