Just use $n!-1,2(n!)-1,\cdots, n(n!)-1$.

See here.....

gghx wrote: Just use $n!-1,2(n!)-1,\cdots, n(n!)-1$. $\forall i: a_i \ge 1402$

gghx wrote: Just use $n!-1,2(n!)-1,\cdots, n(n!)-1$. Following this hint we let $$a_i = i\cdot k! - 1$$and choose $k$ sufficiently large such that $k > n$ and $a_i \geq 1402$ for all $1 \leq i \leq n$. Check that for all $i > j$ $$(a_i, a_j) = (i\cdot k! - 1, j\cdot k! - 1) = ((i - j)\cdot k!, \, j\cdot k! - 1) = (i-j,\, j\cdot k! - 1) = 1$$where the last equality follows from the fact that $i - j < n < k$ and $i-j \mid k!$. Then for $1 \leq i \leq n-1$ we have $$0 < \frac{a_{i}}{a_{i+1}} = \frac{i\cdot k! - 1}{(i+1)\cdot k! - 1} < 1$$which makes $\Big\lfloor \frac{a_{i}}{a_{i+1}} \Big\rfloor = 0$ for all such $i$. Since $$\frac{a_{n}}{a_{1}} = \frac{n\cdot k! - 1}{k! - 1} = n + \frac{n - 1}{k! - 1} \in (n, n+1)$$we get $\Big\lfloor \frac{a_{n}}{a_{1}} \Big\rfloor = n$. So, the LHS of the given equation is $n$. Next, observe that for $2 \leq i \leq n-1$ we have $$\frac{a_{i + 1}}{a_{i}} = \frac{(i + 1)\cdot k! - 1}{i\cdot k! - 1} = 1 + \frac{k! }{i\cdot k! - 1} \in (1, 2)$$which makes $\Big\lfloor \frac{a_{i + 1}}{a_{i}} \Big\rfloor = 1$ for all such $i$. Since $$0 < \frac{a_{1}}{a_{n}} = \frac{k! - 1}{n\cdot k! - 1} < 1$$we get $\Big\lfloor \frac{a_{1}}{a_{n}} \Big\rfloor = 0$. Finally, note that $$\frac{a_{2}}{a_{1}} = \frac{2\cdot k! - 1}{k! - 1} = 2 + \frac{1}{k! - 1} \in (2, 3)$$and this gives us $\Big\lfloor \frac{a_{2}}{a_{1}} \Big\rfloor = 2$. Combining all these we get that the RHS of the given equation is also $n$.

Also it could be solved by using the fact that for each $2\leq N $ there exist a prime number between $N$ and $2N$. (known as Chebyshev's theorem)

Seems like no one mentioned it before. There exists a solution by induction, if someone is interested in, he/she can post the solution by induction.

cefer wrote: $((i - j)\cdot k!, \, j\cdot k! - 1) = (i-j,\, j\cdot k! - 1) $ . What's the reason for that?

AgentC wrote: cefer wrote: $((i - j)\cdot k!, \, j\cdot k! - 1) = (i-j,\, j\cdot k! - 1) $ . What's the reason for that? Because $k!$ divides $j\cdot k!$ which means $(k!, j\cdot k! - 1) = 1$.