

A262480


Number of trivial cWilf equivalence classes in the symmetric group S_n.


4



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
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OFFSET

0,4


COMMENTS

A permutation pattern is cWilf 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 cWilf equivalence classes {123, 321} and {132, 231, 213, 321}.
a(n) is an upper bound of cWilf equivalence classes in the symmetric group S_n.
The numbers of cWilf equivalence classes in S_n are still unknown for large n. Up to 6, they are 1, 1, 2, 7, 25, 92.


LINKS



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.
Dfinite with recurrence: (n3)*a(n) + n*(n3)*a(n1) + (n1)^2*a(n2)  (n2)*(n1)^2*a(n3) = 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*(n3)*a(n1) + (n1)^2*a(n2)  (n2)*(n1)^2*a(n3))/(n3) end:


MATHEMATICA

Join[{1, 1}, RecurrenceTable[{(n3)*a[n] + n*(n3)*a[n1] + (n1)^2*a[n2]  (n2)*(n1)^2*a[n3] == 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=(n1)/2; ((2*n+1)!+2^n*n!)/4, n=n/2; ((2*n)!+2^n*n!)/4)); \\ Michel Marcus, Nov 25 2022


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



