7

Let $n=\prod_p p^{c_p}$, $N\in \mathbb N$ and $$ \alpha_N(n)=\prod_p p^{c_p \bmod N}. $$ The function $\alpha_N$ is multiplicative since $\alpha_N(n)\alpha_N(m)=\alpha_N(nm)$ for co-prime $n$ and $m$ and completely multiplicative for a subset of $\mathbb N$.

Further let $$\log \alpha_N(n)= A_N(n)=\sum_p (c_p \bmod N) \log p. $$ It also reminds me to a sum of an adapted kind of von Mangoldt function, with following definition: $$ \Lambda^\star(n) = \begin{cases} \log p & \text{if }n=p^c \text{ for some prime } p \text{ and integer } c \ge 1 \text{ and } c\bmod N =1, \\ 0 & \text{otherwise.} \end{cases} $$

PS: Since the taste of the question was changed, some of the comments might be misleading. Sorry for that...

Eric Naslund
  • 72,099
draks ...
  • 18,449
  • 1
    I don't think this sort of functions should have or deserve a name. There are pretty weird functions such as the constant function $\alpha_1\equiv 1$, the function $\alpha_2$ where $\alpha_2(1)=1$ and, for an integer $n>1$, $\alpha_2(n)$ is the largest prime divisor of $n$, and the function $\alpha_3$ where $\alpha_3(n)=1$ if $n$ is odd, and $\alpha_3(n)=2$ if $n$ is even. – Batominovski Jun 23 '15 at 20:26
  • 1
    @Batominovski the function $\alpha_3(n)$ has a name! It's called modulus: $\alpha_3(n)=n \bmod 2$...and I also don't see why your other example should be so weird... – draks ... Jun 23 '15 at 20:38
  • 1
    I know that $\alpha_3$ is actually reduction modulo $2$. The point is these functions don't really seem to have any coherent properties. Individually, these functions themselves are not very weird, but, collectively, the assumption that there should be a named property which includes them all is what sounds utterly weird. – Batominovski Jun 23 '15 at 20:45
  • @Batominovski but after all it's a set, as you say. And the set of weird functions (you named it) has already three members... – draks ... Jun 23 '15 at 20:55
  • 1
    Believe me that there are infinitely many such "weird functions." One family can be described as follows. Let $p_1,p_2,\ldots,p_k$ be a collection of pairwise distinct positive prime numbers and let $d$ be a positive integer divided by none of the $p_i$'s. Take $\alpha(n)$ to be the product $d$ and all primes $p_i$'s such that $p_i$ divides $n$ (obviously, $\alpha(1)=d$ in this case). Now, tell me: why should we give the functions that satisfy your functional equation any proper names? They don't seem to have any interesting values. – Batominovski Jun 23 '15 at 21:09
  • @Batominovski in fact I completely changed it. Maybe parts of our discussion doesn't make sense to some reader. Sorry for that... – draks ... Jun 29 '15 at 23:06

1 Answers1

2

I would call this function the $N^{th}$-power free part of $n$, which generalizes the $N=2$ case of the squarefree part of an integer.

Specifically, we have that $$\alpha_N(n)=n\prod_{p^{kN}|n}\left(\frac{1}{p^{N}}\right),$$

and $\alpha_N(n)$ removes all of the $N^{th}$ powers that divide $n$ leaving behind the $N^{th}$-power free part. It will share many properties with the squarefree part. For example, $\alpha_N(n)=n$ for any $N^{th}$-power free integer, which is a set of density $\frac{1}{\zeta(N)}$ in the integers. The average of $\alpha_N(n)$ can be computing using the techniques in the answer Mean Value of a Multiplicative Function close to $n$ in Terms of the Zeta Function. Since $$\mu*\frac{\alpha_{N}(n)}{n}=\begin{cases} 1 & k=0\\ 0 & k\neq0\ \text{mod}\ N\\ -\frac{p^{N}-1}{p^{aN}} & k=aN \end{cases} $$ and we find that $$\sum_{n\leq x}\alpha_N(n)=\frac{x^2}{2}\frac{\zeta(2N)}{\zeta(N)}+O(x\log x).$$ (The error term is $O(x)$ for all $N\geq 3$.) In particular, when $N=2$, we find that the squarefree part of $n$ has average $$\sum_{n\leq x}\alpha_2(n)=\sum_{n\leq x} \text{squarefree}(n)=\frac{x^2\pi^2}{30}+O(x\log x).$$ Similarly, the fourth-powerfree part of $n$ has average $$\sum_{n\leq x}\alpha_4(n)=\sum_{n\leq x}\text{fourth-powerfree-part}(n)=\frac{\pi^4x^2}{210}+O(x).$$

Community
  • 1
Eric Naslund
  • 72,099