A137593 -id:A137593 - cp's OEIS frontend

A007489 a(n) = Sum_{k=1..n} k!.

0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113, 4037913, 43954713, 522956313, 6749977113, 93928268313, 1401602636313, 22324392524313, 378011820620313, 6780385526348313, 128425485935180313, 2561327494111820313, 53652269665821260313, 1177652997443428940313

Offset: 0

N. J. A. Sloane, Robert G. Wilson v

Equals row sums of triangle A143122 starting (1, 3, 9, 33, ...). - Gary W. Adamson, Jul 26 2008
a(n) for n>=4 is never a perfect square. - Alexander R. Povolotsky, Oct 16 2008
Number of cycles that can be written in the form (j,j+1,j+2,...), in all permutations of {1,2,...,n}. Example: a(3)=9 because in (1)(2)(3), (1)(23), (12)(3), (13)(2), (123), (132) we have 3+2+2+1+1+0=9 such cycles. - Emeric Deutsch, Jul 14 2009
Conjectured to be the length of the shortest word over {1,...,n} that contains each of the n! permutations as a factor (cf. A180632) [see Johnston]. - N. J. A. Sloane, May 25 2013
The above conjecture has been disproven for n>=6. See A180632 and the Houston 2014 reference. - Dmitry Kamenetsky, Mar 07 2016
a(n) is also the number of compositions of n if cardinal values do not matter but ordinal rankings do. Since cardinal values do not matter, a sequence of k summands summing to n can be represented as (s(1),...,s(k)), where the s's are positive integers and the numbers in parentheses are the initial ordinal rankings. The number of compositions of these summands are equal to k!, with k ranging from 1 to n. - Gregory L. Simay, Jul 31 2016
When the numbers denote finite permutations (as row numbers of A055089) these are the circular shifts to the left. Compare array A211370 for circular shifts to the left in a broader sense. Compare sequence A001563 for circular shifts to the right. - Tilman Piesk, Apr 29 2017
Since a(n) = (1!+2!+3!+...+n!) = 3(1+3!/3+4!/3+...+n!/3) is a multiple of 3 for n>2, the only prime in this sequence is a(2) = 3. - Eric W. Weisstein, Jul 15 2017
Generalization of 2nd comment: a(n) for n>=4 is never a perfect power (A007916) (Chentzov link). - Bernard Schott, Jan 26 2023

			a(4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33. - _Michael B. Porter_, Aug 03 2016

R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Section B44, Springer 2010.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Carauleanu Marc, Table of n, a(n) for n = 0..212 (first 100 terms from T. D. Noe)
N. N. Chentzov, D. O. Shklarsky, and I. M. Yaglom, The USSR Olympiad Problem Book, Selected Problems and Theorems of Elementary Mathematics, problem 115, pp. 28 and 201-202, Dover publications, Inc., New York, 1993.
R. K. Guy, Letter to N. J. A. Sloane, 1987
Robin Houston, Tackling the Minimal Superpermutation Problem, arXiv:1408.5108 [math.CO], 2014.
Nathaniel Johnston, The minimal superpermutation problem (2013)
Nathaniel Johnston, Non-uniqueness of minimal superpermutations, Discrete Math. 313 (2013), no. 14, 1553--1557. MR3047396
S. Legendre and P. Paclet, On the Permutations Generated by Cyclic Shift , J. Int. Seq. 14 (2011) # 11.3.2.
Hisanori Mishima, Factorizations of many number sequences
Hisanori Mishima, Factorizations of many number sequences
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
Eric Weisstein's World of Mathematics, Factorial
Eric Weisstein's World of Mathematics, Left Factorial
G. Xiao, Sigma Server, Operate on "n!"
Jun Yan, Results on pattern avoidance in parking functions, arXiv:2404.07958 [math.CO], 2024. See p. 5.
Index entries for sequences related to factorial numbers

Equals A003422(n+1) - 1.
Cf. A000142, A000670, A001597, A007916, A137593, A143122, A161128, A180632.
Column k=0 of A120695.

List([1..20],n->Sum([1..n],Factorial)); # Muniru A Asiru, Jan 31 2018

a007489 n = a007489_list !! n
a007489_list = scanl (+) 0 $ tail a000142_list
-- Reinhard Zumkeller, Aug 29 2014

[0] cat [&+[Factorial(i): i in [1..n]]: n in [1..25]]; // Vincenzo Librandi, Sep 02 2016

A007489 := proc(n) local i; add(i!,i=1..n); end proc;

FoldList[Plus, 0, (Range@ 21)! ] (* Robert G. Wilson v, Sep 21 2007 *)
Table[Sum[i!, {i, 1, n}], {n, 0, 21}] (* Zerinvary Lajos, Jul 12 2009 *)
Accumulate[Range[50]!]  (* Harvey P. Dale, Apr 30 2011 *)
Table[Plus@@(Range[n]!), {n, 20}] (* Alonso del Arte, Jul 18 2011 *)

a(n)=sum(k=1,n,k!) \\ Charles R Greathouse IV, Jul 25 2011

a(n) = Sum_{k=1..n} P(n, k)/C(n, k). - Ross La Haye, Sep 21 2004
a(n) = 3*A056199(n) for n>=2. - Philippe Deléham, Feb 10 2007
a(n) = !(n+1)-1=A003422(n+1)-1. - Artur Jasinski, Nov 08 2007 [corrected by Werner Schulte, Oct 20 2021]
Starting (1, 3, 9, 33, 153, ...), = row sums of triangle A137593 - Gary W. Adamson, Jan 28 2008
a(n) = a(n-1) + n! for n >= 1. - Jaroslav Krizek, Jun 16 2009
E.g.f. A(x) satisfies to the differential equation A'(x)=A(x)+x/(1-x)^2+1. - Vladimir Kruchinin, Jan 22 2011
a(0)=0, a(1)=1, a(n) = (n+1)*a(n-1)-n*a(n-2). - Sergei N. Gladkovskii, Jul 05 2012
G.f.: W(0)*x/(2-2*x) , where W(k) = 1 + 1/( 1 - x*(k+2)/( x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
G.f.: x /(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
E.g.f.: exp(x-1)*(Ei(1) - Ei(1-x)) - exp(x) + 1/(1 - x), where Ei(x) is the exponential integral. - Ilya Gutkovskiy, Nov 27 2016
a(n) = sqrt(a(n-1)*a(n+1)-a(n-2)*n*n!), n >= 2. - Gary Detlefs, Oct 26 2020
a(n) ~ n!. - Ridouane Oudra, Jun 11 2025

A136717 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 3-excedances. Equivalently, the number of permutations of the set {3,6,9,...,3n} with k excedances.

Original entry on oeis.org

1, 0, 2, 0, 2, 4, 0, 0, 12, 12, 0, 0, 0, 72, 48, 0, 0, 0, 72, 456, 192, 0, 0, 0, 0, 960, 3120, 960, 0, 0, 0, 0, 0, 10800, 23760, 5760, 0, 0, 0, 0, 0, 10800, 133920, 183600, 34560, 0, 0, 0, 0, 0, 0, 241920, 1572480, 1572480, 241920

Offset: 1

Peter Bala, Jan 23 2008

A permutation (p(1),p(2),...,p(n)) in the symmetric group S_n has a multiplicative 3-excedance at position i, 1 <= i <= n, if 3*p(i) > i. The (n,k)-th entry in this array gives the number of permutations in S_n with k multiplicative 3-excedances.
Compare with A008292, the triangle of Eulerian numbers, which enumerates permutations by the usual excedance number and with A136715, which enumerates permutations by multiplicative 2-excedances.
Let e(p)= |{i | 1 <= i < = n, 3*p(i) > i}| denote the number of multiplicative 3-excedances in the permutation p. This 3-excedance statistic e(p) on the symmetric group S_n is related to a descent statistic as follows. Define a permutation p in S_n to have a multiplicative 3-descent at position i, 1 <= i <= n-1, if p(i) is divisible by 3 and p(i) > p(i+1). For example, the permutation (4,1,6,5,3,2) in S_6 has two multiplicative 3-descents (at position 3 and position 5). Array A136718 records the number of permutations of S_n with k multiplicative 3-descents.
Let d(p) = |{i | 1 <= i <= n-1, p(i) is divisible by 3 & p(i) > p(i+1)}| count the multiplicative 3-descents in the permutation p. Comparison of the recursion relations for the entries of this table with the recursion relations for the entries of A136718 shows that e(p) and d(p) are related by sum {p in S_n} x^e(p) = x^ceiling(2*n/3)* sum {p in S_n} x^d(p). Thus the shifted multiplicative 3-excedance statistic e(p) - ceiling(2*n/3) and the multiplicative 3-descent statistic d(p) are equidistributed on the symmetric group S_n.
(Note: There is also an additive r-excedance statistic on the symmetric group, due to Riordan, where the condition r*p(i) > i is replaced by p(i) >= i+r. See A120434 for the r = 2 case.)
An alternative interpretation of this array is as follows: Let T_n denote the set {3,6,9,...,3n} and let now p denote a bijection p:T_n -> T_n. We say the permutation p has an excedance at position i, 1 <= i <= n, if p(3i) > i. For example, if we represent p in one line notation by the vector (p(3),p(6),...,p(3n)), then the permutation (9,18,3,12,15,6) of T_6 has four excedances in total (at positions 1, 2, 4 and 5). This array gives the number of permutations of the set T_n with k excedances. This is the viewpoint taken in [Jansson].
A137593 = A000012 * this triangular matrix. A137594 = A007318 * this triangular matrix. - Gary W. Adamson, Jan 28 2008

			T(3,3) = 4; the four permutations in S_3 with three multiplicative 3-excedances are (1,2,3), (1,3,2), (2,1,3) and (3,1,2). The remaining two permutations (2,3,1) and (3,2,1) each have two multiplicative 3-excedances.
Equivalently, the four permutations of the set {3,6,9} with 3 excedances are (3,6,9), (3,9,6), (6,3,9) and (9,3,6). The remaining two permutations (6,9,3) and (9,6,3) each have 2 excedances.
Triangle starts
n\k|..1....2....3....4....5....6
---+----------------------------
1..|..1
2..|..0....2
3..|..0....2....4
4..|..0....0...12...12
5..|..0....0....0...72...48
6..|..0....0....0...72..456..192

Fredrik Jansson, Variations on the excedance statistic in permutations, Thesis, Stockholm University, 2006.
S. Kitaev and J. Remmel, Classifying descents according to equivalence mod k, arXiv:math/0604455 [math.CO], 2006.

Cf. A000142 (row sums), A008292, A136718, A136715.

Cf. A137593, A137594.

Recurrence relations (apply proposition 2.2 of [Jansson]):
T(3n,k) = (k+1-2n)*T(3n-1,k) + (5n-k)*T(3n-1,k-1) for n >= 1;
T(3n+1,k) = (k-2n)*T(3n,k) + (5n+2-k)*T(3n,k-1) for n >= 0;
T(3n+2,k) = (k-1-2n)*T(3n+1,k) + (5n+4-k)*T(3n+1,k-1) for n >= 0.
Boundary conditions: T(0,k) = 0 all k; T(n,0) = 0 all n; T(1,1) = 1.
Define the shifted row polynomials R(n,x) by
R(n,x) := x^(1+floor(n/3)-n)* sum {k = n-floor(n/3)..n} T(n,k)*x^k.
The first few values are R(1,x) = x, R(2,x) = 2x, R(3,x) = 2x+4x^2 and R(4,x) = 12x+12x^2.
The recurrence relations yield the identities:
x*d/dx(1/x*R(3n,x)/(1-x)^(3n+1)) = R(3n+1,x)/(1-x)^(3n+2);
x*d/dx(1/x*R(3n+1,x)/(1-x)^(3n+2)) = R(3n+2,x)/(1-x)^(3n+3);
x*d/dx(R(3n+2,x)/(1-x)^(3n+3)) = R(3n+3,x)/(1-x)^(3n+4).
An easy induction argument now gives the Taylor series expansions:
R(3n,x)/(1-x)^(3n+1) = sum {m = 1..inf} m^2*(m+1)*(m+2)^2*(m+3)*...* (m+2n-2)^2*(m+2n-1)*x^m;
R(3n+1,x)/(1-x)^(3n+2) = sum {m = 1..inf} m*((m+1)^2*(m+2)*(m+3)^2*(m+4) *...*(m+2n-1)^2*(m+2n))*x^m.
R(3n+2,x)/(1-x)^(3n+3) = sum {m = 1..inf} m*((m+1)*(m+2)^2*(m+3)*(m+4)^2 *...*(m+2n-1)*(m+2n)^2)*(m+2n+1)*x^m.
For example, for row 6 (n = 2) we have the expansion (72x+456x^2+192x^3)/(1-x)^7 = 72x + 960x^2 + 5400x^3 + ... = (1^2*2*3^2*4)*x + (2^2*3*4^2*5)*x^2 + (3^2*4*5^2*6)*x^3 + ... .

cp's OEIS Frontend

A007489 a(n) = Sum_{k=1..n} k!.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Formula