2016 AMC 12B Problem 25

Write $a_n=2^{\frac{b_n}{19}}.$ The recursion becomes $b_0=0,$ $b_1=1,$ $b_n=b_{n-1}+2b_{n-2},$ solved by $b_n=\tfrac13\bigl(2^n-(-1)^n\bigr).$ The product $a_1\cdots a_k$ is an integer exactly when $b_1+\cdots+b_k$ is divisible by $19.$ Summing the formula for $b_n$ gives $b_1+\cdots+b_k=\dfrac{2^{k+1}-1}{3}$ when $k$ is odd, and $b_1+\cdots+b_k=\dfrac{2^{k+1}-2}{3}$ when $k$ is even. The order of $2$ modulo $19$ is $18$ because $2^9\equiv-1\pmod{19}$ and $2^6\not\equiv1\pmod{19}.$ For odd $k,$ divisibility therefore requires $k+1$ to be divisible by $18,$ first occurring at $k=17.$ For even $k,$ it requires $k$ to be divisible by $18,$ first occurring at $k=18.$ Hence the smallest positive $k$ is $17.$ Thus, the correct answer is A.