Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a  \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\]for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\]for all integers $n \geq 4$. To the nearest integer, what is $\log_{7}(a_{2019})$?
Using the recursive definition, $a_4 = (4  \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)$ or $a_4 = (4^{m})^{k}$ where $m = \frac{1}{\log_{7}(3)}$ and $k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$. Using logarithm rules, we can remove the exponent of the 3 so that $k = \frac{\log_{7}(3)}{\log_{7}(2)}$. Therefore, $a_4 = 4^{\frac{1}{\log_{7}(2)}}$, which is $4  \, \heartsuit \, 2$. We claim that $a_n = n  \, \heartsuit \, 2$ for all $n \geq 3$. We can prove this through induction. Clearly, the base case where $n = 3$ holds. $a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1)  \, \heartsuit \, 2)$ This can be simplified as $a_n = (n^{\log_{n-1}(7)}) \, \diamondsuit \, ((n-1)^{\log_{2}(7)})$. Applying the diamond operation, we can simplify $a_n = n^h$ where $h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}$. By using logarithm rules to remove the exponent of $\log_{7}(n-1)$ and after cancelling, $h = \frac{1}{\log_{7}(2)}$. Therefore, $a_n = n^{\frac{1}{\log_{7}(2)}} = n  \, \heartsuit \, 2$ for all $n \geq 3$, completing the induction. We have $a_{2019} = 2019^{\log_{2}(7)}$. Taking $\log_{2019}$ of both sides gives us ${\log_{2019}(a_{2019})} = {\log_{2}(7)}$. Then, by changing to base $7$ and after cancellation, we arrive at ${\log_{7}(a_{2019})} = {\log_{2}(2019)}$. Because $2^{11} = 2048$ and $2^{10} = 1024$, our answer is $\fbox{11}$.