Erdős Problem #411

declared status 'open'. Formalized: no. Let $g_1=g(n)=n+\phi(n)$ and $g_k(n)=g(g_{k-1}(n))$. For which $n$ and $r$ is it true that $g_{k+r}(n)=2g_k(n)$ for all large $k$? Current best: Selfridge and Weintraub found solutions to $g_{k+9}(n)=9g_k(n)$ and Weintraub found$$g_{k+25}(3114)=729g_k(3114)$$for all $k\geq 6$. Steinerberger [St25] has observed that, for $r=2$, this problem is equivalent to asking for solutions to$$\phi(n)+\phi(n+\phi(n))=n,$$and has shown that if this holds then either the odd part of $n$ is in $\{1,3,5,7,35,47\}$, or is equal to $8m+7$ or $6m+5$, where $8m+7\geq 10^{10}$ is a prime number and $\phi(6m+5)=4m+4$. Cambie conjectures that the only solutions have $r=2$ and $n=2^lp$ for some $l\geq 1$ and $p\in \{2,3,5,7,35,47\}$. Cambie has shown this problem is reducible to the question of which integers $r,t\geq 1$ and primes $p\equiv 7\pmod{8}$ satisfy $g_k(2p^t)=4p^t$, and conjectures there are no solutions to this except when $t=1$ and $p\in \{7,47\}$. Prize: no. OEIS: A383044. Tags: iterated functions, number theory.

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