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The difference between number of even and number of odd Grassmannian permutations of size n.
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%I #30 Jul 30 2022 11:44:35

%S 1,1,0,1,0,3,2,9,8,23,22,53,52,115,114,241,240,495,494,1005,1004,2027,

%T 2026,4073,4072,8167,8166,16357,16356,32739,32738,65505,65504,131039,

%U 131038,262109,262108,524251,524250,1048537,1048536,2097111,2097110,4194261,4194260

%N The difference between number of even and number of odd Grassmannian permutations of size n.

%C A permutation is Grassmann if it has at most one descent. A closed-form formula was proved by J. B. Gil and J. A. Tomasko.

%H Juan B. Gil and Jessica A. Tomasko, <a href="https://arxiv.org/abs/2112.03338">Restricted Grassmannian permutations</a>, arXiv:2112.03338 [math.CO], 2021.

%H Juan B. Gil and Jessica A. Tomasko, <a href="https://doi.org/10.54550/ECA2022V2S4PP6">Restricted Grassmannian permutations</a>, Enum. Combin. Appl. 2 (2022), no. 4, Article #S4PP6.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-4,2).

%F a(n) = 2^(1+floor((n-1)/2))-n.

%F From _Alois P. Heinz_, Jul 28 2022: (Start)

%F G.f.: -(4*x^3-3*x^2-x+1)/((2*x^2-1)*(x-1)^2).

%F a(n) = A000325(n) - A233411(n) = A060546(n) - n = 2^ceiling(n/2) - n.

%F a(n) = A000325(n) - 2*A032085(n) = A000325(n) - 2*A122746(n-2) for n>=2. (End)

%e For n=3, 123, 231, 312 are even Grassmann permutations, and 132, 213 are the odd ones. Hence a(3) = 1.

%t Table[2^Floor[1 + (n - 1)/2] - n, {n, 1, 80}]

%Y Cf. A000325, A001710, A032085, A060546, A122746, A233411.

%Y Bisections give: A005803 (even part), A183155 (odd part).

%K nonn,easy

%O 0,6

%A _Per W. Alexandersson_, Jul 28 2022