Timeline for Why am I unable to find primes of the form $(9n)!+n!+1$?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10, 2019 at 10:33 | comment | added | Mike | I check numbers of the form $(kn)!+n!+1$,It seems there's no special reason can't be primes.It's problem of probility. You bet,from $1$ to $n$, if you find 100 primes of the form $(8n)!+n!+1$, then from $1$ to $n$ you will find at least one prime of the form $(9n)!+n!+1$ | |
| Dec 10, 2019 at 10:27 | comment | added | user142929 |
The only idea, concerning mathematics, that I had in the past about this kind of problems is try to write a new equation using the composition of different arithmetic functions with the purpose to invoke inequalitites/conjectures that satisfy the arithmetic functions that I evoke. For example I cann't find a solution of $\varphi(m)=(9n)!+n!$, where $\varphi(m)$ is the Euler's totient function, for the first few positive integers $n\geq 1$ and $m\geq 1$. See the code for(m=1, 10000, for(n=1, 100, if(eulerphi(m)==(9*n)!+n!,print(m+1)))) in the web Sage Cell Server, choosing as Language GP.
|
|
| Dec 10, 2019 at 0:52 | comment | added | Gerhard Paseman | For $n$ 1 less than a prime, the expression is 0 mod n+1. (This holds for all k.) You need to pick n large enough so that n! + 1 has no prime factors less than 9n. But this is necessary, and not sufficient. Gerhard "A Prime Observation for Compositeness" Paseman, 2019.12.09. | |
| Dec 10, 2019 at 0:34 | history | asked | Maximilian Janisch | CC BY-SA 4.0 |