Hints: The 67th OEIS Problem
Answer to Hint 1: There are $n-1$ consecutive pairs (for $n\ge 2$; for $n=1$ there is no pair to constrain). You need $n-1$ pairwise distinct positive integers $\gcd(a_i,a_{i+1})$.
Would it help if each $\gcd(a_i,a_{i+1})$ were forced to equal a specific integer you choose in advance?
Answer to Hint 2: Yes — a clean strategy is to design $a_i$ so that $\gcd(a_i,a_{i+1})$ is exactly some “marker” $d_i$, and to pick the markers $d_1,\ldots,d_{n-1}$ all different.
A convenient way to control $\gcd(a_i,a_{i+1})$ is to share factors between neighbors. What happens if both $a_i$ and $a_{i+1}$ are divisible by the same prime, but you arrange other factors so that the shared part is easy to read off?
Answer to Hint 3: If you write each term as a product of two factors, you can try to overlap exactly one factor between $a_i$ and $a_{i+1}$, and choose those overlaps to be different from step to step.
Consider odd integers $1,3,5,7,\ldots$ They are spaced two apart. Look at products of two consecutive odd numbers, e.g. something of the form $(2i+1)(2i+3)$. What might $\gcd$ of two neighboring such products pick out?
Answer to Hint 4: Neighboring products share the factor $(2i+3)$: one term is $(2i+1)(2i+3)$ and the next is $(2i+3)(2i+5)$.
Compute $\gcd\bigl((2i+1)(2i+3),\,(2i+3)(2i+5)\bigr)$ using $\gcd$ rules (for example, factor out $2i+3$ and note that consecutive odd numbers are coprime).
Are the results for $i=1,2,\ldots$ all distinct?
Answer to Hint 5: You get $\gcd = 2i+3$ for each $i$, and the values $3,5,7,\ldots$ are strictly increasing, hence distinct.
Taking $a_i = (2i+1)(2i+3)$ for $i=1,\ldots,n$ satisfies the gcd-distinctness condition; check that each $a_i$ lies in the allowed value range for the problem’s constraints.