Increasing sequence containing only finitely many prime numbers Prove the existance of an increasing sequence $(a_n)_{n≥1}$ of positive integers such that
for any $k ≥ 0$, the sequence $k + a_n, n ≥ 1,$ contains only finitely many primes.
 A: Let $p_1 < p_2 < ...$ be the sequence of all primes. $a_1 = 2.$
Inductively, for $n \geqslant 1$, let $ a_{n+1}$ be the least integer greater than $a_n$ that is congruent to $−k \mod p_{k+1}$, for all $k \leqslant n$. We know that such an integer exists by the chinese remainder theorem. We can see that $\forall k \geqslant 0,$
$$k + a_n \equiv 0 \mod p_{k+1} $$ for $n \geqslant k + 1$. $\implies$ at most $k +1$ values in the sequence $k +a_n, n ≥ 1,$ can be prime, as from the $(k + 2)nd$ term onward, the terms of the sequence are nontrivial multiples of $p_{k+1}$, and therefore are composite.
A: Just take $a_n=n!$  or $a_n=n!+2$ to cover the cases $k=0,1$
Clearly $n≥k\implies k\,|\,a_n\implies k\,|\,k+a_n$ so the only possible primes are $\{k+a_1,\cdots,k+a_{k-1}\}$.  This works for $k>1$.
Taking $a_n=n!+2$ effectively replaces $k$ with $k+2$ so we cover $k=0,1$ this way.  Thanks to @didogns for pointing this out.
Note:  you could also take $a_n$ to be the product of the first $n$ primes, by similar logic. Again, add $2$ to cover $k=0,1$.
