Timeline for Projective summation over divisors: must these functions be multiplicative?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2016 at 20:03 | comment | added | Kevin Smith | I think it is too. I noticed that this afternoon but I can't quite see how to prove it, nor how to characterise the general case. | |
| Oct 11, 2016 at 16:53 | comment | added | Greg Martin | I'm fairly sure that $$ a(n) = \begin{cases} 1, &\text{if } n=1, \\ -1, &\text{if $n$ is prime}, \\ 0, &\text{otherwise.} \end{cases} $$ is also a (non-multiplicative) solution for $k=2$. There are probably hybrid solutions between this and $a(n)=\mu(n)$ as well. | |
| Oct 11, 2016 at 8:43 | comment | added | Kevin Smith | Thank you, Greg. I have realised that my expecting solutions to be multiplicative came from starting with the primes and assuming $a(p)$ to be non-zero. | |
| Oct 11, 2016 at 8:34 | vote | accept | Kevin Smith | ||
| Oct 11, 2016 at 6:46 | history | answered | Greg Martin | CC BY-SA 3.0 |