A174072 Number of permutations of length n with no consecutive triples i,i+2,i+4.
1, 1, 2, 6, 24, 114, 674, 4714, 37754, 340404, 3412176, 37631268, 452745470, 5900431012, 82802497682, 1244815252434, 19958707407096, 339960096280062, 6130407887839754, 116675071758609742, 2337186717333367706, 49153251967227002616, 1082860432463176004544
Offset: 0
Keywords
Examples
For n=5 (0,2,4,1,3) is an example of a permutation with an i,i+2,i+4 triple. If we look at 0,2,4 as a block, then we have 3! ways to permute the triple with the remaining 1 & 3. Hence a(5) = 5! - 3! = 114.
Links
- Max Alekseyev, Table of n, a(n) for n = 0..100
- Wayne M. Dymacek, Isaac Lambert and Kyle Parsons, Arithmetic Progressions in Permutations, 2012. [broken link]
Programs
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Maple
b:= proc(s, x, y) option remember; `if`(s={}, 1, add( `if`(x=0 or x-y<>2 or y-j<>2, b(s minus {j}, y, j), 0), j=s)) end: a:= n-> b({$1..n}, 0$2): seq(a(n), n=0..14); # Alois P. Heinz, Apr 13 2021
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Mathematica
b[s_, x_, y_] := b[s, x, y] = If[s == {}, 1, Sum[ If[x == 0 || x - y != 2 || y - j != 2, b[s ~Complement~ {j}, y, j], 0], {j, s}]]; a[n_] := b[Range[n], 0, 0]; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 02 2022, after Alois P. Heinz *)
Extensions
a(0)-a(4) and a(10)-a(11) moved from a duplicate entry based on the Dymacek et al. paper on Apr 13 2021
a(12)-a(22) from Alois P. Heinz, Apr 13 2021
Comments