A non-terminating Erdős 261 greedy orbit has no limiting carry ratio

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. If $g_k\ne0$ for all $k$, then $g_j/(n+j)$ does not converge.

With $x_j=g_j/(n+j)$ and $q_j\in\{0,1\}$ the decision at step $j$, $q_j=\frac{2g_j-g_{j+1}}{n+j+1}=2x_j\frac{n+j}{n+j+1}-x_{j+1}$. If $x_j\to L$ then $q_j\to L$, so $q_j$ is eventually constant, contradicting that both decisions occur infinitely often (Erdos261.greedy_phase_types_infinite).