M. Ryan Julian Jr. - cp's OEIS frontend

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

M. Ryan Julian Jr., Sep 12 2019

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.

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

Diagonal of A326296.

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

a(n) = 6*floor(n/2) + ceil((n-1)^2/2) \\ Andrew Howroyd, Sep 23 2019

Vec((1 - 2*x + 7*x^2 - 4*x^3) / ((1 - x)^3*(1 + x)) + O(x^40)) \\ Andrew Howroyd, Sep 23 2019

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)

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

Original entry on oeis.org

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

M. Ryan Julian Jr., Sep 12 2019

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.

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

Diagonal of A326296.

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

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

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

A326296 Triangle of numbers T(n,k) = 2floor(k/2)(n-k) + ceiling((k-1)^2/2), 1<=k<=n.

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 0, 5, 4, 5, 0, 7, 6, 9, 8, 0, 9, 8, 13, 12, 13, 0, 11, 10, 17, 16, 19, 18, 0, 13, 12, 21, 20, 25, 24, 25, 0, 15, 14, 25, 24, 31, 30, 33, 32, 0, 17, 16, 29, 28, 37, 36, 41, 40, 41, 0, 19, 18, 33, 32, 43, 42, 49, 48, 51, 50, 0, 21, 20, 37, 36, 49, 48, 57, 56, 61, 60, 61

Offset: 1

M. Ryan Julian Jr., Sep 10 2019

T(n,k) gives the maximum number of inversions in a permutation on n symbols containing a single k-cycle and (n-k) other fixed points.
T(n,n) = A000982(n).
T(n,n-1) = A097063(n).

			Triangle begins:
0;
0, 1;
0, 3, 2;
0, 5, 4, 5;
0, 7, 6, 9, 8;
0, 9, 8, 13, 12, 13;
0, 11, 10, 17, 16, 19, 18;
0, 13, 12, 21, 20, 25, 24, 25;
0, 15, 14, 25, 24, 31, 30, 33, 32;
0, 17, 16, 29, 28, 37, 36, 41, 40, 41;
0, 19, 18, 33, 32, 43, 42, 49, 48, 51, 50;
0, 21, 20, 37, 36, 49, 48, 57, 56, 61, 60, 61;
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Diagonals give A000982, A097063, A326657, A326658.
Columns give A000004, A005408, A005843, A016813, A008586, A016921, A008588, A017077, A008590.
Row sums give A000330.

T(n,k) = {2*floor(k/2)*(n-k) + ceil((k-1)^2/2)} \\ Andrew Howroyd, Sep 10 2019

T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2).

T(n,k) = 2*floor(k/2)*(n-k) + binomial(k,2) - ceiling(k/2) + 1.

cp's OEIS Frontend

User: M. Ryan Julian Jr.

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

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

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A326296 Triangle of numbers T(n,k) = 2floor(k/2)(n-k) + ceiling((k-1)^2/2), 1<=k<=n.

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cp's OEIS Frontend

User: M. Ryan Julian Jr.

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

Views

Author

Keywords

Comments

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Crossrefs

Programs

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

PARI

Formula

A326296 Triangle of numbers T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2), 1<=k<=n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A326296 Triangle of numbers T(n,k) = 2floor(k/2)(n-k) + ceiling((k-1)^2/2), 1<=k<=n.