Timeline for Is it possible to show that :for $n \geq 1:\sigma(n!-1) $ never be prime and why $\sigma(n!-1)\bmod 10 $ at most is $0$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2016 at 21:07 | comment | added | user99666 | and i don't know why the downvote to my modest question | |
| Dec 1, 2016 at 21:06 | comment | added | Greg Hurst | @YoussraElYossraYoussra Wow, don't know how I missed those... | |
| Dec 1, 2016 at 21:05 | comment | added | user99666 | @ChipHurst : 4!-1 and 32!-1 also are primes | |
| Nov 28, 2016 at 19:02 | comment | added | Jan-Christoph Schlage-Puchta | The probability that a random integer $n\leq x$ is prime is $\sim\frac{1}{\log x}$, so if we treat $n!-1$ as a random number, then the probability that $n!-1$ is prime is $\sim\frac{1}{n\log n}$. Since $n!-1$ is not divisible by any prime $\leq n$, one would rather expect a density like $\frac{1}{n}$, thus one would expect infinitely many $n$ for which $n!-1$ is prime. The probability that a random number $n\leq x$ is a cube is $x^{-2/3}$, which is much smaller and leads to the conjecture that there are only finitely many solutions of $n!-1=m^3$. | |
| Nov 27, 2016 at 4:29 | comment | added | Gerhard Paseman | Sorry, that's "if $n!-1$ turns out to be a(n odd) prime power". Gerhard "Minused It By That Much" Paseman, 2016.11.26. | |
| Nov 27, 2016 at 4:18 | comment | added | Gerhard Paseman | More generally, $\sigma(a^{2k+1})$ is composite for $k \gt 0$ and prime $a \gt 1$, and $a^{2k}$ is ruled out for congruential reasons. So $\sigma(n!-1)$ is composite for $n \gt 2$, even if $n!$ turns out to be a(n odd) prime power. I doubt that the quantity has only finitely many exceptions to being 0 mod 5. Gerhard "Win Some And Lose Some" Paseman, 2016.11.26. | |
| Nov 26, 2016 at 23:13 | comment | added | Greg Hurst | $n!-1$ can indeed be prime. For example, $469!-1$ is prime (see here). | |
| Nov 26, 2016 at 22:17 | comment | added | Pat Devlin | I like the argument, but I don't like the "since cubes of primes are rare" part. Primes are already rare, yet I expect $n!-1$ is prime infinitely often. | |
| Nov 26, 2016 at 22:15 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |