Let $$$m$$$ be an arbitrary positive integer. We construct an array $$${a}$$$ where $$$\forall n\in\mathbb{N}_+,\sum_{k\mid n}a_k=m^n$$$. Please prove: $$$\forall n\in\mathbb{N}_+,n\mid a_n$$$.
My first thought is to use Mobius inversion to translate the definition of $$${a}$$$ into $$$a_n=\sum_{k\mid n}m^k\mu(\frac nk)$$$. Then I tried to prove: Let $$$n=p_1^{c_1}p_2^{c^2}\cdots p_k^{c_k}$$$ be the unique factorization of $$$n$$$, for all $$$1\le i\le k$$$ the condition $$$p_i^{c_i}\mid a_n$$$ holds. But I made a little mistake. I forgot the other powers of $$$p_i$$$.
So who can prove this statement? Please comment here, thanks!
Is there an English translation?
$$$\forall n\in\mathbb{N}_+,\sum_{k\mid n}a_k=m^n$$$: For every positive integer $$$n$$$, the sum of $$$a_k$$$ where $$$k$$$ is a divisor of $$$n$$$ is $$$m^n$$$.
Oh ok, I see. So if $$$m = 1$$$, the array would be like $$$[1, 0, 0, 0, 0, 0, \cdots]$$$. I'm not sure if this is helpful, but I think that that looks pretty similar to the mobius function.
$$$a_k$$$ is exactly the numbers of arrays $$$b_1,b_2,\cdots,b_k\in [1,m]\cap \mathbb Z$$$ where no $$$d|k,d<k$$$ is a period of $$$b$$$, and is therefore a multiple of $$$k$$$.