Pick $a_1, \dots, a_M$ to be a arithmetic progression of primes with $M \ge 2019,$ guaranteed to exist by the Green-Tao Theorem. Then $d(a_i) = 2,$ so $\gcd(k, d(a_1)d(a_2)\dots d(a_{2019})) = \gcd(k, 2^{2019}) = 1.$ Is such a solution permitted? Solving the problem without any heavy machinery would be much harder.

Start with the increasing arithmetic progression $1+i(2019!)$ for $0 \le i \le 2018.$ It is easy to see the terms are pairwise relatively prime. Let $p_1,\dots,p_k$ be the primes dividing the sequence and $q_1,\dots,q_l$ the primes dividing $k$. Since the terms have no divisors in common the exponents of $p_i$ in the factorizations are all $0$ except for one $e_i$. For each $j$ there exists a residue $r_j$ such that none of $0+r_j,e_i+r_j$ are $-1 \pmod{q_j}$ as $q_j$ is odd and thus at least $3$. Then, by the Chinese remainder theorem there is a solution $M_i \equiv r_j \pmod{q_j}$ for each $j$. Therefore the arithmetic progression $(1+i(2019!))\displaystyle\prod_{i=1}^k p_i^{M_i}$ for $0 \le i \le 2018$ satisfies the conditions as the exponents in the prime factorizations of the terms are all $M_i,e_i+M_i$ for $1 \le i \le k$ which we constructed to be different from $-1 \pmod{q_j}$ for each $q_j$ meaning all the $d(a_i)$'s have no prime divisors in common with $k$ as desired.

Problem is very easy if you know Green-Tao Theorem. As skrublord420 stated.

inmemories wrote: Problem is very easy if you know Green-Tao Theorem. As skrublord420 stated. Not sure it was allowed.

Hamel wrote: inmemories wrote: Problem is very easy if you know Green-Tao Theorem. As skrublord420 stated. Not sure it was allowed. It was allowed.