23. Define the operations: $a \bigcirc b=a^{\log _{7} b}, a \otimes b=a^{\frac{1}{\log ^{6} b}}$, where $a, b \in \mathbf{R}$, and the sequence $\left\{a_{n}\right\}(n \geqslant 4)$ satisfies:
$$
a_{3}=3 \otimes 2, a_{n}=(n \otimes(n-1)) \bigcirc a_{n-1} \text {. }
$$

Then the integer closest to $\log _{7} a_{2019}$ is ( ).
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
23. D.

From the problem, we know
$$
\begin{array}{l}
a_{3}=3^{\frac{1}{\log _{7} 2}}, a_{n}=\left(n^{\frac{1}{\log _{7}(n-1)}}\right)^{\log _{7} a_{n-1}}=n^{\log _{n-1} a_{n-1}} . \\
\text { Then } \log _{n} a_{n}=\log _{n-1} a_{n-1}=\cdots \\
=\log _{3} a_{3}=\frac{1}{\log _{7} 2}=\log _{2} 7 \\
\Rightarrow a_{n}=n^{\log _{2} 7} . \\
\text { Therefore, } \log _{7} a_{2019}=\log _{7} 2019^{\log _{2} 7} \\
=\frac{\log _{7} 2019}{\log _{7} 2}=\log _{2} 2019 \\
<\log _{2} 2048=11 .
\end{array}
$$

Therefore, the integer closest to $\log _{7} a_{2019}$ is 11.