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A382819 Number of Grassmannian permutations on [n] of order dividing 3.

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%I A382819 #24 Apr 12 2025 16:29:04
%S A382819 1,1,1,3,5,7,12,17,22,31,40,49,63,77,91,111,131,151,178,205,232,267,
%T A382819 302,337,381,425,469,523,577,631,696,761,826,903,980,1057,1147,1237,
%U A382819 1327,1431,1535,1639,1758,1877,1996,2131,2266,2401,2553,2705,2857,3027,3197,3367,3556,3745,3934
%N A382819 Number of Grassmannian permutations on [n] of order dividing 3.
%H A382819 Kassie Archer and Aaron Geary, <a href="https://arXiv.org/abs/2406.09369">Descents in powers of permutations</a>, arXiv:2406.09369 [math.CO], 2024.
%H A382819 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,2,-4,2,-1,2,-1).
%F A382819 a(n) = binomial(floor(n/3)+3,3) + binomial(floor((n-1)/3)+3,3) + binomial(floor((n-2)/3)+3,3) - n.
%F A382819 a(n) = 2*a(n-1) - a(n-2) + n/3 + 1 for n mod 3 = 0
%F A382819 a(n) = 2*a(n-1) - a(n-2) for n mod 3 <> 0.
%F A382819 a(n) ~ n^3/54. - _Stefano Spezia_, Apr 06 2025
%F A382819 G.f.: -(x^7-2*x^4+x-1)/((x^2+x+1)^2*(x-1)^4). - _Alois P. Heinz_, Apr 06 2025
%e A382819 For n = 4 there are 5 Grassmannian permutations whose cubes are the identity permutation: 1234, 3124, 1423, 2314, 1342, so a(4) = 5.
%Y A382819 Cf. A000325 (Grassmannian permutations), A001470 (permutations of order dividing 3).
%K A382819 nonn,easy
%O A382819 0,4
%A A382819 _Aaron Geary_, Apr 05 2025