A165961 Number of circular permutations of length n without 3-sequences.
1, 5, 20, 102, 627, 4461, 36155, 328849, 3317272, 36757822, 443846693, 5800991345, 81593004021, 1228906816941, 19733699436636, 336554404751966, 6075478765948135, 115734570482611885, 2320148441078578447, 48827637296350480457, 1076313671861962141616
Offset: 3
Keywords
Examples
For n=4 the a(4)=5 solutions are (0,1,3,2), (0,2,1,3), (0,2,3,1), (0,3,1,2) and (0,3,2,1).
References
- Wayne M. Dymacek, Isaac Lambert and Kyle Parsons, Arithmetic Progressions in Permutations, http://math.ku.edu/~ilambert/CN.pdf, 2012. - From N. J. A. Sloane, Sep 15 2012 [broken link]
Links
- Michael De Vlieger, Table of n, a(n) for n = 3..450
- Wayne M. Dymacek and Isaac Lambert, Permutations Avoiding Runs of i, i+1, i+2 or i, i-1, i-2, Journal of Integer Sequences, Vol. 14 (2011), Article 11.1.6.
- Kyle Parsons, Arithmetic progressions in permutations, Thesis, 2011.
Crossrefs
A column of A216718. - N. J. A. Sloane, Sep 15 2012
Programs
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Maple
d[0] := 1: for n to 51 do d[n] := n*d[n-1]+(-1)^n end do: a := proc (n) options operator, arrow: sum(binomial(n-k, k)*d[n-k-1], k = 0 .. floor((1/2)*n)) end proc: seq(a(n), n = 3 .. 23); # Emeric Deutsch, Sep 07 2010
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Mathematica
a[n_] := Sum[Binomial[n-k, k] Subfactorial[n-k-1], {k, 0, n/2}]; a /@ Range[3, 21] (* Jean-François Alcover, Oct 29 2019 *)
Formula
Let b(n) be the sequence A002628. Then for n > 5, this sequence satisfies a(n) = b(n-1) - b(n-3) + a(n-3).
a(n) = Sum_{k=0..n/2} binomial(n-k,k)*d(n-k-1), where d(j)=A000166(j) are the derangement numbers. - Emeric Deutsch, Sep 07 2010
Extensions
More terms from Emeric Deutsch, Sep 07 2010
Edited by N. J. A. Sloane, Apr 04 2011
Comments