Timeline for For any prime $p$, is there $C$ such that if $x\ge C$, then all but one integer among $x+1, x+2, \dots, x+p$ has Greatest Prime Factor $> p$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2013 at 11:38 | comment | added | Jim White | I ask because I've always been interested in investigating an extension to Lehmer's method as described above, but have had no particular motivation to do so. | |
| Jan 12, 2013 at 3:12 | comment | added | Jim White | I'm kind of preoccupied with a couple of other questions, so I don't quite know what you mean by "patterns like A213523". This could just be attention-deficit on my part, but can you explain in more detail what you are looking for? | |
| Jan 11, 2013 at 21:28 | comment | added | Larry Freeman | Hi Dr. Memory, My interest in primarily in understanding the context behind the Sylvester-Schur Theorem: mathoverflow.net/questions/111823/… For example, I am especially interested in patterns like this: oeis.org/A213253 Cheers, -Larry | |
| Jan 11, 2013 at 2:40 | comment | added | Jim White | I meant $(pS_1, pS_1 + p)$. You can't edit comments! | |
| Jan 11, 2013 at 2:38 | comment | added | Jim White | Ah, but then again, we might still get lucky, for perhaps our $S_m$ is in fact always $(pS_1, pS_1 + p). | |
| Jan 11, 2013 at 2:20 | comment | added | Jim White | Correction: $S_2 < S_1$ was only true if $gcd(S, S+2)=1$. Clearly any $S, S+1$ will correspond to smooth pairs $(kS, kS + k)$ for any $k$ you like, so there is no avoiding the complications (wrt the Lehmer method) described above. | |
| Jan 11, 2013 at 0:54 | history | edited | Jim White | CC BY-SA 3.0 |
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| Jan 11, 2013 at 0:45 | history | edited | Jim White | CC BY-SA 3.0 |
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| Jan 11, 2013 at 0:40 | history | answered | Jim White | CC BY-SA 3.0 |