Problem #291: Let $n\geq 1$ and define $L_n$ to be the least common...

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Progress

Steinerberger has observed that the answer to the second question is trivially yes: for example, any $n$ which begins with a $2$ in base $3$ has $3\mid (a_n,L_n)$.

More generally, if the leading digit of $n$ in base $p$ is $p-1$ then $p\mid (a_n,L_n)$. There is in fact a necessary and sufficient condition: a prime $p\leq n$ divides $(a_n,L_n)$ if and only if $p$ divides the numerator of $1+\cdots+\frac{1}{k}$, where $k$ is the leading digit of $n$ in base $p$. This can be seen by writing\[a_n = \frac{L_n}{1}+\cdots+\frac{L_n}{n}\]and observing that the right-hand side is congruent to $1+\cdots+1/k$ modulo $p$. (The previous claim about $p-1$ follows immediately from Wolstenholme's theorem.)

This leads to a heuristic prediction (see for example a preprint of Shiu [Sh16]) of $\asymp\frac{x}{\log x}$ for the number of $n\in [1,x]$ such that $(a_n,L_n)=1$. In particular, there should be infinitely many $n$, but the set of such $n$ should have density zero. Unfortunately this heuristic is difficult to turn into a proof.

Wu and Yan [WuYa22] have proved, conditional on $\frac{1}{\log p}$ being linearly independent over $\mathbb{Q}$ for any finite collection of primes $p$ (itself a consequence of Schanuel's conjecture), that the set of $n$ for which $(a_n,L_n)>1$ has upper density $1$.