Are there infinitely many $n$ with $\binom{2n}{n}$ coprime to $105$? (Erdős #376)

Let $\binom{2n}{n}$ be the central binomial coefficient. Are there infinitely many positive integers $n$ with $\gcd\!\left(\binom{2n}{n},\,105\right)=1$ (equivalently, $\binom{2n}{n}$ divisible by none of $3,5,7$)? By Kummer's theorem a prime $p$ does not divide $\binom{2n}{n}$ iff every base-$p$ digit of $n$ is at most $(p-1)/2$. So the condition is: $n$ has all base-$3$ digits in $\{0,1\}$, all base-$5$ digits in $\{0,1,2\}$, and all base-$7$ digits in $\{0,1,2,3\}$, simultaneously.

Acceptance. FULLY RESOLVES is a proof (count unbounded, or bounded), ideally via a finite-automaton / digit-DP argument over the simultaneous base-$3,5,7$ digit constraints. PARTIAL and highly informative: compute $\#\{n\le N:\gcd(\binom{2n}{n},105)=1\}$ for large $N$ using the digit characterisation (an $O(\log N)$-state digit-DP, no big binomials), report the largest known such $n$ and the counting-function growth. Provide the enumeration code.