I suggested in a comment, to look for colorings that are the same for each <i>a</i>, except for a fixed permutation. Suppose the colors are given by function c(x), for x ≥ 1. And a permutation function p(i,j), where i,j ∈ [1..n]. Where i is the index of the permutation, and j the index of the permutation of the color. If n = 4 and the coloring starts with abcd, then p(1,1) = a, p(1,2) = b, p(1,3) = c, p(1,4) = d, p(2,1) = b, p(2,2) = d. If the coloring is the same for each <i>a</i>, then we get the following equation: c(an) = p(c(a),c(n)) From this it follows, that the permutation must p(i,j)=p(j,i). More important, is to reverse this: <b>Lemma 1:</b>If p(i,j) = p(j,i) and p(x,y) = p(v,w) for xy = vw, then a coloring can be constructed. <b>Lemma 2:</b>For each n a permutation exists such that p(i,j) = p(j,i) and p(x,y) = p(v,w) for xy = vw. To see how this works, for n = 4, we start the following permutation matrix:<br/> abcd<br/> bd..<br/> c...<br/> d...<br/> Fill in the second row:<br/> abcd<br/> bdac<br/> ca..<br/> dc..<br/> And complete it:<br/> abcd<br/> bdac<br/> cadb<br/> dcba<br/> From this construct the coloring: abcdxaxcdxxbxxxa By each prime number > n (the x values in the coloring), you can choose a color, but you have to continue with the permutation of that color. Lemma 1 is trivial to prove, because it follows that c(xy) is c(yx). I haven't proven lemma 2 yet. For n is 1,2,3,4, there is only 1 permutation matrix. For n = 5 there are a few. The condition p(x,y) = p(v,w) for xy = vw becomes relative weaker for larger n. So, I don't think it is a problem to create such permutation matrix (also given the number of different Sudokus). Of course, lemma 2 needs to be proven properly.