A091940 Given n colors, sequence gives number of ways to color the vertices of a square such that no edge has the same color on both of its vertices.
0, 2, 18, 84, 260, 630, 1302, 2408, 4104, 6570, 10010, 14652, 20748, 28574, 38430, 50640, 65552, 83538, 104994, 130340, 160020, 194502, 234278, 279864, 331800, 390650, 457002, 531468, 614684, 707310, 810030, 923552, 1048608, 1185954, 1336370, 1500660
Offset: 1
Examples
a(4) = 84 since there are 84 different ways to color the vertices of a square with 4 colors such that no two vertices that share an edge are the same color. There are 4 possible colors for the first vertex and 3 for the second vertex. For the third vertex, divide into two cases: the third vertex can be the same color as the first vertex, and then the fourth vertex has 3 possible colors (4 * 3 * 1 * 3 = 36 colorings). Or the third vertex can be a different color from the first vertex, and then the fourth vertex has 2 possible colors (4 * 3 * 2 * 2 = 48 colorings). So there are a total of 36 + 48 = 84. - _Michael B. Porter_, Jul 24 2016
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
- OEIS Wiki, Colorings of square grid graphs
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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Maple
a := n -> (n-1)+(n-1)^4; for n to 35 do a(n) end do; # Rainer Rosenthal, Dec 03 2006
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Mathematica
Table[2Binomial[n, 2] + 12Binomial[n, 3] + 24Binomial[n, 4], {n, 35}] (* Robert G. Wilson v, Mar 16 2004 *) Table[(n-1)^4+(n-1),{n,1,60}] (* Vladimir Joseph Stephan Orlovsky, May 12 2011 *)
Formula
a(n) = 2*C(n,2) + 12*C(n,3) + 24*C(n,4) = n*(n-1)*(n^2-3*n+3).
a(n) = (n-1) + (n-1)^4. - Rainer Rosenthal, Dec 03 2006
G.f.: 2*x^2*(1+4*x+7*x^2)/(1-x)^5. a(n) = 2*A027441(n-1). - R. J. Mathar, Sep 09 2008
For n > 1, a(n) = floor(n^7/(n^3-1)). - Gary Detlefs, Feb 10 2010
E.g.f.: exp(x)*x^2*(1 + x)^2. - Stefano Spezia, Oct 08 2022
Extensions
More terms from Robert G. Wilson v, Mar 16 2004
Comments