Prove that for every sufficiently large $n$, we can write the positive integer $n$ in the form $n = a_1+ a_2+...+a_{2018}$ with $: (a_i , a_j ) = 1 $ Prove that for every sufficiently large $n$, we can write the positive integer $n$ in the form $n = a_1+ a_2+...+a_{2018}$
In there $: (a_i , a_j ) = 1 $ for all $i<j$ and $a_i>2018$ for all $i$
Instead of trying with $2018 ,$ I tried with $ 4 , 5 ,6 .$ And as a result, I get the prediction that: for every large enough $ n,$ we can construct $n$ as the sum of $2017$ primes and some number. (I'm pretty sure this prediction is correct since I tested over $100$ numbers.)
But I have not proven my prediction yet. I hope to get help from everyone. Thanks very much !
 A: I enjoyed this interesting problem. I construct the solution in the form:
$a_k = {b_k}^{c_k} (1\leq k \leq 2017), a_{2018}=n-\sum_{k=1}^{2017} a_k$.
Define $b_1=3, b_2=5$.
For $3\leq k$, first define $d_{k-1}$ as the least positive integer such that ${b_{k-1}}^{d_{k-1}} \equiv 1 \pmod{\prod_{j=1}^{k-2} b_j}$, and then define $b_k$ by choosing any prime satisfying following:
(1) coprime to ${b_{k-1}}^{d_{k-1}} - 1$
(2) coprime to $\prod_{j=1}^{k-1} {b_k}$
(Explicitly, $d_2=2$ and I can choose $b_3=7$ because $5^2 \equiv 1 \pmod 3$ and $7$ is coprime to $5^2-1$ and $15$. Then, $d_3$ will be $4$ as $7^4 \equiv 1 \pmod {15}$ and so on.)
By the condition (2), $b_k$ are coprime each other, so I only need to adjust $a_{2018}$ to be coprime to $b_k$.
First, let $c_k$ be the least positive integer such that ${b_k}^{c_k} > 2018$.
Then, adjust $c_k$ as follows:
If $3|a_{2018}$, add $1$ to $c_2$. This changes $a_{2018} \bmod 3$.
If $5|a_{2018}$, add $1$ to $c_1$. This preserves $a_{2018} \bmod 3$ and changes $a_{2018} \bmod 5$. 
For each $3 \leq k$, if $b_k|a_{2018}$, add $d_{k-1}$ to $c_{k-1}$.
This preserves $a_{2018} \bmod b_{k-1}$.
By the setting of $d_{k-1}$ this also preserves $a_{2018}\bmod b_j$ for $j\leq k-2$.
However, by the condition (1) for defining $b_k$, this changes $a_{2018} \bmod b_k$.
So, performing the procedure consecutively, I can make $a_{2018}$ coprime to all $b_{k}$.
The change of $c_k$ is bounded, so I can assure $a_{2018}>2018$ for sufficiently large $n$.
(Aside:
If you can use Dirichlet's theorem, setting $b_k$ be the prime satisfying $b_k \equiv 1 \pmod{ \prod_{j=1}^{k-1} {b_j} }$ will make the discussion simpler because $d_k=1$.)
