$a_{[n/1]}+a_{[n/2]}+...+a_{[n/n]}=1$

Question

The sequence $a_n$ satisfy $$a_{[n/1]}+a_{[n/2]}+...+a_{[n/n]}=1,$$ for all $n \in \Bbb N$. (the subscript $[n/k]$ is the integer part of $n/k$)

$Proof:$for any $k>0$,$$\lim_{n \rightarrow \infty} \frac{a_n}{n^{1/2+k}} = 0$$

Thanks!

Answer 1

Here is a proof of Arthur's observation :

Let $m_n = a_n - a_{n-1}$ and $m_1 = 1$

Suppose $n>1$. We have $0 = 1 - 1 = \sum_{k=1}^{n-1} a_{\left[\frac{n}k\right]} - a_{\left[\frac{n-1}k\right]} + a_1$. If $k$ divides $n$ then $\left[\frac{n}k\right] = \left[\frac {n-1}k\right] + 1$. If not, then $\left[\frac{n}k\right] = \left[\frac{n-1}k\right]$. Hence we get $0 = \sum_{n = kd, d>1} a_d - a_{d-1} + a_1 = \sum_{d \mid n} m_d$.
For $n=1$, $\sum_{d \mid n} m_d = m_1 = 1$.

Here, we recognize that this recurrence is exactly the one satisfied by the Möbius sequence : $m(n) = \mu(n)$, and so $a_n$ is the Mertens sequence.

Now your question is to show that $\forall \epsilon > 0, \sum_{k=1}^n \mu(k) = o(n^{1/2+\epsilon})$, which is equivalent to the Riemann Hypothesis.
Good luck trying to prove this.