Integrality of a certain quantity $\sum_{i =1}^n a_{\lfloor \frac{n}{i}\rfloor }=n^{10}, $ Problem :A sequence $a_1,a_2,\dots$ satisfy 
$$
\sum_{i =1}^n a_{\lfloor \frac{n}{i}\rfloor }=n^{10},
$$
for every $n\in\mathbb{N}$.
Let $c$ be a positive integer. Prove that, for every positive integer $n$,
$$
\frac{c^{a_n}-c^{a_{n-1}}}{n}
$$
is an integer.
I  try :

Note that the required proposition is true if both $a_n\geqslant n$ and $\phi (n)\mid a_n-a_{n-1}$ is true for all positive integer $n>1$.
  From the condition given, we get that $a_1=1$ and 
  \begin{align*}
(n+1)^{10}-n^{10}-1& =\sum_{j=1}^{n}{\left( a_{\lfloor \frac{n+1}{j}\rfloor }-a_{\lfloor \frac{n}{j}\rfloor }\right) }\\
& =\sum_{\substack{j\mid n+1 \\1\leqslant j<n+1}}{\left( a_{\frac{n+1}{j}} -a_{\frac{n+1}{j}-1} \right) } \\
& =\sum_{\substack{d\mid n+1 \\d>1}}{ (a_d-a_{d-1})} .
\end{align*}
  Following work  I can't 

 A: You are in the right direction. Differencing yields
$$\begin{eqnarray}
n^{10} - (n-1)^{10} &=& a_1 +\sum_{1\leq i \leq n-1} (a_{\lfloor \frac{n}{i}\rfloor }-a_{\lfloor \frac{n-1}{i}\rfloor })\\
&=&a_1 + \sum_{i|n, i<n} (a_{\frac{n}{i} }-a_{\frac{n}{i}-1 })\\
&=&a_1 + \sum_{i|n, i>1} (a_{i }-a_{i-1 }) = \sum_{i|n} (a_{i }-a_{i-1 }) 
\end{eqnarray}$$ if we let $a_0 = 0$. Define $b_n = a_n - a_{n-1}$ and $p(n) = n^{10} - (n-1)^{10}$. By Mobius inversion formula, (see https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula) we have
$$
b_n = \sum_{j|n} p(j)\mu(\frac{n}{j}).
$$ What we show first is that $\phi(n) $ divides each $b_n$ for all $n\geq 1.$ To this end, we show the following claim: for each $r\geq 0$, $\sum_{j|n} j^r\mu(\frac{n}{j})$ has $\phi(n)$ as its factor. Proof goes like this. Let $$c_n = \sum_{j|n} j^r\mu(\frac{n}{j}).$$ Observe that $c_1 = 1$, hence $\phi(1) | c_1$. We first show $\phi(n)|c_n$ for $n = p^k$ where $p$ is a prime. This follows easily since
$$
c_{p^k} = p^{rk} - p^{r(k-1)},
$$ and
$$
\phi(p^k) = p^k - p^{k-1}.
$$ (Note that $x-y | x^r - y^r$.) Next we show that $c_n$ is multiplicative, that is, for $p,q$ such that $(p,q)=1$, it holds that $c_{pq} = c_pc_q$. This also follows easily from
$$\begin{eqnarray}
c_{pq}  &=&\sum_{j|pq} j^r\mu(\frac{pq}{j}) \\
&=& \sum_{n|p, m|q} (nm)^r\mu(\frac{pq}{nm})\\
&=&\sum_{n|p, m|q} n^r\mu(\frac{p}{n})\cdot m^r\mu(\frac{q}{m})\\
&=& \sum_{n|p} n^r\mu(\frac{p}{n})\cdot \sum_{m|q}m^r\mu(\frac{q}{m})\\
&=& c_p\cdot c_q.
\end{eqnarray}$$ Since $\phi(n)$ is also multiplicative, these two facts prove the claim.
So far we've shown that for every monomial $j^r$, it holds that $\phi(n) |\sum_{j|n} j^r\mu(\frac{n}{j})$. Thus it holds for any polynomial with integer coefficients, and especially for $p(j)$. This shows
$$
\phi(n) \:|\: b_n =  \sum_{j|n} p(j)\mu(\frac{n}{j}),
$$ as we wanted.
It remains to show $a_n \geq n$. Assume $c\cdot j^{10} \leq a_j \leq j^{10}$ for $j=1,2,\ldots,n-1$ for some $0<c<1$. Then we have
$$\begin{eqnarray}
a_n &=& n^{10} - \sum_{2\leq i \leq n} a_{\lfloor \frac{n}{i}\rfloor }\\
&\geq & n^{10} - n^{10}\sum_{2\leq i \leq n} \frac{1}{i^{10}} \\
&\geq & n^{10} (1-\sum_{2\leq i <\infty} \frac{1}{i^{10}}),
\end{eqnarray}$$ and $a_n \leq n^{10}$. If we let $c = 1-\sum_{i=2}^{\infty}\frac{1}{i^{10}}\in (0.99,1)$, then by induction hypothesis, it holds for every $n\in \mathbb{N}$ once if we prove it for $n=1$. But this is obviously true, since $a_1 = 1$. Finally, we see that
$$
a_n \geq c\cdot n^{10} \geq (c\cdot 2^9)\cdot n >500n
$$ for all $n\geq 2$, establishing $a_n \geq n$.
A: Something maybe helpful:
Denote $b_n=a_n-a_{n-1}$ and $f(n)=n^{10}-(n-1)^{10}$.
For any prime number $p$, we have
$$b_p=a_{p}-a_{p-1}=p^{10}-(p-1)^{10}-1=f(p)-1$$
Notice that for any $n=p^m$ where $m$ is a positive integer :
$$b_{p^m}=f(p^m)-1-\sum_{k=1}^{m-1}{b_{p^k}}$$
then by mathematical induction, we can get that
$$b_{p^m}=f(p^m)-f(p^{m-1})=(p^m)^{10}-(p^m-1)^{10}-(p^{m-1})^{10}+(p^{m-1}-1)^{10}$$
For two different prime number $p,q$, we have
$$b_{pq}+b_p+b_q=f(pq)-1$$
which implies that
$$b_{pq}=f(pq)-f(p)-f(q)+1$$
then by mathematical induction, we can get that
$$b_{p_1p_2\cdots p_k}=\sum^{k}_{i=1}(-1)^{k-i}\sum_{1\leq k_1<k_2<\cdots<k_i\leq k}{f(p_{k_1}p_{k_2}\cdots p_{k_i})}+(-1)^{k}$$
