Erdős 261 greedy carry: a subtracting step sends the deficit $D$ to $2D+2$

Let $g_j$ be the greedy carry for Erdős Problem 261 started at $(n,n)$: $g_0=n$, and $g_{j+1}=2g_j-(n+j+1)$ if $2g_j\ge n+j+1$ (a subtracting step), otherwise $g_{j+1}=2g_j$ (a doubling step). The greedy algorithm for $n/2^n=\sum a_k/2^{a_k}$ terminates iff some $g_j=0$. Theorem. Write $D_j=(n+j)-g_j$. If step $j$ is subtracting ($2g_j\ge n+j+1$), then

This corrects the recursion $D\mapsto 2D+1$ stated in an earlier version of the phase calculus; it is the elementary step behind the closed form for subtracting blocks (subtracting_block).