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Claim. With $g(n)=n+\phi(n)$ and $g_k$ its $k$-th iterate, the shift-two case of Problem 411 asks for the $n$ with $g_{k+2}(n)=2g_k(n)$ for all large $k$. Steinerberger's preprint [St25] (library card) shows, in its Section 2.1, that the relation holds from step $k$ on exactly when $m=g_k(n)$ solves

$$\phi(m)+\phi(m+\phi(m))=m,$$

because such an $m$ is even and $g(2m)=2g(m)$ for even $m$ carries the doubling to every later step. Its Theorem puts every solution $m$ of the equation into one of two branches: the odd part of $m$ lies in $\{1,3,5,7,35,47\}$, or $m=2^\ell(8t+7)$ or $m=2^\ell(6t+5)$ with $8t+7\ge10^{10}$ prime and $\phi(6t+5)=4t+4$. The first branch is settled by the preprint: the powers of $2$ and the numbers $3\cdot2^\ell$ are solutions (Section 2.2), the orbits $7\cdot2^\ell\to5\cdot2^{\ell+1}\to 7\cdot2^{\ell+1}$ and $47\cdot2^\ell\to35\cdot2^{\ell+1}\to47\cdot2^{\ell+1}$ give the other four families (Section 2.10), and $1$ and $2$ are not solutions (Section 2.1), so the first-branch solutions are exactly

$$\{2^a s: s\in\{1,3,5,7,35,47\},\ a\ge a_s\},\qquad a_1=2,\quad a_s=1\ (s\ne1).$$

The shift-two solutions $n$ of the problem are the $n$ whose orbit reaches such an $m$; they include $n$ with other odd parts, such as $n=18$ (orbit $18\to24\to32\to48$) and $n=22$ (orbit $22\to32\to48$), which satisfy the relation from $k=1$. The two solutions the site records, $n=10$ and $n=94$, are first-branch solutions with $a=1$. The preprint's computer search (Section 2.10) finds no prime $8t+7\le10^{10}$ with $\phi(6t+5)=4t+4$ other than $7$ and $47$ ($t=0,5$), the primes behind the first-branch odd parts $7,5$ and $47,35$; it relates the second branch to whether $\phi(q)=\tfrac23(q+1)$ has infinitely many solutions ($q=5,35,1295,1679615$ are known). The site's commentary reports the reduction and the two branches and credits them to [St25].

Covers. The first branch of the reduction for $r=2$: the solutions $m$ of $\phi(m)+\phi(m+\phi(m))=m$ with odd part in $\{1,3,5,7,35,47\}$ are exactly the six doubling families above, each of which satisfies $g_{k+2}(m)=2g_k(m)$ for every $k\ge0$. It does not settle $r=2$: the second branch is open, and which $n$ reach a solution of the equation is not determined; it says nothing about other shifts $r$, and the problem's classification over all $n$ and $r$ stays open.

Depends on. No page of this wiki.

Later claim of the same result. A partial proof claim by Alateng Pan, posted on the site's proof-claims tab on 2026-09-08 under the user name Tonycollatz (proof claim 287), asserts the first-branch classification above, with the thresholds $a_1=2$ and $a_s=1$, by the six base cases $n=4,6,10,14,70,94$ and the doubling step; its forum summary names it the first branch of Steinerberger's reduction, and the write-up, Erdős–Graham Problem #411: a construction of six solution families for the r=2 case (Zenodo record 21991040, fourth version of 2026-08-18, CC BY 4.0), credits [St25] with the reduction and the branches. The record's three earlier versions, of 2026-08-11, 2026-08-14 and 2026-08-15, were titled as a complete proof of the $r=2$ case and claimed it; the fourth version retitles the result to the first branch and leaves the second branch open. The submission states that the system DeepSeek was used for language polishing and formatting only, with the mathematics the author's own. The result being the one this page records, the claim is disclosed here and gets no page of its own.

Standing. An arXiv preprint (v1 of 2025-04-10) with no journal reference on its arXiv record and no formalization; the site's commentary reports the result but labels the problem OPEN (page last edited 28 October 2025; the later claim carries no comments), and the community database notes a partial $r=2$ result without changing the problem's open status, so no acceptance evidence is listed and the claim stays claimed.