A382819 Number of Grassmannian permutations on [n] of order dividing 3.
1, 1, 1, 3, 5, 7, 12, 17, 22, 31, 40, 49, 63, 77, 91, 111, 131, 151, 178, 205, 232, 267, 302, 337, 381, 425, 469, 523, 577, 631, 696, 761, 826, 903, 980, 1057, 1147, 1237, 1327, 1431, 1535, 1639, 1758, 1877, 1996, 2131, 2266, 2401, 2553, 2705, 2857, 3027, 3197, 3367, 3556, 3745, 3934
Offset: 0
Examples
For n = 4 there are 5 Grassmannian permutations whose cubes are the identity permutation: 1234, 3124, 1423, 2314, 1342, so a(4) = 5.
Links
- Kassie Archer and Aaron Geary, Descents in powers of permutations, arXiv:2406.09369 [math.CO], 2024.
- Index entries for linear recurrences with constant coefficients, signature (2,-1,2,-4,2,-1,2,-1).
Formula
a(n) = binomial(floor(n/3)+3,3) + binomial(floor((n-1)/3)+3,3) + binomial(floor((n-2)/3)+3,3) - n.
a(n) = 2*a(n-1) - a(n-2) + n/3 + 1 for n mod 3 = 0
a(n) = 2*a(n-1) - a(n-2) for n mod 3 <> 0.
a(n) ~ n^3/54. - Stefano Spezia, Apr 06 2025
G.f.: -(x^7-2*x^4+x-1)/((x^2+x+1)^2*(x-1)^4). - Alois P. Heinz, Apr 06 2025