Abstract: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 ${p_i}^{a_i+1} < 2p_1p_2p_3\cdots p_k$.
From: Joshua Zelinsky [view email]
[v1]
Mon, 25 May 2020 13:55:08 UTC (7 KB)
[v2]
Mon, 6 Nov 2023 16:01:37 UTC (8 KB)
[v3]
Tue, 10 Dec 2024 19:17:15 UTC (9 KB)
[v4]
Tue, 16 Jun 2026 18:28:22 UTC (11 KB)
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。