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COMMENT
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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 re-order everything so that these numbers are non-decreasing (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)).
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MATHEMATICA
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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]}]]; A091969[n_] := a[n, 2^(n-1), 2^(n-1)]
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