Must a primitive non-deficient number have a component not much larger than its radical?
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arXiv
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| Natura: | Preprint |
| Pubblicazione: |
2020
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| _version_ | 1866913606346473472 |
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| author | Zelinsky, Joshua |
| author_facet | Zelinsky, Joshua |
| contents | Let $n$ be a primitive non-deficient number where $n=p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$ where $p_1, p_2 \cdots p_k$ are distinct primes. We prove that there exists an $i$ such that $$p_i^{a_i+1} < 2k(p_1p_2p_3\cdots p_k).$$ We conjecture that in fact one can always find an $i$ such that ${p_i}^{a_i+1} < p_1p_2p_3\cdots p_k$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2005_12115 |
| institution | arXiv |
| publishDate | 2020 |
| record_format | arxiv |
| spellingShingle | Must a primitive non-deficient number have a component not much larger than its radical? Zelinsky, Joshua Number Theory 11A25, 11N64 Let $n$ be a primitive non-deficient number where $n=p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$ where $p_1, p_2 \cdots p_k$ are distinct primes. We prove that there exists an $i$ such that $$p_i^{a_i+1} < 2k(p_1p_2p_3\cdots p_k).$$ We conjecture that in fact one can always find an $i$ such that ${p_i}^{a_i+1} < p_1p_2p_3\cdots p_k$. |
| title | Must a primitive non-deficient number have a component not much larger than its radical? |
| topic | Number Theory 11A25, 11N64 |
| url | https://arxiv.org/abs/2005.12115 |