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A382819
Number of Grassmannian permutations on [n] of order dividing 3.
0
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
0,4
LINKS
Kassie Archer and Aaron Geary, Descents in powers of permutations, arXiv:2406.09369 [math.CO], 2024.
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
EXAMPLE
For n = 4 there are 5 Grassmannian permutations whose cubes are the identity permutation: 1234, 3124, 1423, 2314, 1342, so a(4) = 5.
CROSSREFS
Cf. A000325 (Grassmannian permutations), A001470 (permutations of order dividing 3).
Sequence in context: A241515 A070334 A137700 * A325267 A241544 A208716
KEYWORD
nonn,easy
AUTHOR
Aaron Geary, Apr 05 2025
STATUS
approved