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
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 $a_n \geq n$.