cp's OEIS Frontend

A graph is uniquely 3-colorable if there is a unique partition of its vertex set into 3 independent sets. This implies that every proper 3-coloring of the graph has this partition as its set of color classes.

A (cyclic) Gray code is a listing of the binary n-tuples in a cyclic sequence so that adjacent elements differ in exactly one bit position. So we can describe a Gray code just by listing the bit that gets changed at each step. This gives us a sequence of 2^n numbers, each of which lies in {0..n-1}.

Let c_i be the number of times that the bit in position i is changed. This gives us a sequence (c0,c1,...,c_{n-1}), called the transition count of the code, such that each number is even and the sum is 2^n.

In addition, if we assume that we reorder everything so that these numbers are nondecreasing (c_0 <= c_1 <= ... <= c_{n-1}) then there is an additional condition c_0 + c_1 + ... + c_{j-1} >= 2^j by noting that all 2^j patterns must occur in the j least-flipped bit positions. Then a(n) is the number of sequences satisfying these conditions.

For example, for n=4 there are 4 possible sequences: 2 2 4 8, 2 2 6 6, 2 4 4 6 and 4 4 4 4 and indeed there are cyclic Gray codes with each possible transition count sequence, so a(4) = 4.

Additional comments from Rob Pratt: set b_i = c_i / 2. Let a(n,s,p) be the number of solutions to b_0 + b_1 + ... + b_{n-1} = s, 1 <= b_0 <= b_1 <= ... <= b_{n-1} <= p and b_0 + b_1 + ... + b_{j-1} >= 2^j for j = 1 to n.

Then a(n,s,p) satisfies the following recursion (written in Mathematica syntax). a[1, s_, p_] := a[1, s, p] = If[1 <= s <= p, 1, 0]; a[n_, s_, p_] := a[n, s, p] = If[s < 2^(n - 1), 0, Sum[a[n - 1, s - k, Min[p, k]], {k, 1, Min[p, s]}]]; we want to compute a(n,2^(n-1),2^(n-1)).

Equals A066037(n)/(n!/2). See A006069, A003042, A066037 for more information.

Could A002632 be an erroneous version of this sequence?

The old entry with this sequence number was a duplicate of A057475.