Timeline for There seem to be only few primes of the form ${n\choose k}+1$ if $k\geq 3$ is odd
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2021 at 15:30 | vote | accept | Roland Bacher | ||
| Aug 8, 2021 at 15:30 | comment | added | Roland Bacher | The last statement is somtimes not true $\{6\choose 5}+1=7=n+1$ is prime but does not divide $(n+1)!$. This situation should nowever be the only exception! | |
| Aug 8, 2021 at 15:12 | history | edited | user334725 | CC BY-SA 4.0 |
added 392 characters in body
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| Aug 8, 2021 at 15:08 | comment | added | Roland Bacher | You are right. That proves that these sets are finite. Write it as an answer so I can accept it! | |
| Aug 8, 2021 at 15:05 | comment | added | user334725 | Note also that all the "small sets" have $(n+1)$ dividing $k!$. | |
| Aug 8, 2021 at 14:59 | comment | added | user334725 | When $k$ is odd, writing $f_k(n)={n\choose k}+1$ you have $f_k(-1)=0$ as a polynomial evaluation, and removing the denominator gives you the extra $k!$ in integers. I.e., $3!f_3(n)=n(n-1)(n-2)+6$, so $$3!f_3(-1)=(-1)(-2)(-3)+6=-6+6=0.$$ As $(-1)$ is a root of $f_k(n)$, then $(n+1)$ divides it (again worrying about denominators). | |
| Aug 8, 2021 at 14:56 | review | First posts | |||
| Aug 8, 2021 at 15:18 | |||||
| Aug 8, 2021 at 14:55 | history | answered | user334725 | CC BY-SA 4.0 |