A326658 a(n) = 6*floor(n/2) + ceiling((n-1)^2/2).

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

Offset: 0

Author

M. Ryan Julian Jr., Sep 12 2019

Keywords

• nonn
• easy

Comments

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.

Sequence is a diagonal of A326296.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Diagonal of A326296.

Programs

Mathematica

Table[6*Floor[n/2] + Ceiling[(n - 1)^2/2], {n, 80}] (* Wesley Ivan Hurt, Sep 13 2019 *)

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

Formula

a(n) = 6*floor(n/2) + ceiling((n-1)^2/2).

a(n) = A326296(3 + n, n) for n > 0.

From Colin Barker, Sep 13 2019: (Start)

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)