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A326658
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a(n) = 6*floor(n/2) + ceiling((n-1)^2/2).
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2
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1, 0, 7, 8, 17, 20, 31, 36, 49, 56, 71, 80, 97, 108, 127, 140, 161, 176, 199, 216, 241, 260, 287, 308, 337, 360, 391, 416, 449, 476, 511, 540, 577, 608, 647, 680, 721, 756, 799, 836, 881, 920, 967, 1008, 1057, 1100, 1151, 1196, 1249, 1296, 1351, 1400, 1457, 1508, 1567, 1620, 1681
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OFFSET
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0,3
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COMMENTS
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a(n) gives the maximum number of inversions in a permutation on n + 3 symbols consisting of a single n-cycle and 3 fixed points.
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LINKS
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FORMULA
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a(n) = 6*floor(n/2) + ceiling((n-1)^2/2).
a(n) = A326296(3 + n, n) for n > 0.
G.f.: (1 - 2*x + 7*x^2 - 4*x^3) / ((1 - x)^3*(1 + x)).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n >= 4.
a(n) = (-3 + 7*(-1)^n + 8*n + 2*n^2) / 4.
(End)
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MATHEMATICA
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Table[6*Floor[n/2] + Ceiling[(n - 1)^2/2], {n, 80}] (* Wesley Ivan Hurt, Sep 13 2019 *)
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PROG
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(PARI) a(n) = 6*floor(n/2) + ceil((n-1)^2/2) \\ Andrew Howroyd, Sep 23 2019
(PARI) Vec((1 - 2*x + 7*x^2 - 4*x^3) / ((1 - x)^3*(1 + x)) + O(x^40)) \\ Andrew Howroyd, Sep 23 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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