Timeline for Test for pair of odd primes $(p, 2p^2-1)$
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| May 4 at 16:12 | vote | accept | Notamathematician | ||
| May 4 at 7:13 | comment | added | Denis Shatrov | If $k$ is prime but $2k^2 - 1$ is composite, then $B_{2k} > 6k$. Your formula for $B_i$ holds if $i \le k$, and $B_i$ is still even. Thus $B_{2k} \ge 6k$. $B_{2k} = 6k$ if and only if $\gcd(2k^2 - 1, u) = 1$ for all $2k \le u \le 3k - 1$. Let $2k^2 - 1 = ab$. Then $a < 3k/2$ or $b < 3k/2$. If $a \le k$, then there is a multiple of $a$ in the range $[2k, 3k - 1]$. If $k < a < 3k/2$, then $2a$ is in this range. | |
| May 3 at 23:56 | history | rollback | Max Alekseyev |
Rollback to Revision 1
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| May 3 at 23:26 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 143 characters in body; added 67 characters in body
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| May 3 at 23:00 | history | answered | Max Alekseyev | CC BY-SA 4.0 |