A262480 Number of trivial c-Wilf equivalence classes in the symmetric group S_n.

1, 1, 1, 2, 8, 32, 192, 1272, 10176, 90816, 908160, 9980160, 119761920, 1556766720, 21794734080, 326918753280, 5230700052480, 88921859604480, 1600593472880640, 30411275148656640, 608225502973132800, 12772735543856332800, 281000181964839321600, 6463004184741681561600, 155112100433800357478400, 3877802510833236993638400

Offset

0,4

Comments

A permutation pattern is c-Wilf equivalent to its complement and reverse and therefore we can get trivial equivalence classes based on complement and reverse. a(3) = 2 because there are two trivial c-Wilf equivalence classes {123, 321} and {132, 231, 213, 321}.

a(n) is an upper bound of c-Wilf equivalence classes in the symmetric group S_n.

The numbers of c-Wilf equivalence classes in S_n are still unknown for large n. Up to 6, they are 1, 1, 2, 7, 25, 92.

Links

Table of n, a(n) for n=0..25.

Christian Bean, Émile Nadeau, Jay Pantone, and Henning Ulfarsson, Permutations avoiding bipartite partially ordered patterns have a regular insertion encoding, arXiv:2312.07716 [math.CO], 2023.

Adrian Duane and Jeffrey Remmel, Minimal overlapping patterns in colored permutations, Electron. J. Combin. 18 (2011) #P25.

Brian Koichi Nakamura, Computational methods in permutation patterns, Ph. D. dissertation at Rutgers University, 2013.

Formula

a(0) = a(1) = 1, a(2*n) = ((2*n)!+(2*n)!!)/4, a(2*n+1) = ((2*n+1)!+(2*n)!!)/4, for n >= 1.

D-finite with recurrence: -(n-3)*a(n) + n*(n-3)*a(n-1) + (n-1)^2*a(n-2) - (n-2)*(n-1)^2*a(n-3) = 0 for n >= 5. - Georg Fischer, Nov 25 2022

Maple

a := proc(n) option remember; if n < 5 then return [1, 1, 1, 2, 8][n+1] fi;
(n*(n-3)*a(n-1) + (n-1)^2*a(n-2) - (n-2)*(n-1)^2*a(n-3))/(n-3) end:
seq(a(n), n = 0..25); # Peter Luschny, Nov 25 2022

Mathematica

Join[{1, 1}, RecurrenceTable[{-(n-3)*a[n] + n*(n-3)*a[n-1] + (n-1)^2*a[n-2] - (n-2)*(n-1)^2*a[n-3] == 0, a[2]==1, a[3]==2, a[4]==8}, a, {n, 2, 25}]] (* Georg Fischer, Nov 25 2022 *)

Prog

(PARI) a(n) = if(n<=1, 1, if (n%2, n=(n-1)/2; ((2*n+1)!+2^n*n!)/4, n=n/2; ((2*n)!+2^n*n!)/4)); \\ Michel Marcus, Nov 25 2022

Crossrefs

Cf. A000165, A000142.

Sequence in context: A294506 A206303 A048855 * A062797 A369645 A134751

Adjacent sequences: A262477 A262478 A262479 * A262481 A262482 A262483

Keyword

nonn

Author

Ran Pan, Sep 24 2015

Status

approved