On $\frac{1}{n!}$ in Cantor sets

This paper proves that the only elements of the form $1/n!$contained in the middle-third Cantor set are$1$and$1/5!$, resolving a recent question by Jiang and extending the result to show that any missing-digit set contains only finitely many such elements.

The Gist

Imagine you have a very strange, dusty shelf called the Cantor Set.

To build this shelf, you start with a long wooden plank (the number line from 0 to 1). You chop out the middle third, leaving two smaller planks. Then, you chop the middle third out of those planks. You keep doing this forever. What's left is the Cantor Set: a collection of tiny, scattered dust motes that somehow still add up to a specific shape.

Now, imagine you have a giant bag of special marbles. These marbles are labeled with numbers like $1/1!$,$1/2!$,$1/3!$,$1/4!$, and so on. These are "factorial fractions." They get smaller and smaller very quickly:

• $1/1! = 1$
• $1/2! = 1/2$
• $1/3! = 1/6$
• $1/4! = 1/24$
• $1/5! = 1/120$
• ...and so on.

The Big Question:
If you dump all these factorial marbles onto your dusty Cantor shelf, which ones will actually land on the wood and stay there? Which ones will fall through the gaps?

For a long time, mathematicians knew that the marble labeled 1 (the very first one) and the marble labeled 1/120 (which is $1/5!$) would land safely on the shelf. But nobody knew if any of the other thousands of tiny marbles ($1/6!$,$1/7!$, etc.) would also fit.

The Discovery:
The authors of this paper, Kehao Lin, Yufeng Wu, and Siyu Yang, acted like detectives. They wanted to solve the mystery: "Are there any other factorial marbles that fit on the Cantor shelf?"

They proved a surprising answer: No.
The only two marbles that fit are 1 and 1/120. Every single other factorial marble, no matter how small, falls through the cracks of the Cantor Set.

How Did They Solve It? (The Detective Work)

To understand their method, let's use an analogy of secret codes and patterns.

1. The Language of the Shelf:
   The Cantor Set has a strict rule for how numbers are written. In our normal math, we write numbers in base 10 (using digits 0-9). The Cantor Set only allows numbers written in "base 3" (using digits 0, 1, and 2), but with a twist: it strictly forbids the digit '1'.

   • If a number needs a '1' in its base-3 code to be written down, it doesn't belong on the shelf. It falls through the gap.

2. The Factorial Problem:
   The team looked at the numbers $1/n!$. They realized that as$n$ gets bigger, these numbers become incredibly complex. Their "base-3 codes" become long, messy strings of digits.

   The researchers used a powerful mathematical tool (called Korobov's Lemma) which acts like a pattern detector. This tool tells us how often a specific digit (like '1') appears in the code of a fraction.

   • The Logic: If a number belongs to the Cantor Set, the digit '1' must never appear in its code.
   • The Conflict: The pattern detector showed that for very large factorials ($n \ge 21$), the digit '1' must appear in the code. It's mathematically impossible for the code to avoid the forbidden digit once the number gets that big.

3. The "Too Big" Threshold:
   They calculated a specific cutoff point (around $n=21$). They proved that for any factorial larger than this, the "pattern detector" screams: "This number definitely contains a forbidden '1'!"

   So, they didn't need to check every single number up to infinity. They just had to check the small ones (from $n=1$ to $n=20$) manually.

   • They checked $1/1!$ (Fits! No '1's in the code).
   • They checked $1/2!$through$1/4!$ (Fails! They have '1's).
   • They checked $1/5!$ (Fits! No '1's).
   • They checked $1/6!$through$1/20!$ (All fail!).

The Bigger Picture

This paper didn't just solve the puzzle for the standard Cantor Set. The authors showed that this logic works for any "missing-digit" shelf.

Imagine a shelf that only allows digits 0 and 4 in base 5, or 0, 2, and 7 in base 10. The authors proved that for any of these weird shelves, there are only a finite number of factorial marbles that will ever fit. You can always write a computer program to find them all, and the list will never be infinite.

Summary in Plain English

• The Puzzle: Which tiny fractions of the form $1/n!$ fit inside a specific fractal shape (the Cantor Set)?
• The Answer: Only two of them: 1 and 1/120.
• The Method: They proved that for large numbers, the mathematical "fingerprint" of these fractions inevitably contains a forbidden digit, making it impossible for them to fit.
• The Takeaway: Even in the infinite, chaotic world of fractals, there are strict rules that limit how many special numbers can hide inside.

Technical Summary

1. Problem Statement

The paper addresses a specific question regarding the intersection of the set of reciprocals of factorials with the middle-third Cantor set ($C$).

• Context: Determining the intersection $S \cap K$ between a sequence $S$ and a fractal set $K$ (specifically Cantor sets) is a difficult problem with connections to the open Erdős similarity problem.
• Specific Question: Recently, Jiang et al. [2] asked to determine the exact set:
  $$\left\{ \frac{1}{n!} : n \in \mathbb{N} \right\} \cap C$$
  where $C$ is the standard middle-third Cantor set. It was already known that $1$and$1/5!$ belong to this intersection. The goal was to determine if there are any other elements.
• Generalization: The authors extend this problem to missing-digit sets $K_{m,D}$, defined as numbers in base $m$ whose digits belong to a subset $D \subset \{0, 1, \dots, m-1\}$ (where $1 < |D| < m$). The goal is to characterize the intersection of${1/n!}$ with any such set.

2. Methodology

The authors employ a combination of p-adic valuation, multiplicative order theory, and digit distribution estimates in base expansions.

A. Reduction to Coprime Denominators

Since the middle-third Cantor set is invariant under the map $x \mapsto 3x \pmod 1$, if $1/n! \in C$, then$1/M_n \in C$, where$M_n$is the part of$n!$ coprime to 3.
$$n! = 3^{\nu_3(n!)} M_n$$
The problem reduces to analyzing when $1/M_n$can be represented using only digits in$D={0, 2}$ in base 3.

B. Digit Distribution and Korobov's Lemma

The core of the proof relies on the distribution of digits in the $m$-ary expansion of rational numbers.

• Purely Periodic Expansions: For a reduced fraction $r/q$ with $\gcd(q, m)=1$, the base-$m$ expansion is purely periodic with period length $L = \text{ord}_q(m)$.
• Korobov's Estimate: The authors utilize a theorem by Korobov [4] which bounds the deviation of the count of a specific digit $i$ in one period from the expected uniform distribution:
  $$\left| N_i\left(\frac{r}{q}\right) - \frac{\text{ord}_q(m)}{m} \right| \le 2 \cdot \text{ord}_{\text{rad}(q)}(m)$$
  where $N_i(r/q)$ is the number of occurrences of digit $i$ in the period.

C. The Contradiction Argument

If $r/q \in K_{m,D}$, then any digit $i \notin D$ must have a count of zero ($N_i(r/q) = 0$).

1. Applying Korobov's bound with $N_i = 0$ yields an upper bound on the period length:
   $$\text{ord}_q(m) \le 2m \cdot \text{ord}_{\text{rad}(q)}(m)$$
2. The authors analyze the specific case where $q = M_n$ (the factorial part coprime to $m$).
3. They prove that for sufficiently large $n$, the multiplicative order $\text{ord}_{M_n}(m)$ grows significantly faster than the bound $2m \cdot \text{ord}_{\text{rad}(M_n)}(m)$.
   • This is achieved by comparing the 2-adic (or $p$-adic) valuations of the orders.
   • Specifically, they show that for $n \ge 21$ (in the base-3 case), the inequality required for the intersection to exist fails.

3. Key Contributions and Results

Theorem 1.2 (Main Result)

The authors prove that the intersection of the reciprocals of factorials with the middle-third Cantor set is exactly:
$$\left\{ \frac{1}{n!} : n \in \mathbb{N} \right\} \cap C = \left\{ 1, \frac{1}{5!} \right\}$$
This definitively answers Question 1.1 posed by Jiang et al.

Theorem 1.4 (Generalization)

The result is generalized to any missing-digit set $K_{m,D}$.

• Finiteness: The intersection $\{1/n!\} \cap K_{m,D}$ is finite for any valid base $m$ and digit set $D$.
• Effectiveness: The set of solutions can be effectively determined. There exists a computable threshold $n_0$ such that for all $n \ge n_0$, $1/n! \notin K_{m,D}$. Therefore, one only needs to check a finite range$1 \le n < n_0$.

Algorithm 1

The paper provides a concrete algorithm to compute this intersection for any given $(m, D)$:

1. Select an odd prime $p$ coprime to $m$.
2. Compute the multiplicative order $d = \text{ord}_p(m)$ and the valuation $t = \nu_p(m^d - 1)$.
3. Determine a threshold $n_0$ satisfying a specific logarithmic inequality derived from the growth rates of orders.
4. Exhaustively check $1 \le n < n_0$to see if$1/M_n \in K_{m,D}$.

Table 1 Examples

The authors provide computed examples for various bases:

• $(3, \{0, 1\})$: Intersection is $\{1/2!, 1/3!, 1/4!, 1/6!\}$.
• $(3, \{0, 2\})$: Intersection is $\{1, 1/5!\}$ (The main result).
• $(4, \{0, 3\})$: Intersection is $\{1\}$.
• $(5, \{0, 4\})$: Intersection is $\{1, 1/6!\}$.

4. Significance

• Resolution of an Open Question: The paper settles a specific question raised in recent literature regarding the Erdős similarity problem context, providing a definitive "yes/no" answer for the specific sequence $1/n!$ in the Cantor set.
• Quantitative Finiteness: While previous work (Schleischitz, Shparlinski) established that rational numbers of the form $a/q^k$ in missing-digit sets are finite, this paper provides a constructive and effective method to find them for the specific sequence $1/n!$.
• Methodological Innovation: The paper successfully adapts Korobov's estimate on digit distribution to factorial denominators, bridging analytic number theory (multiplicative orders, Legendre's formula) with fractal geometry.
• Computational Utility: The provided algorithm allows researchers to compute these intersections for arbitrary missing-digit sets, moving from theoretical existence to practical calculation.

In summary, the paper demonstrates that the "sparsity" of the factorial sequence combined with the "rigidity" of missing-digit constraints forces the intersection to be extremely small and computable, with the middle-third Cantor set containing only two such points.