Outside the Borwein–Loring numbers, the Erdős 261 greedy carry cannot stop within its first two phases

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 $g_j=0$. Write $c_j=2^{j+1}-j-2$ (so $c_0=0$, $c_{j+1}=2c_j+j+1$; these are the Borwein–Loring numbers). Theorem. Suppose $c_j<n<c_{j+1}$ (so $n$ is not a Borwein–Loring number) and that the $t$ steps after step $j$ are all doubling steps, i.e. $g_{j+s+1}=2g_{j+s}$ for $s<t$. Then

So the orbit cannot reach $0$ during its initial subtracting run, during the following doubling run, or at the first subtracting step after it. Writing $r=n-c_j$, after the doubling run $g_{j+t}=2^t r$, and termination at the next step would mean $r(2^{t+1}-1)=2^{j+1}+t-1$, which is impossible (for $t=0$ it would force $n=c_{j+1}$). Every terminating $n$ other than the Borwein–Loring numbers therefore needs a later phase.