Let WLOG $a_{21}$ be the largest number. By plugging $(a_{21}, a_i)$ for $i=1,2,...,20$ and supposing otherwise we obtain $lcm(a_1, a_2,..., a_{20})/a_{21}$, so $2^{20} \leq a_{21}$, size contradiction. @below, ok you're right. We have to prove that $a_k/a_{k+1}$ (otherwise $F(a_k, a_{k+1})$ is between $a_k$ and $a_{k+1}$)

VicKmath7 wrote: Let WLOG $a_{21}$ be the largest number. By plugging $(a_{21}, a_i)$ for $i=1,2,...,20$ and supposing otherwise we obtain $lcm(a_1, a_2,..., a_{20})/a_{21}$, so $2^{20} \leq a_{21}$, size contradiction. $\mathrm{lcm}$ of 20 different numbers can be less than $2^{20}$. for example, the 30 divisors of $2^5\cdot 3^4=2592$