smc

Let $b_i=19{\text{log}}_2{a}_i$. Then $b_0=0,b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\ge 2$. The characteristic polynomial of this linear recurrence is $x^2-x-2=0$, which has roots $2$ and $-1$. Therefore, $b_n=k_1{2}^n+k_2(-1{)}^n$ for constants to be determined $k_1,k_2$. Using the fact that $b_0=0,b_1=1,$ we can solve a pair of linear equations for $k_1,k_2$: $k_1+k_2=0$ $2k_1-k_2=1$. Thus $k_1=\frac{1}{3}$, $k_2=-\frac{1}{3}$, and $b_n=\frac{2^n-(-1{)}^n}{3}$. Now, $a_1{a}_2\cdots a_k=2^{\frac{(b_1+b_2+\cdots +b_k)}{19}}$, so we are looking for the least value of $k$ so that $b_1+b_2+\cdots +b_k\equiv 0(\mathrm{mod}19)$. Note that we can multiply all $b_i$ by three for convenience, as the $b_i$ are always integers, and it does not affect divisibility by $19$. Now, for all even $k$ the sum (adjusted by a factor of three) is $2^1+2^2+\cdots +2^k=2^{k+1}-2$. The smallest $k$ for which this is a multiple of $19$ is $k=18$ by Fermat's Little Theorem, as it is seen with further testing that $2$ is a primitive root $(\mathrm{mod}19)$. Now, assume $k$ is odd. Then the sum (again adjusted by a factor of three) is $2^1+2^2+\cdots +2^k+1=2^{k+1}-1$. The smallest $k$ for which this is a multiple of $19$ is $k=17$, by the same reasons. Thus, the minimal value of $k$ is $\boxed{17}$.