Jtdylktuy

Problem :A sequence $a_1,a_2,dots$ satisfy
$$
sum_{i =1}^n a_{lfloor frac{n}{i}rfloor }=n^{10},
$$
for every $ninmathbb{N}$.
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 $a_ngeqslant n$ 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{jmid n+1 \1leqslant j<n+1}}{left( a_{frac{n+1}{j}} -a_{frac{n+1}{j}-1} right) } \
& =sum_{substack{dmid n+1 \d>1}}{ (a_d-a_{d-1})} .
end{align*}
Following work I can't

You are in the right direction. Differencing yields
$$begin{eqnarray}
n^{10} - (n-1)^{10} &=& a_1 +sum_{1leq i leq n-1} (a_{lfloor frac{n}{i}rfloor }-a_{lfloor frac{n-1}{i}rfloor })\
&=&a_1 + sum_{i|n, i<n} (a_{frac{n}{i} }-a_{frac{n}{i}-1 })\
&=&a_1 + sum_{i|n, i>1} (a_{i }-a_{i-1 }) = sum_{i|n} (a_{i }-a_{i-1 })
end{eqnarray}$$ 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 $ngeq 1.$ To this end, we show the following claim: for each $rgeq 0$, $sum_{j|n} j^rmu(frac{n}{j})$ has $phi(n)$ as its factor. Proof goes like this. Let $$c_n = sum_{j|n} j^rmu(frac{n}{j}).$$ 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
$$begin{eqnarray}
c_{pq} &=&sum_{j|pq} j^rmu(frac{pq}{j}) \
&=& sum_{n|p, m|q} (nm)^rmu(frac{pq}{nm})\
&=&sum_{n|p, m|q} n^rmu(frac{p}{n})cdot m^rmu(frac{q}{m})\
&=& sum_{n|p} n^rmu(frac{p}{n})cdot sum_{m|q}m^rmu(frac{q}{m})\
&=& c_pcdot c_q.
end{eqnarray}$$ 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 $phi(n) |sum_{j|n} j^rmu(frac{n}{j})$. 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 $ccdot j^{10} leq a_j leq j^{10}$ for $j=1,2,ldots,n-1$ for some $0<c<1$. Then we have
$$begin{eqnarray}
a_n &=& n^{10} - sum_{2leq i leq n} a_{lfloor frac{n}{i}rfloor }\
&geq & n^{10} - n^{10}sum_{2leq i leq n} frac{1}{i^{10}} \
&geq & n^{10} (1-sum_{2leq i <infty} frac{1}{i^{10}}),
end{eqnarray}$$ 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 $nin mathbb{N}$ once if we prove it for $n=1$. But this is obviously true, since $a_1 = 1$. Finally, we see that
$$
a_n geq ccdot n^{10} geq (ccdot 2^9)cdot n >500n
$$ for all $ngeq 2$, establishing $a_n geq n$.

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
$$b_{p_1p_2cdots p_k}=sum^{k}_{i=1}(-1)^{k-i}sum_{1leq k_1<k_2<cdots<k_ileq k}{f(p_{k_1}p_{k_2}cdots p_{k_i})}+(-1)^{k}$$