A232568 Number of non-equivalent binary n X n matrices with three pairwise nonadjacent 1's.
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
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]
Links
- Heinrich Ludwig, Table of n, a(n) for n = 2..1001
- Index entries for linear recurrences with constant coefficients, signature (3,1,-11,6,14,-14,-6,11,-1,-3,1).
Programs
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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
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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
Comments