Let $G_n$ be the complete graph whose vertices are the $2^n$ $n$-bit strings. Let $H_n$ denote the Hamiltonian path through $G_n$ that uses the maximum number of edges that correspond to a single bit transition $0\mapsto 1$. What is this maximum number? And is there an algorithm that generates this Hamiltonian path?
The classical Gray code is a Hamiltonian path that uses only edges with $0\mapsto 1$ or $1 \mapsto 0$ transitions (single bit flips in both directions). I want to use as many single bit flips as possible in one direction ($0\to 1$) and have no restrictions on other transitions.
Here is an example for $n=4$ that uses 10 $(0\mapsto 1)$ transitions $(\rightarrow)$ and 6 others $(\Rightarrow)$:
$0000\rightarrow 0001 \rightarrow 0011 \rightarrow 0111 \rightarrow 1111 \Rightarrow 0010\rightarrow 0110 \rightarrow 1110 \Rightarrow 0100 \rightarrow 0101 \rightarrow 1101 \Rightarrow 1000 \rightarrow 1001 \rightarrow 1011 \Rightarrow 1010 \Rightarrow 1100 \Rightarrow 0000$