# $a_{[n/1]}+a_{[n/2]}+…+a_{[n/n]}=1$

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!

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What are you asking? Find the sequence? – vonbrand Mar 25 '13 at 10:01
@vonbrand:To proof the limit equal to $0$. – Next Mar 25 '13 at 10:02
I calculated the first 16 terms, and that yielded this sequence: oeis.org/A002321. This might be the right one, it might not, but they do at least agree that far. – Arthur Mar 25 '13 at 10:30

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.

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