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| Format: | Preprint |
| Published: |
2025
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| Online Access: | https://arxiv.org/abs/2504.08023 |
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| _version_ | 1866912320409567232 |
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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. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2504_08023 |
| 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 |