A033312 a(n) = n! - 1.

0, 0, 1, 5, 23, 119, 719, 5039, 40319, 362879, 3628799, 39916799, 479001599, 6227020799, 87178291199, 1307674367999, 20922789887999, 355687428095999, 6402373705727999, 121645100408831999, 2432902008176639999, 51090942171709439999, 1124000727777607679999

Offset: 0

Author

N. J. A. Sloane. This sequence appeared in the 1973 "Handbook", but was then dropped from the database. Resubmitted by Eric W. Weisstein. Entry revised by N. J. A. Sloane, Jun 12 2012

Keywords

• nonn
• easy
• changed

Comments

a(n) gives the index number in any table of permutations of the entry in which the last n + 1 items are reversed. - Eugene McDonnell (eemcd(AT)mac.com), Dec 03 2004

a(n), n >= 1, has the factorial representation [n - 1, n - 2, ..., 1, 0]. The (unique) factorial representation of a number m from {0, 1, ... n! - 1} is m = sum(m_j(n)*j!, j = 0 .. n - 1) with m_j(n) from {0, 1, .., j}, n>=1. This is encoded as [m_{n-1},m_{n-2},...,m+1,m_0] with m_0=0. This can be interpreted as (D. N.) Lehmer code for the lexicographic rank of permutations of the symmetric group S_n (see the W. Lang link under A136663). The Lehmer code [n - 1, n - 2, ..., 1, 0] stands for the permutation [n, n - 1, ..., 1] (the last in lexicographic order). - Wolfdieter Lang, May 21 2008

For n >= 3: a(n) = numbers m for which there is one iteration {floor (r / k)} for k = n, n - 1, n - 2, ... 2 with property r mod k = k - 1 starting at r = m. For n = 5: a(5) = 119; floor (119 / 5) = 23, 119 mod 5 = 4; floor (23 / 4) = 5, 23 mod 4 = 3; floor (5 / 3) = 1, 5 mod 3 = 2; floor (1 / 2) = 0; 1 mod 2 = 1. - Jaroslav Krizek, Jan 23 2010

For n = 4, define the sum of all possible products of 1, 2, 3, 4 to be 1 + 2 + 3 + 4 add 1*2 + 1*3 + 1*4 add 2*3 + 2*4 + 3*4 add 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 add 1*2*3*4. The sum of this is 119 = (4 + 1)! - 1. For n = 5 I get the sum 719 = (5 + 1)! - 1. The proof for the general case seems to follow by induction. - J. M. Bergot, Jan 10 2011

Examples

			G.f. = x^2 + 5*x^3 + 23*x^4 + 119*x^5 + 719*x^6 + 5039*x^7 + 40319*x^8 + ...
		

References

• Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 181, p. 92.
• Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 6, 1969, p. 3, 1993.
• Problem 598, J. Rec. Math., 11 (1978), 68-69.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..300
• Jonathan Beagley and Lara Pudwell, Colorful Tilings and Permutations, Journal of Integer Sequences, Vol. 24 (2021), Article 21.10.4.
• The IMO Compendium, Problem 6, 1st Canadian Mathematical Olympiad 1969.
• Stéphane Legendre and Philippe Paclet, On the Permutations Generated by Cyclic Shift , J. Int. Seq. 14 (2011) # 11.3.2.
• Gerard P. Michon, Wilson's Theorem.
• Hisanori Mishima, Factorizations of many number sequences.
• Hisanori Mishima, Factorizations of many number sequences.
• Michael Penn, Make it look like a simple calculus problem., YouTube video, 2021.
• Andrew Walker, Factors of n! +- 1.
• Eric Weisstein's World of Mathematics, Factorial.
• Eric Weisstein's World of Mathematics, Permutation Pattern.
• Index entries for sequences related to factorial base representation.
• Index entries for sequences related to factorial numbers.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A000142, A001563 (first differences), A002582, A002982, A038507 (factorizations), A054415, A056110, A331373.

Row sums of A008291.

Programs

Magma

[Factorial(n)-1: n in [0..25]]; // Vincenzo Librandi, Jul 20 2011

Mathematica

FoldList[#1*#2 + #2 - 1 &, 0, Range[19]] (* Robert G. Wilson v, Jul 07 2012 *)
Range[0, 19]! - 1 (* Alonso del Arte, Jan 24 2013 *)

Maxima

A033312(n):= n!-1$
makelist(A033312(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

PARI

a(n)=n!-1 \\ Charles R Greathouse IV, Jul 19 2011

Formula

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

a(n) = a(n - 1)*(n - 1) + a(n - 1) + n - 1, a(0) = 0. - Reinhard Zumkeller, Feb 03 2003

a(0) = a(1) = 0, a(n) = a(n - 1) * n + (n - 1) for n >= 2. - Jaroslav Krizek, Jan 23 2010

E.g.f.: 1/(1 - x) - exp(x). - Sergei N. Gladkovskii, Jun 29 2012

0 = 1 + a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(+3 + a(n+1)) + a(n+2)*(-1) for n>=0. - Michael Somos, Feb 24 2017

Sum_{n>=2} 1/a(n) = A331373. - Amiram Eldar, Nov 11 2020