A326657 a(n) = 4*floor(n/2) + ceiling((n-1)^2/2).

1, 0, 5, 6, 13, 16, 25, 30, 41, 48, 61, 70, 85, 96, 113, 126, 145, 160, 181, 198, 221, 240, 265, 286, 313, 336, 365, 390, 421, 448, 481, 510, 545, 576, 613, 646, 685, 720, 761, 798, 841, 880, 925, 966, 1013, 1056, 1105, 1150, 1201, 1248, 1301, 1350, 1405, 1456, 1513, 1566, 1625, 1680

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 + 2 symbols consisting of a single n-cycle and 2 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

PARI

a(n) = 4*floor(n/2) + ceil((n-1)^2/2) \\ Andrew Howroyd, Sep 23 2019

PARI

Vec((1 - 2*x + 5*x^2 - 2*x^3) / ((1 - x)^3*(1 + x)) + O(x^40)) \\ Andrew Howroyd, Sep 23 2019

Formula

a(n) = 4*floor(n/2) + ceiling((n-1)^2/2).

a(n) = A326296(2 + n, n) for n > 0.

From Colin Barker, Sep 15 2019: (Start)

G.f.: (1 - 2*x + 5*x^2 - 2*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) = (-1 + 5*(-1)^n + 4*n + 2*n^2) / 4.

(End)

E.g.f.: (1/4)*(5*exp(-x) + exp(x)*(-1 + 6*x + 2*x^2)). - Stefano Spezia, Sep 16 2019 after Colin Barker