cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A302912 Determinant of n X n matrix whose main diagonal consists of the first n 8-gonal numbers and all other elements are 1's.

Original entry on oeis.org

1, 7, 140, 5460, 349440, 33196800, 4381977600, 766846080000, 171773521920000, 47924812615680000, 16294436289331200000, 6631835569757798400000, 3183281073483743232000000, 1779454120077412466688000000, 1145968453329853628547072000000
Offset: 1

Views

Author

Muniru A Asiru, Apr 15 2018

Keywords

Examples

			The matrix begins:
  1   1   1   1   1   1   1 ...
  1   8   1   1   1   1   1 ...
  1   1  21   1   1   1   1 ...
  1   1   1  40   1   1   1 ...
  1   1   1   1  65   1   1 ...
  1   1   1   1   1  96   1 ...
  1   1   1   1   1   1 133 ...
		

Crossrefs

Cf. A000567 (octagonal numbers).
Cf. Determinant of n X n matrix whose main diagonal consists of the first n k-gonal numbers and all other elements are 1's: A000142 (k=2), A067550 (k=3), A010791 (k=4, with offset 1), A302909 (k=5), A302910 (k=6), A302911 (k=7), this sequence (k=8), A302913 (k=9), A302914 (k=10).

Programs

  • Maple
    d:=(i,j)->`if`(i<>j,1,i*(3*i-2)):
    seq(LinearAlgebra[Determinant](Matrix(n,d)),n=1..16);
  • Mathematica
    nmax = 20; Table[Det[Table[If[i == j, i*(3*i - 2), 1], {i, 1, k}, {j, 1, k}]], {k, 1, nmax}] (* Vaclav Kotesovec, Apr 16 2018 *)
    Table[FullSimplify[3^(n+1) * Gamma[n] * Gamma[n + 4/3] / (4*Gamma[1/3])], {n, 1, 15}] (* Vaclav Kotesovec, Apr 16 2018 *)RecurrenceTable[{a[n+1] == a[n] * n * (3*n + 4), a[1] == 1}, a, {n, 1, 20}] (* Vaclav Kotesovec, Apr 16 2018 *)
  • PARI
    a(n) = matdet(matrix(n, n, i, j, if (i!=j, 1, i*(3*i-2)))); \\ Michel Marcus, Apr 16 2018

Formula

From Vaclav Kotesovec, Apr 16 2018: (Start)
a(n) = 3^(n+1) * Gamma(n) * Gamma(n + 4/3) / (4*Gamma(1/3)).
a(n) ~ Pi * 3^(n+1) * n^(2*n + 1/3) / (2 * Gamma(1/3) * exp(2*n)).
a(n+1) = a(n) * n*(3*n + 4).
(End)