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.

A232568 Number of non-equivalent binary n X n matrices with three pairwise nonadjacent 1's.

Original entry on oeis.org

0, 6, 40, 210, 681, 1919, 4443, 9481, 18206, 33164, 56570, 92996, 146175, 223565, 330981, 479779, 678508, 943586, 1287036, 1731654, 2293765, 3004011, 3883935, 4973645, 6300906, 7917064, 9857198, 12185816, 14946491, 18218969, 22056585, 26556551, 31783320
Offset: 2

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Author

Heinrich Ludwig, Nov 28 2013

Keywords

Comments

Also: Number of non-equivalent ways to place three non-attacking wazirs on an n X n board.
Two matrix elements are considered adjacent if the difference of their row indices is 1 and the column indices are equal, or vice versa (von Neumann neighborhood).
Counted for this sequence are equivalence classes induced by the dihedral group D_4. If equivalent matrices are being destinguished, the number of matrices is A172226(n).

Examples

			There are a(3) = 6 non-equivalent 3 X 3 matrices with three pairwise nonadjacent 1's (and no other 1's):
  [1 0 0]    [1 0 1]    [1 0 0]    [1 0 1]   [1 0 1]   [0 1 0]
  |0 1 0|    |0 0 0|    |0 0 1|    |0 0 0|   |0 1 0|   |1 0 1|
  [0 0 1]    [1 0 0]    [0 1 0]    [0 1 0]   [0 0 0]   [0 0 0]
		

Crossrefs

Programs

  • Maple
    A232568:=n->(n^6-15*n^4+28*n^3+29*n^2-76*n-15-((n+1) mod 2)*(8*n^3-21*n^2+40*n-63))/48; seq(A232568(n), n=2..50); # Wesley Ivan Hurt, Dec 06 2013
  • Mathematica
    Table[(n^6-15n^4+28n^3+29n^2-76n-15-Mod[n+1,2](8n^3-21n^2+40n-63))/48, {n, 2, 50}] (* Wesley Ivan Hurt, Dec 06 2013 *)

Formula

a(n) = (n^6 - 15*n^4 + 20*n^3 + 50*n^2 - 116*n + 48)/48 if n is even; a(n) = (n^6 - 15*n^4 + 28*n^3 + 29*n^2 - 76*n - 15)/48 if n is odd.
G.f.: x^3*(x^9-4*x^8+x^7+12*x^6+9*x^5-70*x^4-77*x^3-84*x^2-22*x-6) / ((x-1)^7*(x+1)^4). - Colin Barker, Dec 06 2013
a(n) = (n^6 - 15n^4 + 28n^3 + 29n^2 - 76n - 15 - ((n+1) mod 2) * (8n^3 - 21n^2 + 40n - 63))/48. - Wesley Ivan Hurt, Dec 06 2013