Let $n=\prod_{p|n}p^{\alpha_p}$. While considering arithmetic functions of the projection $$\textrm{rad}_k(n)=\prod_{p|n}p^{\textrm{min}(\alpha_p,k)},$$ a basic question has arisen that is more difficult to answer than I had supposed.

Let $a,f:\mathbb{N}\to\mathbb{C}$ and define $$Af(n)=\sum_{d|n}a(d)f(d)$$

so that $$A^2f(n)=\sum_{d|n}a(d)\sum_{e|d}a(e)f(e)$$ and so on. The question is simply this:

If $A^{k+1}=A$ for some $k\in\mathbb{N}$ and $a(1)=1$, does it follow that $a(n)$ is multiplicative?

This is certainly the case if $k=1$ or $k=2$ in which cases $a(n)$ is $\delta(n)$ and $\mu(n)$ respectively. Multiplicative solutions also exist for every $k$; these are the functions $$a(n)=\delta_k(n)z^{\Omega(n)}$$ where $z$ is a primitive $k$th root of unity and $\delta_k(n)=0,1$ depending on whether $n$ is divisible by a $k$th power or not. There are $\phi(k)$ primitive roots, and an affirmative answer to the question would mean that accounts for all solutions.

Interchanging the order of summation, the statement says that $$a(m)=a(m)\sum_{d_1|n}a(d_1m)\cdots\sum_{d_k|d_{k-1}}a(d_km)$$ for all $m,n$ from which one can infer that $$a^{k+1}(n)=a(n)$$ for all $n$ and $$a(m)a(n)=a(mn)$$ whenever $a(m)$, $a(n)$ and $a(mn)$ are all non-zero.

From the stated property of $A$ we have that $A^k$ is a projection and when $a(n)$ is multiplicative that $$A^kf(n)=f(\textrm{rad}_{k-1}(n)).$$

Is every solution as such?

Let $n=\prod_{p|n}p^{\alpha_p}$. While considering arithmetic functions of the projection $$\textrm{rad}_k(n)=\prod_{p|n}p^{\textrm{min}(\alpha_p,k)},$$ a basic question has arisen that is more difficult to answer than I had supposed.

Let $a,f:\mathbb{N}\to\mathbb{C}$ and define $$Af(n)=\sum_{d|n}a(d)f(d)$$

so that $$A^2f(n)=\sum_{d|n}a(d)\sum_{e|d}a(e)f(e)$$ and so on. The question is simply this:

If $A^{k+1}=A$ for some $k\in\mathbb{N}$ and $a(1)=1$, does it follow that $a(n)$ is multiplicative?

Multiplicative solutions exist for every $k$; these are the functions $$a(n)=\delta_k(n)z^{\Omega(n)}$$ where $z$ is a primitive $k$th root of unity and $\delta_k(n)=0,1$ depending on whether $n$ is divisible by a $k$th power or not. There are $\phi(k)$ primitive roots, and an affirmative answer to the question would mean that accounts for all solutions.

Interchanging the order of summation, the statement says that $$a(m)=a(m)\sum_{d_1|n}a(d_1m)\cdots\sum_{d_k|d_{k-1}}a(d_km)$$ for all $m,n$ from which one can infer that $$a^{k+1}(n)=a(n)$$ for all $n$ and $$a(m)a(n)=a(mn)$$ whenever $a(m)$, $a(n)$ and $a(mn)$ are all non-zero.

From the stated property of $A$ we have that $A^k$ is a projection and when $a(n)$ is multiplicative that $$A^kf(n)=f(\textrm{rad}_{k-1}(n)).$$

Is every solution as such?

Let $n=\prod_{p|n}p^{\alpha_p}$. While considering arithmetic functions of the projection $$\textrm{rad}_k(n)=\prod_{p|n}p^{\textrm{min}(\alpha_p,k)},$$ a basic question has arisen that is more difficult to answer than I had supposed.

Let $a,f:\mathbb{N}\mapsto\mathbb{C}$ and define $$Af(n)=\sum_{d|n}a(d)f(d)$$

so that $$A^2f(n)=\sum_{d|n}a(d)\sum_{e|d}a(e)f(e)$$ and so on. The question is simply this:

If $A^{k+1}=A$ for some $k\in\mathbb{N}$ and $a(1)=1$, does it follow that $a(n)$ is multiplicative?

This is certainly the case if $k=1$ or $k=2$ in which cases $a(n)$ is $\delta(n)$ and $\mu(n)$ respectively. Multiplicative solutions also exist for every $k$; these are the functions $$a(n)=\delta_k(n)z^{\Omega(n)}$$ where $z$ is a primitive $k$th root of unity and $\delta_k(n)=0,1$ depending on whether $n$ is divisible by a $k$th power or not. There are $\phi(k)$ primitive roots, and an affirmative answer to the question would mean that accounts for all solutions.

Interchanging the order of summation, the statement says that $$a(m)=a(m)\sum_{d_1|n}a(d_1m)\cdots\sum_{d_k|d_{k-1}}a(d_km)$$ for all $m,n$ from which one can infer that $$a^{k+1}(n)=a(n)$$ for all $n$ and $$a(m)a(n)=a(mn)$$ whenever $a(m)$, $a(n)$ and $a(mn)$ are all non-zero.

From the stated property of $A$ we have that $A^k$ is a projection and when $a(n)$ is multiplicative that $$A^kf(n)=f(\textrm{rad}_{k-1}(n)).$$

Is every solution as such?

Let $n=\prod_{p|n}p^{\alpha_p}$. While considering arithmetic functions of the projection $$\textrm{rad}_k(n)=\prod_{p|n}p^{\textrm{min}(\alpha_p,k)},$$ a basic question has arisen that is more difficult to answer than I had supposed.

Let $a,f:\mathbb{N}\to\mathbb{C}$ and define $$Af(n)=\sum_{d|n}a(d)f(d)$$

so that $$A^2f(n)=\sum_{d|n}a(d)\sum_{e|d}a(e)f(e)$$ and so on. The question is simply this:

If $A^{k+1}=A$ for some $k\in\mathbb{N}$ and $a(1)=1$, does it follow that $a(n)$ is multiplicative?

This is certainly the case if $k=1$ or $k=2$ in which cases $a(n)$ is $\delta(n)$ and $\mu(n)$ respectively. Multiplicative solutions also exist for every $k$; these are the functions $$a(n)=\delta_k(n)z^{\Omega(n)}$$ where $z$ is a primitive $k$th root of unity and $\delta_k(n)=0,1$ depending on whether $n$ is divisible by a $k$th power or not. There are $\phi(k)$ primitive roots, and an affirmative answer to the question would mean that accounts for all solutions.

Interchanging the order of summation, the statement says that $$a(m)=a(m)\sum_{d_1|n}a(d_1m)\cdots\sum_{d_k|d_{k-1}}a(d_km)$$ for all $m,n$ from which one can infer that $$a^{k+1}(n)=a(n)$$ for all $n$ and $$a(m)a(n)=a(mn)$$ whenever $a(m)$, $a(n)$ and $a(mn)$ are all non-zero.

From the stated property of $A$ we have that $A^k$ is a projection and when $a(n)$ is multiplicative that $$A^kf(n)=f(\textrm{rad}_{k-1}(n)).$$

Is every solution as such?