A non-terminating Erdős 261 greedy orbit has infinitely many doubling and infinitely many subtracting steps

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, the term $(n+j+1)/2^{n+j+1}$ is used), otherwise $g_{j+1}=2g_j$ (a doubling step). Then $g_j/2^{n+j}$ is what remains of $n/2^n$ after the terms used so far, and the greedy algorithm terminates iff some $g_j=0$. Theorem. If $g_k\ne0$ for all $k$, then there are infinitely many $k$ with $2g_k<n+k+1$ (doubling) and infinitely many $k$ with $2g_k\ge n+k+1$ (subtracting).

If the steps were eventually all doubling, $g$ would grow like $2^t$ while the threshold grows linearly. If they were eventually all subtracting, the deficit $d_k=(n+k)-g_k$ would satisfy $d_{k+1}=2d_k+2$, so it grows exponentially, contradicting $0<g_k$, i.e. $d_k<n+k$.