Timeline for On finite products of $\frac{p+4}{p+2}$ with $p$ prime
Current License: CC BY-SA 4.0
22 events
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| Sep 2 at 8:33 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 31 at 18:30 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 31 at 14:04 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 27 at 14:05 | vote | accept | Zhi-Wei Sun | ||
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| Aug 26 at 20:53 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 26 at 7:47 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 25 at 8:34 | comment | added | Daniel Weber | Let us continue this discussion in chat. | |
| Aug 25 at 8:26 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 25 at 8:21 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 25 at 7:31 | comment | added | Deyi Chen | If $r=5$, then $p_5\leq 8467$. | |
| Aug 25 at 7:28 | comment | added | Daniel Weber | With $\omega(n) \leq 4$ there's no solution for $p_r \leq 10^5$ | |
| Aug 25 at 7:26 | comment | added | Daniel Weber | Update: it's minimal for $p_r \leq 69109$ | |
| Aug 25 at 5:27 | comment | added | Daniel Weber | 675790721971 is the minimal value for $p_r < 10^4$ | |
| Aug 25 at 5:13 | comment | added | Deyi Chen | $\omega(a(21))\leq 5$ since $\prod_{i=1}^{6}(41i-2)>3\times10^{12}.$ I think it can be proven that $\omega(a(21))=5.$ | |
| Aug 25 at 5:08 | comment | added | Daniel Weber | I found a smaller value for $a(21)$: $675790721971 = 113 \times 157 \times 271 \times 367 \times 383$ | |
| Aug 25 at 5:03 | comment | added | Deyi Chen | @ Daniel Weber According to the remark，$p_r\leq m<(8(2n-1))^{2^r-1}$ | |
| Aug 25 at 4:45 | comment | added | Daniel Weber | Are you able to give an upper bound to $p_r$? | |
| Aug 24 at 18:45 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 24 at 18:38 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 24 at 12:03 | history | edited | Deyi Chen | CC BY-SA 4.0 |
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| Aug 24 at 9:40 | review | Low quality posts | |||
| Aug 24 at 9:50 | |||||
| Aug 24 at 9:22 | history | answered | Deyi Chen | CC BY-SA 4.0 |