A057693 Number of permutations on n letters that have only cycles of length 3 or less.
1, 1, 2, 6, 18, 66, 276, 1212, 5916, 31068, 171576, 1014696, 6319512, 41143896, 281590128, 2007755856, 14871825936, 114577550352, 913508184096, 7526682826848, 64068860545056, 561735627038496, 5068388485760832, 47026385852423616, 447837548306401728
Offset: 0
Keywords
Examples
For example, a(4)=18 since there are 6 permutations with cycles of length 4 to exclude from the 24 permutations on 4 letters, namely (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3) and (1 4 3 2).
References
- Dennis P. Walsh, The number of permutations with only small cycles, preprint.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..300
- P. L. Krapivsky, J. M. Luck, Coverage fluctuations in theater models, arXiv:1902.04365 [cond-mat.stat-mech], 2019.
- I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637, 2013 and J. Int. Seq. 17 (2014) #14.1.1
- R. Petuchovas, Asymptotic analysis of the cyclic structure of permutations, arXiv:1611.02934 [math.CO], p. 6, 2016.
- Dennis P. Walsh, Derivation of the sequence
Programs
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Maple
a:= proc(n) option remember; `if`(n<3, n!, a(n-1) +(n-1)*a(n-2) +(n-1)*(n-2)*a(n-3)) end: seq(a(n), n=0..30); # Alois P. Heinz, Jun 06 2013 -
Mathematica
nn=20;Range[0,nn]!CoefficientList[Series[Exp[ x + x^2/2 + x^3/3],{x,0,nn}],x] (* Geoffrey Critzer, Oct 28 2012 *)
Formula
a(n) = sum(binomial((n, 3 * j) * (3 * j)! * (1/3)^j/j! * sum(binomial(n-3 * j, 2 * k) * (2 * k)! * (1/2)^k/k!, k=0..floor((n-3 * j)/2)), j=0..floor(n/3)))
E.g.f.: exp( x + (x^2)/2 + (x^3)/3 ). Replacing 3 by "length k or less" in the definition of the sequence the E.g.f. is exp( x + (x^2)/2 + ... + (x^k)/k ). - Sharon Sela (sharonsela(AT)hotmail.com), May 16 2002
a(n) = a(n-1)+(n-1)*a(n-2)+(n-1)(n-2)*a(n-3). Generally, for n-permutations that have only cycles of length k or less the recurrence is: a(n)=Sum_i=0...k-1;P(n-1,i)*a(n-i-1) where P(x,i) is the falling factorial. - Geoffrey Critzer, May 23 2009
a(n) ~ n^(2*n/3)*exp(-2*n/3-5/18+5/6*n^(1/3)+1/2*n^(2/3))/sqrt(3) * (1 + 31/(324*n^(1/3)) + 302669/(1049760*n^(2/3))). - Vaclav Kotesovec, Aug 15 2013
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