Runs of the Erdős 261 greedy carry have logarithmic length

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). Equivalently $(N,R)\mapsto(N+1,\,2R\bmod(N+1))$ with $N_j=n+j$, $R_j=g_j$; the greedy algorithm terminates iff some $g_j=0$. Theorem. Let $N=n+j\ge2$ and $h=\lceil\log_2 N\rceil$. Then the $h+2$ steps $j,\dots,j+h+1$ are not all subtracting, and, if $g_j\ge1$, they are not all doubling. Hence every maximal run of equal decisions starting at $N$ has length at most $\lceil\log_2 N\rceil+1$.

A doubling run gives $g_{j+L}=2^Lg_j\ge4N>N+L$, and a subtracting run gives $D_{j+L}=2^LD_j\ge8N>N+L+2$, contradicting $0\le g\le N$. (Close to Borwein–Loring 1990, Prop. 2 / Cor. 2 on strings of 0s and 1s.)