A208528 Number of permutations of n>1 having exactly 3 points P on the boundary of their bounding square.

0, 4, 16, 72, 384, 2400, 17280, 141120, 1290240, 13063680, 145152000, 1756339200, 22992076800, 323805081600, 4881984307200, 78460462080000, 1339058552832000, 24186745110528000, 460970906812416000, 9245027631071232000, 194632160654131200000

Offset: 2

Author

David Nacin, Feb 27 2012

Keywords

• nonn

Comments

A bounding square for a permutation of n is the square with sides parallel to the coordinate axis containing (1,1) and (n,n), and the set of points P of a permutation p is the set {(k,p(k)) for 0

a(n) is the number of permutations of n symbols that 3-commute with a transposition (see A233440 for definition): a permutation p of {1,...,n} has exactly three points on the boundary of their bounding square if and only if p 3-commutes with transposition (1, n). - Luis Manuel Rivera Martínez, Feb 27 2014

Examples

			a(3) = 4 because {(1,1),(2,3),(3,2)}, {(1,3),(2,1),(3,2)}, {(1,2),(2,3),(3,1)} and {(1,2),(2,1),(3,3)} each have three points on the bounding square.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 2..200
• Emeric Deutsch, Permutations and their bounding squares, Math Magazine, 85(1) (2012), p. 63.
• Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.

Crossrefs

Cf. A098916, A208529.

Cf. A001620, A091725, A099285.

Programs

Mathematica

Table[(4n-8)(n-2)!, {n, 2, 10}]

Python

import math
def a(n):
    return (4*n-8)*math.factorial(n-2)

Formula

a(n) = (4*n-8) * (n-2)!

From Amiram Eldar, May 17 2022: (Start)

Sum_{n>=3} 1/a(n) = (Ei(1) - gamma)/4, where Ei(1) = A091725 and gamma = A001620.

Sum_{n>=3} (-1)^(n+1)/a(n) = (gamma - Ei(-1))/4, where Ei(-1) = -A099285. (End)

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