Conjecture: Given any $k\in\Bbb{N}$ there exists $k$ consecutive composite integers with the property that reciprocal of each composite integer is Cantor points then $k=1$.
$x$ is a Cantor point iff $x\in\mathcal{C}$$($Cantor's middle-third set$)$ i.e $x=\sum_{n\in\Bbb{N}} \frac{a_n}{3^n}$ where $a_n\in\{0, 2\}$.In other words base-$3$ expansion or ternary expansion of $x$ contains only $0$ and $2$ and if contain $1$ then it can be replaced by a strings of $2$'s.
Attempt:
For $k\ge 1$ , the set $S_k=\{(k+1)!+j:2\le j\le k+1\}$ is a set of $k$ consecutive composite integers.
For $k=1$ , $(1+1)! +2=4$ and $\frac{1}{4}=(0.\overline{02})_3$
For $k= 2$ , $S_2=\{8,9\}$
But $\frac{1}{8}=(0.\overline{01})_3\notin\mathcal{C}$ and $\frac{1}{9}=(0.{01})_3\in\mathcal{C}$
$1)$Can we construct a different set $S_k$ which satisfy all requirements? Or $2)$ there is no $m, n\in\Bbb{N}$ consecutive composite number such that both $\frac{1}{m}, \frac{1}{n}\in\mathcal{C}$ ?
Answer to the second question is negative$[(m, n)=(9, 10) ]$
No three consecutive integers have their reciprocals in the Cantor set.
Proof: Suppose $n \equiv 2 \pmod {3}$. Then $\frac{1}{n}$ has a purely periodic base-$3$ expansion: $$\frac{1}{n} = \frac{P}{3^r} + \frac{P}{3^{2r}} + \frac{P}{3^{3r}}+ \ldots = \frac{P}{3^r-1}$$ for some positive integers $r$ and $P < 3^r$. But then $$P = \frac{3^r-1}{n} \equiv 1 \pmod{3},$$ so the base $3$ expansion of $\frac{1}{n}$ has a $1$ in it. So for $n\equiv 2\pmod{3}$, $\frac{1}{n}$ is not in the Cantor set.