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Bibliographic Details
Main Author: Steinerberger, Stefan
Format: Preprint
Published: 2025
Subjects:
Online Access:https://arxiv.org/abs/2504.08023
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author Steinerberger, Stefan
author_facet Steinerberger, Stefan
contents Erdős and Graham define $g(n) = n + ϕ(n)$ and the iterated application $g_k(n) = g(g_{k-1}(n))$. They ask for solutions of $g_{k+r}(n) = 2 g_{k}(n)$ and observe $g_{k+2}(10) = 2 g_{k}(10)$ and $g_{k+2}(94) = 2 g_{k}(94)$. We show that understanding the case $r = 2$ is equivalent to understanding all solutions of the equation $ϕ(n) + ϕ(n + ϕ(n)) = n$ and find the explicit solutions $ n = 2^{\ell} \cdot \left\{1,3,5,7,35,47\right\}$. This list of solutions is possibly complete: any other solution derives from a number $n=2^{\ell} p$ where $p \geq 10^{10}$ is a prime satisfying $ϕ((3p-1)/4) = (p+1)/2$. Primes with this property seem to be very rare and maybe no such prime exists.
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institution arXiv
publishDate 2025
record_format arxiv
spellingShingle On an iterated arithmetic function problem of Erdos and Graham
Steinerberger, Stefan
Number Theory
Erdős and Graham define $g(n) = n + ϕ(n)$ and the iterated application $g_k(n) = g(g_{k-1}(n))$. They ask for solutions of $g_{k+r}(n) = 2 g_{k}(n)$ and observe $g_{k+2}(10) = 2 g_{k}(10)$ and $g_{k+2}(94) = 2 g_{k}(94)$. We show that understanding the case $r = 2$ is equivalent to understanding all solutions of the equation $ϕ(n) + ϕ(n + ϕ(n)) = n$ and find the explicit solutions $ n = 2^{\ell} \cdot \left\{1,3,5,7,35,47\right\}$. This list of solutions is possibly complete: any other solution derives from a number $n=2^{\ell} p$ where $p \geq 10^{10}$ is a prime satisfying $ϕ((3p-1)/4) = (p+1)/2$. Primes with this property seem to be very rare and maybe no such prime exists.
title On an iterated arithmetic function problem of Erdos and Graham
topic Number Theory
url https://arxiv.org/abs/2504.08023