Timeline for On the largest prime factor and the largest component of an odd perfect number
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 5 at 17:39 | vote | accept | Pascal Ochem | ||
| S Feb 4 at 15:46 | history | suggested | CommunityBot | CC BY-SA 4.0 |
mentioning that $p$ itself should be small.
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| Feb 4 at 8:06 | review | Suggested edits | |||
| S Feb 4 at 15:46 | |||||
| S Feb 4 at 5:27 | history | suggested | J. W. Tanner |
Added perfect-numbers tag
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| Feb 4 at 1:09 | review | Suggested edits | |||
| S Feb 4 at 5:27 | |||||
| Feb 3 at 19:19 | answer | added | JoshuaZ | timeline score: 3 | |
| Feb 3 at 4:11 | comment | added | Daniel Weber | We have $\Psi(x, \log^2 x) = x^{\frac12 + o(1)}$, which is also the number of values of $p^a + 1 \leq x$ for even $a$, so heuristically by the birthday paradox we should expect to find a collision around this $\log^2 x$. For $10^{62}$ this is around 20500. I don't have any idea how we could locate this collision, though | |
| Feb 3 at 4:04 | comment | added | Daniel Weber | If we look at $p^a + 1 = p_1^{a_1} p_2^{a_2} \cdots$ then from the abc conjecture we should have $p^{a-1} \lesssim p_1 p_2 \cdots$, and additionally it can't contain any primes which are $3 \pmod 4$, so you get a lower bound of about 317. | |
| Feb 3 at 2:36 | history | asked | Pascal Ochem | CC BY-SA 4.0 |