solution: Let $\frac{a_2}{a_1} = \frac{b}{a},$ where $a$ and $b$ are relatively prime positive integers, and $a < b.$  Then $a_2 = \frac{b}{a} \cdot a_1,$ and
  \[a_3 = \frac{a_2^2}{a_1} = \frac{(b/a \cdot a_1)^2}{a_1} = \frac{b^2}{a^2} \cdot a_1.\]This implies $a_1$ is divisible by $a^2.$  Let $a_1 = ca^2$; then $a_2 = cab,$ $a_3 = cb^2,$
  \begin{align*}
  a_4 &= 2a_3 - a_2 = 2cb^2 - cab = cb(2b - a), \\
  a_5 &= \frac{a_4^2}{a_3} = \frac{[cb(2b - a)]^2}{(cb^2)} = c(2b - 2a)^2, \\
  a_6 &= 2a_5 - a_4 = 2c(2b - a)^2 - cb(2b - a) = c(2b - a)(3b - 2a), \\
  a_7 &= \frac{a_6^2}{a_5} = \frac{[c(2b - a)(3b - 2a)]^2}{c(2b - a)^2} = c(3b - 2a)^2, \\
  a_8 &= 2a_7 - a_6 = 2c(3b - 2a)^2 - c(2b - a)(3b - 2a) = c(3b - 2a)(4b - 3a), \\
  a_9 &= \frac{a_8^2}{a_7} = \frac{[c(3b - 2a)(4b - 3a)]^2}{[c(3b - 2a)^2} = c(4b - 3a)^2,
  \end{align*}and so on.
  
  More generally, we can prove by induction that
  \begin{align*}
  a_{2k} &= c[(k - 1)b - (k - 2)a][kb - (k - 1)a], \\
  a_{2k + 1} &= c[kb - (k - 1)a]^2,
  \end{align*}for all positive integers $k.$
  
  Hence, from $a_{13} = 2016,$
  \[c(6b - 5a)^2 = 2016 = 2^5 \cdot 3^2 \cdot 7 = 14 \cdot 12^2.\]Thus, $6b - 5a$ must be a factor of 12.
  
  Let $n = 6b - 5a.$  Then $a < a + 6(b - a) = n,$ and
  \[n - a = 6b - 6a = 6(b - a),\]so $n - a$ is a multiple of 6.  Hence,
  \[6 < a + 6 \le n \le 12,\]and the only solution is $(a,b,n) = (6,7,12).$  Then $c = 14,$ and $a_1 = 14 \cdot 6^2 = \boxed{504}.$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  A strictly increasing sequence of positive integers $a_1$, $a_2$, $a_3$, $\dots$ has the property that for every positive integer $k$, the subsequence $a_{2k-1}$, $a_{2k}$, $a_{2k+1}$ is geometric and the subsequence $a_{2k}$, $a_{2k+1}$, $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$. Find $a_1$.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 504
extra_info.question: A strictly increasing sequence of positive integers $a_1$, $a_2$, $a_3$, $\dots$ has the property that for every positive integer $k$, the subsequence $a_{2k-1}$, $a_{2k}$, $a_{2k+1}$ is geometric and the subsequence $a_{2k}$, $a_{2k+1}$, $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$. Find $a_1$.
extra_info.split: test