

A000142


Factorial numbers: n! = 1*2*3*4*...*n (order of symmetric group S_n, number of permutations of n letters).
(Formerly M1675 N0659)


2664



1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

The earliest publication that discusses this sequence appears to be the Sepher Yezirah [Book of Creation], circa AD 300. (See Knuth, also the Zeilberger link.)  N. J. A. Sloane, Apr 07 2014
For n >= 1, a(n) is the number of n X n (0,1) matrices with each row and column containing exactly one entry equal to 1.
This sequence is the BinomialMean transform of A000354. (See A075271 for definition.)  John W. Layman, Sep 12 2002 [This is easily verified from the Paul Barry formula for A000354, by interchanging summations and using the formula: Sum_k (1)^k C(ni, k) = KroneckerDelta(i,n).  David Callan, Aug 31 2003]
Number of distinct subsets of T(n1) elements with 1 element A, 2 elements B, ..., n  1 elements X (e.g., at n = 5, we consider the distinct subsets of ABBCCCDDDD and there are 5! = 120).  Jon Perry, Jun 12 2003
n! is the smallest number with that prime signature. E.g., 720 = 2^4 * 3^2 * 5.  Amarnath Murthy, Jul 01 2003
a(n) is the permanent of the n X n matrix M with M(i, j) = 1.  Philippe Deléham, Dec 15 2003
Given n objects of distinct sizes (e.g., areas, volumes) such that each object is sufficiently large to simultaneously contain all previous objects, then n! is the total number of essentially different arrangements using all n objects. Arbitrary levels of nesting of objects are permitted within arrangements. (This application of the sequence was inspired by considering leftover moving boxes.) If the restriction exists that each object is able or permitted to contain at most one smaller (but possibly nested) object at a time, the resulting sequence begins 1,2,5,15,52 (Bell Numbers?). Sets of nested wooden boxes or traditional nested Russian dolls come to mind here.  Rick L. Shepherd, Jan 14 2004
Stirling transform of [2, 2, 6, 24, 120, ...] is A052856 = [2, 2, 4, 14, 76, ...].
Stirling transform of [1, 2, 6, 24, 120, ...] is A000670 = [1, 3, 13, 75, ...].
Stirling transform of [0, 2, 6, 24, 120, ...] is A052875 = [0, 2, 12, 74, ...].
Stirling transform of [1, 1, 2, 6, 24, 120, ...] is A000629 = [1, 2, 6, 26, ...].
Stirling transform of [0, 1, 2, 6, 24, 120, ...] is A002050 = [0, 1, 5, 25, 140, ...].
Stirling transform of (A165326*A089064)(1...) = [1, 0, 1, 1, 8, 26, 194, ...] is [1, 1, 2, 6, 24, 120, ...] (this sequence). (End)
First Eulerian transform of 1, 1, 1, 1, 1, 1... The first Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} e(n, k)s(k), where e(n, k) is a firstorder Eulerian number [A008292].  Ross La Haye, Feb 13 2005
Conjecturally, 1, 6, and 120 are the only numbers which are both triangular and factorial.  Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
n! is the nth finite difference of consecutive nth powers. E.g., for n = 3, [0, 1, 8, 27, 64, ...] > [1, 7, 19, 37, ...] > [6, 12, 18, ...] > [6, 6, ...].  Bryan Jacobs (bryanjj(AT)gmail.com), Mar 31 2005
a(n+1) = (n+1)! = 1, 2, 6, ... has e.g.f. 1/(1x)^2.  Paul Barry, Apr 22 2005
Write numbers 1 to n on a circle. Then a(n) = sum of the products of all n  2 adjacent numbers. E.g., a(5) = 1*2*3 + 2*3*4 + 3*4*5 + 4*5*1 +5*1*2 = 120.  Amarnath Murthy, Jul 10 2005
The number of chains of maximal length in the power set of {1, 2, ..., n} ordered by the subset relation.  Rick L. Shepherd, Feb 05 2006
The number of circular permutations of n letters for n >= 0 is 1, 1, 1, 2, 6, 24, 120, 720, 5040, 40320, ...  Xavier Noria (fxn(AT)hashref.com), Jun 04 2006
a(n) is the number of deco polyominoes of height n (n >= 1; see definitions in the Barcucci et al. references).  Emeric Deutsch, Aug 07 2006
a(n) is the number of partition tableaux of size n. See Steingrimsson/Williams link for the definition.  David Callan, Oct 06 2006
Consider the n! permutations of the integer sequence [n] = 1, 2, ..., n. The ith permutation consists of ncycle(i) permutation cycles. Then, if the Sum_{i=1..n!} 2^ncycle(i) runs from 1 to n!, we have Sum_{i=1..n!} 2^ncycle(i) = (n+1)!. E.g., for n = 3 we have ncycle(1) = 3, ncycle(2) = 2, ncycle(3) = 1, ncycle(4) = 2, ncycle(5) = 1, ncycle(6) = 2 and 2^3 + 2^2 + 2^1 + 2^2 + 2^1 + 2^2 = 8 + 4 + 2 + 4 + 2 + 4 = 24 = (n+1)!.  Thomas Wieder, Oct 11 2006
a(n) is the number of set partitions of {1, 2, ..., 2n  1, 2n} into blocks of size 2 (perfect matchings) in which each block consists of one even and one odd integer. For example, a(3) = 6 counts 123456, 123645, 142356, 142536, 162345, 162534.  David Callan, Mar 30 2007
Consider the multiset M = [1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...] = [1, 2, 2, ..., n x 'n'] and form the set U (where U is a set in the strict sense) of all subsets N (where N may be a multiset again) of M. Then the number of elements U of U is equal to (n+1)!. E.g. for M = [1, 2, 2] we get U = [[], [2], [2, 2], [1], [1, 2], [1, 2, 2]] and U = 3! = 6. This observation is a more formal version of the comment given already by Rick L. Shepherd, Jan 14 2004.  Thomas Wieder, Nov 27 2007
For n >= 1, a(n) = 1, 2, 6, 24, ... are the positions corresponding to the 1's in decimal expansion of Liouville's constant (A012245).  Paul Muljadi, Apr 15 2008
Triangle A144107 has n! for row sums (given n > 0) with right border n! and left border A003319, the INVERTi transform of (1, 2, 6, 24, ...).  Gary W. Adamson, Sep 11 2008
a(n) is also the number of orderdecreasing full transformations (of an nchain).
a(n1) is also the number of nilpotent orderdecreasing full transformations (of an nchain). (End)
n! is also the number of optimal broadcast schemes in the complete graph K_{n}, equivalent to the number of binomial trees embedded in K_{n} (see Calin D. Morosan, Information Processing Letters, 100 (2006), 188193).  Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Let S_{n} denote the nstar graph. The S_{n} structure consists of n S_{n1} structures. This sequence gives the number of edges between the vertices of any two specified S_{n+1} structures in S_{n+2} (n >= 1).  K.V.Iyer, Mar 18 2009
Chromatic invariant of the sun graph S_{n2}.
It appears that a(n+1) is the inverse binomial transform of A000255.  Timothy Hopper (timothyhopper(AT)hotmail.co.uk), Aug 20 2009
a(n) is also the determinant of a square matrix, An, whose coefficients are the reciprocals of beta function: a{i, j} = 1/beta(i, j), det(An) = n!.  Enrique Pérez Herrero, Sep 21 2009
The asymptotic expansions of the exponential integrals E(x, m = 1, n = 1) ~ exp(x)/x*(1  1/x + 2/x^2  6/x^3 + 24/x^4 + ...) and E(x, m = 1, n = 2) ~ exp(x)/x*(1  2/x + 6/x^2  24/x^3 + ...) lead to the factorial numbers. See A163931 and A130534 for more information.  Johannes W. Meijer, Oct 20 2009
a(n) = G^n where G is the geometric mean of the first n positive integers.  Jaroslav Krizek, May 28 2010
Increasing colored 12 trees with choice of two colors for the rightmost branch of nonleaves.  Wenjin Woan, May 23 2011
The sequence 1!, (2!)!, ((3!)!)!, (((4!)!)!)!, ..., ((...(n!)!)...)! (n times) grows too rapidly to have its own entry. See Hofstadter.
The e.g.f. of 1/a(n) = 1/n! is BesselI(0, 2*sqrt(x)). See AbramowitzStegun, p. 375, 9.3.10.  Wolfdieter Lang, Jan 09 2012
Number of permutations of elements 1, 2, ..., n + 1 with a fixed element belonging to a cycle of length r does not depend on r and equals a(n).  Vladimir Shevelev, May 12 2012
a(n) is the number of fixed points in all permutations of 1, ..., n: in all n! permutations, 1 is first exactly (n1)! times, 2 is second exactly (n1)! times, etc., giving (n1)!*n = n!.  Jon Perry, Dec 20 2012
For m >= 3, a(m2) is the number hp(m) of acyclic Hamiltonian paths in a simple graph with m vertices, which is complete except for one missing edge. For m < 3, hp(m)=0.  Stanislav Sykora, Jun 17 2014
a(n) is the number of increasing forests with n nodes.  Brad R. Jones, Dec 01 2014
The factorial numbers can be calculated by means of the recurrence n! = (floor(n/2)!)^2 * sf(n) where sf(n) are the swinging factorials A056040. This leads to an efficient algorithm if sf(n) is computed via prime factorization. For an exposition of this algorithm see the link below.  Peter Luschny, Nov 05 2016
Treeshelves are ordered (plane) binary (012) increasing trees where the nodes of outdegree 1 come in 2 colors. There are n! treeshelves of size n, and classical Françon's bijection maps bijectively treeshelves into permutations.  Sergey Kirgizov, Dec 26 2016
Satisfies Benford's law [Diaconis, 1977; BergerHill, 2017]  N. J. A. Sloane, Feb 07 2017
a(n) = Sum((d_p)^2), where d_p is the number of standard tableaux in the Ferrers board of the integer partition p and summation is over all integer partitions p of n. Example: a(3) = 6. Indeed, the partitions of 3 are [3], [2,1], and [1,1,1], having 1, 2, and 1 standard tableaux, respectively; we have 1^2 + 2^2 + 1^2 = 6.  Emeric Deutsch, Aug 07 2017
a(n) is the nth derivative of x^n.  Iain Fox, Nov 19 2017
a(n) is the number of maximum chains in the ndimensional Boolean cube {0,1}^n in respect to the relation "precedes". It is defined as follows: for arbitrary vectors u, v of {0,1}^n, such that u = (u_1, u_2, ..., u_n) and v = (v_1, v_2, ..., v_n), "u precedes v" if u_i <= v_i, for i=1, 2, ..., n.  Valentin Bakoev, Nov 20 2017
a(n) is the number of shortest paths (for example, obtained by Breadth First Search) between the nodes (0,0,...,0) (i.e., the allzeros vector) and (1,1,...,1) (i.e., the allones vector) in the graph H_n, corresponding to the ndimensional Boolean cube {0,1}^n. The graph is defined as H_n = (V_n, E_n), where V_n is the set of all vectors of {0,1}^n, and E_n contains edges formed by each pair adjacent vectors.  Valentin Bakoev, Nov 20 2017
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = sigma(gcd(i,j)) for 1 <= i,j <= n.  Bernard Schott, Dec 05 2018
a(n) is also the number of inversion sequences of length n. A length n inversion sequence e_1, e_2, ..., e_n is a sequence of n integers such that 0 <= e_i < i.  Juan S. Auli, Oct 14 2019
The term "factorial" ("factorielle" in French) was coined by the French mathematician Louis François Antoine Arbogast (17591803) in 1800. The notation "!" was first used by the French mathematician Christian Kramp (17601826) in 1808.  Amiram Eldar, Apr 16 2021
Also the number of signotopes of rank 2, i.e., mappings X:{{1..n} choose 2}>{+,} such that for any three indices a < b < c, the sequence X(a,b), X(a,c), X(b,c) changes its sign at most once (see FelsnerWeil reference).  Manfred Scheucher, Feb 09 2022


REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 125; also p. 90, ex. 3.
Diaconis, Persi, The distribution of leading digits and uniform distribution mod 1, Ann. Probability, 5, 1977, 7281,
Douglas R. Hofstadter, Fluid concepts & creative analogies: computer models of the fundamental mechanisms of thought, Basic Books, 1995, pages 4446.
A. N. Khovanskii. The Application of Continued Fractions and Their Generalizations to Problem in Approximation Theory. Groningen: Noordhoff, Netherlands, 1963. See p. 141 (10.19).
D. E. Knuth, The Art of Computer Programming, Vol. 3, Section 5.1.2, p. 23. [From N. J. A. Sloane, Apr 07 2014]
J.M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 693 pp. 90, 297, Ellipses Paris 2004.
A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 19861992.
R. W. Robinson, Counting arrangements of bishops, pp. 198214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
Sepher Yezirah [Book of Creation], circa AD 300. See verse 52.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Stanton and D. White, Constructive Combinatorics, Springer, 1986; see p. 91.
Carlo Suares, Sepher Yetsira, Shambhala Publications, 1976. See verse 52.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 102.


LINKS

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].


FORMULA

Sum_{i=0..n} (1)^i * i^n * binomial(n, i) = (1)^n * n!.  Yong Kong (ykong(AT)curagen.com), Dec 26 2000
Sum_{i=0..n} (1)^i * (ni)^n * binomial(n, i) = n!.  Peter C. Heinig (algorithms(AT)gmx.de), Apr 10 2007
The sequence trivially satisfies the recurrence a(n+1) = Sum_{k=0..n} binomial(n,k) * a(k)*a(nk).  Robert FERREOL, Dec 05 2009
Dfinite with recurrence: a(n) = n*a(n1), n >= 1. n! ~ sqrt(2*Pi) * n^(n+1/2) / e^n (Stirling's approximation).
a(0) = 1, a(n) = subs(x = 1, (d^n/dx^n)(1/(2x))), n = 1, 2, ...  Karol A. Penson, Nov 12 2001
a(n) = Sum_{k=0..n} (1)^(nk)*A000522(k)*binomial(n, k) = Sum_{k=0..n} (1)^(nk)*(x+k)^n*binomial(n, k).  Philippe Deléham, Jul 08 2004
a(n) = Sum_{i=1..n} ((1)^(i1) * sum of 1..n taken n  i at a time)  e.g., 4! = (1*2*3 + 1*2*4 + 1*3*4 + 2*3*4)  (1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4) + (1 + 2 + 3 + 4)  1 = (6 + 8 + 12 + 24)  (2 + 3 + 4 + 6 + 8 + 12) + 10  1 = 50  35 + 10  1 = 24.  Jon Perry, Nov 14 2005
a(n) = (n1)*(a(n1) + a(n2)), n >= 2.  Matthew J. White, Feb 21 2006
1 / a(n) = determinant of matrix whose (i,j) entry is (i+j)!/(i!(j+1)!) for n > 0. This is a matrix with Catalan numbers on the diagonal.  Alexander Adamchuk, Jul 04 2006
For n >= 2, a(n2) = (1)^n*Sum_{j=0..n1} (j+1)*Stirling1(n,j+1).  Milan Janjic, Dec 14 2008
G.f.: 1/(1xx^2/(13x4x^2/(15x9x^2/(17x16x^2/(19x25x^2... (continued fraction), hence Hankel transform is A055209.
G.f. of (n+1)! is 1/(12x2x^2/(14x6x^2/(16x12x^2/(18x20x^2... (continued fraction), hence Hankel transform is A059332. (End)
a(n) = Product_{p prime} p^(Sum_{k > 0} floor(n/p^k)) by Legendre's formula for the highest power of a prime dividing n!.  Jonathan Sondow, Jul 24 2009
It appears that a(n) = (1/0!) + (1/1!)*n + (3/2!)*n*(n1) + (11/3!)*n*(n1)*(n2) + ... + (b(n)/n!)*n*(n1)*...*2*1, where a(n) = (n+1)! and b(n) = A000255.  Timothy Hopper, Aug 12 2009
a(n) = A_{n}(1) where A_{n}(x) are the Eulerian polynomials.  Peter Luschny, Aug 03 2010
a(n) = n*(2*a(n1)  (n1)*a(n2)), n > 1.  Gary Detlefs, Sep 16 2010
1/a(n) = Sum_{k=1..n+1} (2)^k*(n+k+2)*a(k)/(a(2*k+1)*a(n+1k)).  Groux Roland, Dec 08 2010
a(n) = Product_{p prime, p <= n} p^(Sum_{i >= 1} floor(n/p^i)).
The infinitary analog of this formula is: a(n) = Product_{q terms of A050376 <= n} q^((n)_q), where (n)_q denotes the number of those numbers <= n for which q is an infinitary divisor (for the definition see comment in A037445). (End)
The terms are the denominators of the expansion of sinh(x) + cosh(x).  Arkadiusz Wesolowski, Feb 03 2012
G.f.: 1 / (1  x / (1  x / (1  2*x / (1  2*x / (1  3*x / (1  3*x / ... )))))).  Michael Somos, May 12 2012
G.f. 1 + x/(G(0)x) where G(k) = 1  (k+1)*x/(1  x*(k+2)/G(k+1)); (continued fraction, 2step).  Sergei N. Gladkovskii, Aug 14 2012
G.f.: W(1,1;x)/(W(1,1;x)  x*W(1,2;x)), where W(a,b,x) = 1  a*b*x/1! + a*(a+1)*b*(b+1)*x^2/2!  ... + a*(a+1)*...*(a+n1)*b*(b+1)*...*(b+n1)*x^n/n! + ...; see [A. N. Khovanskii, p. 141 (10.19)].  Sergei N. Gladkovskii, Aug 15 2012
G.f.: A(x) = 1 + x/(G(0)  x) where G(k) = 1 + (k+1)*x  x*(k+2)/G(k+1); (continued fraction).
Let B(x) be the g.f. for A051296, then A(x) = 2  1/B(x). (End)
G.f.: 1 + x*(G(0)  1)/(x1) where G(k) = 1  (2*k+1)/(1x/(x  1/(1  (2*k+2)/(1x/(x  1/G(k+1) ))))); (continued fraction).  Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*(1  G(0))/(sqrt(x)x) where G(k) = 1  (k+1)*sqrt(x)/(1sqrt(x)/(sqrt(x)1/G(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Jan 25 2013
G.f.: 1 + x/G(0) where G(k) = 1  x*(k+2)/( 1  x*(k+1)/G(k+1) ); (continued fraction).  Sergei N. Gladkovskii, Mar 23 2013
a(n) = det(S(i+1, j), 1 <= i, j <=n ), where S(n,k) are Stirling numbers of the second kind.  Mircea Merca, Apr 04 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1  x*(k+1)/(x*(k+1) + 1/G(k+1))); (continued fraction).  Sergei N. Gladkovskii, May 24 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1  1/(1  1/(2*x*(k+1)) + 1/G(k+1))); (continued fraction).  Sergei N. Gladkovskii, May 29 2013
G.f.: G(0), where G(k) = 1 + x*(2*k+1)/(1  x*(2*k+2)/(x*(2*k+2) + 1/G(k+1))); (continued fraction).  Sergei N. Gladkovskii, Jun 07 2013
a(n) = P(n1, floor(n/2)) * floor(n/2)! * (n  (n2)*((n+1) mod 2)), where P(n, k) are the kpermutations of n objects, n > 0.  Wesley Ivan Hurt, Jun 07 2013
G.f.: 1 + x/Q(0), m=+2, where Q(k) = 1  2*x*(2*k+1)  m*x^2*(k+1)*(2*k+1)/( 1  2*x*(2*k+2)  m*x^2*(k+1)*(2*k+3)/Q(k+1) ); (continued fraction).  Sergei N. Gladkovskii, Sep 24 2013
a(n) = round(Sum_{k>=1} log(k)^n/k^2), for n>=1, which is related to the nth derivative of the Riemann zeta function at x=2 as follows: round((1)^n * zeta^(n)(2)). Also see A073002.  Richard R. Forberg, Dec 30 2014
a(n) ~ Sum_{j>=0} j^n/e^j, where e = A001113. When substituting a generic variable for "e" this infinite sum is related to Eulerian polynomials. See A008292. This approximation of n! is within 0.4% at n = 2. See A255169. Accuracy, as a percentage, improves rapidly for larger n.  Richard R. Forberg, Mar 07 2015
a(n) = Product_{k=1..n} (C(n+1, 2)C(k, 2))/(2*k1); see Masanori Ando link.  Michel Marcus, Apr 17 2015
Sum_{n>=0} a(n)/(a(n + 1)*a(n + 2)) = Sum_{n>=0} 1/((n + 2)*(n + 1)^2*a(n)) = 2  exp(1)  gamma + Ei(1) = 0.5996203229953..., where gamma = A001620, Ei(1) = A091725.  Ilya Gutkovskiy, Nov 01 2016
a(2^n) = 2^(2^n  1) * 1!! * 3!! * 7!! * ... * (2^n  1)!!. For example, 16! = 2^15*(1*3)*(1*3*5*7)*(1*3*5*7*9*11*13*15) = 20922789888000.  Peter Bala, Nov 01 2016
a(n) = sum(prod(B)), where the sum is over all subsets B of {1,2,...,n1} and where prod(B) denotes the product of all the elements of set B. If B is a singleton set with element b, then we define prod(B)=b, and, if B is the empty set, we define prod(B) to be 1. For example, a(4)=(1*2*3)+(1*2)+(1*3)+(2*3)+(1)+(2)+(3)+1=24.  Dennis P. Walsh, Oct 23 2017
Sum_{n >= 0} 1/(a(n)*(n+2)) = 1.  Multiplying the denominator by (n+2) in Jaume Oliver Lafont's entry above creates a telescoping sum.  Fred Daniel Kline, Nov 08 2020
O.g.f.: Sum_{k >= 0} k!*x^k = Sum_{k >= 0} (k+y)^k*x^k/(1 + (k+y)*x)^(k+1) for arbitrary y.  Peter Bala, Mar 21 2022
E.g.f.: 1/(1 + LambertW(x*exp(x))) = 1/(1x), see A258773. (1/x)*substitute(z = x*exp(x), z*(d/dz)LambertW(z)) = 1/(1  x). See A075513. Proof: Use the compositional inverse (x*exp(x))^[1] = LambertW(z). See A000169 or A152917, and Richard P. Stanley: Enumerative Combinatorics, vol. 2, p. 37, eq. (5.52).  Wolfdieter Lang, Oct 17 2022
1/a(n) = (e/(2*Pi*n)*Integral_{x=oo..oo} cos(xn*arctan(x))/(1+x^2)^(n/2) dx). Proof: take the real component of Laplace's integral for 1/Gamma(x).
a(n) = Integral_{x=0..1} e^(t)*LerchPhi(1/e, n, t) dt. Proof: use the relationship Gamma(x+1) = Sum_{n >= 0} Integral_{t=n..n+1} e^(t)t^x dt = Sum_{n >= 0} Integral_{t=0..1} e^((t+n))(t+n)^x dt and interchange the order of summation and integration.
Conjecture: a(n) = 1/(2*Pi)*Integral_{x=oo..oo}(n+i*x+1)!/(i*x+1)(n+i*x1)!/(i*x1)dx. (End)
a(n) = floor(b(n)^n / (floor(((2^b(n) + 1) / 2^n)^b(n)) mod 2^b(n))), where b(n) = (n + 1)^(n + 2) = A007778(n+1). Joint work with Mihai Prunescu.  Lorenzo Sauras Altuzarra, Oct 18 2023


EXAMPLE

There are 3! = 1*2*3 = 6 ways to arrange 3 letters {a, b, c}, namely abc, acb, bac, bca, cab, cba.
Let n = 2. Consider permutations of {1, 2, 3}. Fix element 3. There are a(2) = 2 permutations in each of the following cases: (a) 3 belongs to a cycle of length 1 (permutations (1, 2, 3) and (2, 1, 3)); (b) 3 belongs to a cycle of length 2 (permutations (3, 2, 1) and (1, 3, 2)); (c) 3 belongs to a cycle of length 3 (permutations (2, 3, 1) and (3, 1, 2)).  Vladimir Shevelev, May 13 2012
G.f. = 1 + x + 2*x^2 + 6*x^3 + 24*x^4 + 120*x^5 + 720*x^6 + 5040*x^7 + ...


MAPLE

A000142 := n > n!; seq(n!, n=0..20);
spec := [ S, {S=Sequence(Z) }, labeled ]; seq(combstruct[count](spec, size=n), n=0..20);
# Maple program for computing cycle indices of symmetric groups
M:=6: f:=array(0..M): f[0]:=1: print(`n= `, 0); print(f[0]); f[1]:=x[1]: print(`n= `, 1); print(f[1]); for n from 2 to M do f[n]:=expand((1/n)*add( x[l]*f[nl], l=1..n)); print(`n= `, n); print(f[n]); od:
with(combstruct):ZL0:=[S, {S=Set(Cycle(Z, card>0))}, labeled]: seq(count(ZL0, size=n), n=0..20); # Zerinvary Lajos, Sep 26 2007


MATHEMATICA

RecurrenceTable[{a[n] == n*a[n  1], a[0] == 1}, a, {n, 0, 22}] (* Ray Chandler, Jul 30 2015 *)


PROG

(Axiom) [factorial(n) for n in 0..10]
(Magma) a:= func< n  Factorial(n) >; [ a(n) : n in [0..10]];
(Haskell)
a000142 :: (Enum a, Num a, Integral t) => t > a
a000142 n = product [1 .. fromIntegral n]
a000142_list = 1 : zipWith (*) [1..] a000142_list
(Python)
for i in range(1, 1000):
y = i
for j in range(1, i):
y *= i  j
print(y, "\n")
(Python)
import math
for i in range(1, 1000):
math.factorial(i)
print("")
(Scala) (1: BigInt).to(24: BigInt).scanLeft(1: BigInt)(_ * _) // Alonso del Arte, Mar 02 2019


CROSSREFS

Cf. A000165, A001044, A001563, A003422, A009445, A010050, A012245, A033312, A034886, A038507, A047920, A048631.
Factorial base representation: A007623.
Cf. A001013 (multiplicative closure).
For factorials with initial digit d (1 <= d <= 9) see A045509, A045510, A045511, A045516, A045517, A045518, A282021, A045519; A045520, A045521, A045522, A045523, A045524, A045525, A045526, A045527, A045528, A045529.


KEYWORD

core,easy,nonn,nice


AUTHOR



STATUS

approved



