TITLE: "Anti-Gray codes" that maximize the number of bits that change at each step
QUESTION [16 upvotes]: Let $N$ be a non-negative integer and let $S$ be some ordering of the $2^N$ distinct $N$-tuples of binary digits.  We can look at the number of bits that change from each element to the next, and add this up for each element and its successor:
$$T(S) = \sum_{i=1}^{2^N} H(S_{i-1}, S_i)$$
where $H(x,y)$ is the Hamming distance between tuples $x$ and $y$, the number of positions in which they differ.  It's well known that this quantity $T(S)$ is minimized for certain  sequences ("Gray codes") in which each element of $S$ differs from its successor in exactly one bit.  So we have
$$m(N) = \min_{\text{all }S} T(S) = 2^N - 1$$
My question is:

What is the corresponding maximum value $$M(N) = \max_{\text{all }S} T(S)?$$

Small cases for $N\le3$ are easily seen to be 0, 1, 5, 18.  I did not see anything evident in OEIS about this.
A sequence $S$ contains $2^N-1$ transitions between elements, and it's tempting to conjecture that when $N>0$, half of these (rounded up) can change all $N$ bits and the other half (rounded down) can change all but one bit.  For example, when $N=3$ an "anti-Gray" sequence which does this is:
    000
    111
    100
    011
    110
    001
    010
    101

If this can be done for all $N$, then $$\begin{align}M(N) & = N\cdot2^{N-1} + (N-1)\cdot(2^{N-1}-1) \\& = N2^N - 2^{N-1}-N+1;
\end{align}$$ certainly this is true for $N\le 3$.  Is it true in general?
(I solved this problem before I finished writing  up the question, but I am posting it anyway per earlier discussions [1][2][3].)

REPLY [2 votes]: I would like to claim and prove the following:
Claim: 
"It is possible to construct Anti-Gray codes of length 'n'such that the consecutive elements in the sequence differ by exactly 'n-k' bits for all k >= 1"   
Proof:
I try to prove the above using induction on 'k'.
As a base inductive step, I would prove there is an anti-gray sequence of length 'n' which differ by exactly 'n-1' bits.
I show a construction of such a sequence using Gray codes of length 'n'. It is understood that Gray codes can be constructed for any 'n' such that the bits differ by exactly 1 bit. To construct an Anti-Gray sequence out of this Gray code sequence, we would "negate" every "alternate term" in the Gray code sequence. Let us consider terms '2t' and '2t-1' in this new sequence. Before, they were part of a Gray code sequence, and thus differed in exactly 1-bit. Since we have reversed all bits in '2t' (for ex), 2t and 2t-1 now differ in exactly 'n-1' bits. The same argument holds true for 2t and 2t+1. This construction is efficient, and takes exactly log(N) bits for representation.
Induction Hypothesis:
We assume the claim is true for all 'k' < K.
By Induction Hypothesis, there exists an Anti-Graty sequence:
   a1, a2, a3, ..., aN 
such that consequtive terms differ exactly by 'n-k' bits.
To construct a sequence which would differ in 'n-(k+1)' bits, we simply add a leading '0' or a '1' to every term in this sequence:
   0a1, 0a2, 0a3, ... 0aN.
This new sequence has now length 'n+1' and the consecutive elements differ by one less bit.
Hence by induction, the claim is true for all 'k'>=1 and the proof is complete.
Cheers!