A069480 Determinant of rank n matrix of 1..n^2 filled successively along antidiagonals.

1, -2, -5, 6, 13, -12, -25, 20, 41, -30, -61, 42, 85, -56, -113, 72, 145, -90, -181, 110, 221, -132, -265, 156, 313, -182, -365, 210, 421, -240, -481, 272, 545, -306, -613, 342, 685, -380, -761, 420, 841, -462, -925, 506, 1013, -552, -1105, 600, 1201, -650, -1301, 702, 1405, -756, -1513

Offset

1,2

Comments

The sorted absolute values of this sequence begin: 1, 2, 5, 6, 12, 13, 20, 25, 30, 41, 42, 56, 61, 72, 85, 90, 110, 113, 132, 145, 156, 181, 182, ...

References

On comp.soft-sys.math.mathematica (Mar 20 2002), Mark R. Diamond asked for a nice way to generate matrices 'along the diagonals' of the form (see example). The coding is from Wouter Meeussen.

Links

Stefano Spezia, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (0,-3,0,-3,0,-1).

Formula

a(n) = n*(n+2)/4 when n == 0 (mod 4), (n^2+1)/2 when n == 1 (mod 4), -n*(n+2)/4 when n == 2 (mod 4), -(n^2+1)/2, n == 3 (mod 4).

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

From Stefano Spezia, Apr 25 2021: (Start)

E.g.f.: ((2 - x)*x*cos(x) + (2 - 3*x - 2*x^2)*sin(x))/4.

a(n) = (n*(n + 2)*cos(n*Pi/2) + 2*(n^2 + 1)*sin(n*Pi/2))/4.

a(n) = -3*a(n-2) - 3*a(n-4) - a(n-6) for n > 6. (End)

Example

a(4) = 6 = 4*6/4 = |1 3 6 10 | 2 5 9 13 | 4 8 12 15 | 7 11 14 16|.

Mathematica

f[n_] := Table[(i + j - 1)((i + j - 1) - 1)/2 + 1 + (j - 1) - Mod[i + j - 1, n, 1]^2 Quotient[i + j - 1, n, 1], {i, n}, {j, n}]; Table[ Det[ f[n]], {n, 1, 50}]
LinearRecurrence[{0, -3, 0, -3, 0, -1}, {1, -2, -5, 6, 13, -12}, 60] (* Harvey P. Dale, Aug 14 2021 *)

Crossrefs

Sequence in context: A277012 A277022 A232603 * A354025 A100613 A070911

Adjacent sequences: A069477 A069478 A069479 * A069481 A069482 A069483

Keyword

easy,sign

Author

Robert G. Wilson v and Marc LeBrun, Mar 26 2002

Status

approved