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Elementary Proofs of Two Congruences for Partitions with Odd Parts Repeated at Most Twice

This paper provides two elementary proofs for the congruences $a(4n+2) \equiv 0 \pmod{2}$ and $a(4n+3) \equiv 0 \pmod{2}$ , where $a(n)$ counts partitions of $n$ with odd parts repeated at most twice, thereby fulfilling a request made by Merca.

Original authors: James A. Sellers

Published 2026-06-29

📖 5 min read🧠 Deep dive

CC BY 4.0

Original authors: James A. Sellers

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

Imagine you have a giant bag of numbered blocks. Your job is to build "towers" (which mathematicians call partitions) by stacking these blocks so that their total sum equals a specific number, say $n$ .

Usually, you can stack the blocks however you like. But in this paper, we are playing a very specific game with a special rule: You can only use "odd" numbered blocks (1, 3, 5, etc.) at most twice in a single tower. You can use "even" numbered blocks (2, 4, 6, etc.) as many times as you want.

Let's call the number of different towers you can build for a number $n$ as $a(n)$ .

The Mystery

A mathematician named Merca recently looked at this game and noticed something strange. He found that if you try to build towers for numbers that look like $4n + 2$ (like 2, 6, 10, 14...) or $4n + 3$ (like 3, 7, 11, 15...), the total number of ways to build them is always an even number.

In math terms, he proved:

$a(4n + 2)$ is divisible by 2.
$a(4n + 3)$ is divisible by 2.

Merca proved this using a very powerful, automated computer-like method. While correct, it was like solving a puzzle by brute force; it didn't explain why the answer was even. He asked for a "classical" or "elementary" proof—a simple, logical explanation that a human could follow without a supercomputer.

James Sellers, the author of this paper, says: "Challenge accepted!" He provides two simple ways to prove this.

Method 1: The "Magic Filter" (Generating Functions)

Think of a generating function as a magical machine that takes a list of numbers and turns them into a giant algebraic recipe (a polynomial). If you expand this recipe, the number of times a specific term appears tells you how many towers exist for that number.

Merca's original recipe for this problem was incredibly complex and messy, like a 20-page instruction manual with hundreds of steps.

Sellers' first proof is like finding a simplified version of that manual.

He takes the complex recipe and uses a few clever algebraic tricks (called "dissections") to slice it up.
He separates the recipe into two parts: one for even numbers and one for odd numbers.
When he looks specifically at the parts for $4n+2$ and $4n+3$ , he discovers something amazing: The entire recipe for these specific numbers has a "2" sitting right in front of it.

The Analogy: Imagine you are baking cookies. The original recipe says, "Mix flour, sugar, eggs, and a secret ingredient." Sellers' proof shows that for the specific batch of cookies labeled "4n+2," the recipe actually says, "Take 2 batches of this mixture."
If you have 2 batches, you automatically have an even number of cookies. No matter how you count them, the total is even. This proves the rule simply by looking at the structure of the recipe.

Method 2: The "Square Hunt" (Theta Functions)

The second proof is more like a detective story involving squares.

Sellers connects the tower-building game to a special mathematical object called a Theta function. Think of this function as a flashlight that only shines on numbers that can be written in a very specific shape: $3k^2 + 2k$ (where $k$ is a whole number).

If a number fits this shape, the flashlight shines (the value is 1).
If it doesn't fit, the flashlight stays off (the value is 0).

The proof then links the tower counts ( $a(n)$ ) to these flashlight numbers. It turns out that the number of towers for $n$ is related to a sum of these flashlight numbers.

The Detective Work:
Sellers asks: "Can the numbers $4n + 2$ or $4n + 3$ ever fit the shape $3k^2 + 2k$ ?"

He does a quick calculation and realizes that if $4n + 2$ fit the shape, it would have to be a number that, when multiplied by 3 and added to 1, becomes a perfect square.
However, he checks the math and finds that $4n + 2$ (and $4n + 3$ ) always result in numbers that cannot be perfect squares (they leave a remainder of 3 or 2 when divided by 4, and squares never do that).

The Conclusion:
Since the flashlight never turns on for these specific numbers (the value is always 0), the sum that determines the number of towers becomes zero (or even).
The Analogy: Imagine you are trying to find a specific key in a giant pile of keys. The "key" you are looking for only exists if the number is a perfect square. You check the numbers 2, 6, 10, 14... and realize none of them are perfect squares. Therefore, the key doesn't exist. If the key doesn't exist, the "count" of keys is zero. Since zero is an even number, the rule holds true.

Summary

The paper doesn't just say "it's true because a computer said so." Instead, it offers two clear, human-readable reasons:

The Recipe Method: The mathematical formula for these numbers literally has a "2" multiplied in front of it, guaranteeing an even result.
The Square Method: The numbers in question ( $4n+2$ and $4n+3$ ) are mathematically "forbidden" from being perfect squares, which forces the count of partitions to be even.

Both methods confirm Merca's observation using simple, classical logic rather than complex automation.

Technical Summary: Elementary Proofs of Two Congruences for Partitions with Odd Parts Repeated at Most Twice

Problem Statement
The paper addresses a specific problem in partition theory regarding the auxiliary function $a(n)$ , which counts the number of partitions of an integer $n$ where odd parts are repeated at most twice (with no restrictions on even parts). This function is equivalent to counting partitions of $n$ where no part is congruent to 3 modulo 6. In a recent study on overpartitions, Merca established two congruences for this function:

$a(4n + 2) \equiv 0 \pmod 2$
$a(4n + 3) \equiv 0 \pmod 2$

While Merca provided proofs for these results using automated techniques based on the work of Radu and Smoot, he noted that these methods offered no insight into the underlying structure of the results and requested a "classical" or elementary proof. The goal of this paper is to fulfill that request by providing such proofs.

Methodology
The author, James A. Sellers, employs two distinct elementary approaches to prove the congruences, avoiding the complex automated machinery used in previous work.

Simplified Generating Function Identities (Section 2):
The first method relies on the dissection of generating functions. The author utilizes two elementary lemmas regarding the dissection of $f_3/f_1$ and $1/f_1^2$ (citing da Silva and Sellers). By applying these lemmas to the generating function of $a(n)$ , defined as $\sum a(n)q^n = f_3 / (f_1 f_6)$ , the author performs successive 2-dissections. This process isolates the terms corresponding to $a(4n+2)$ and $a(4n+3)$ , yielding significantly simpler generating function identities than those previously derived by Merca.
Theta Function and Modular Arithmetic (Section 3):
The second method links the generating function of $a(n)$ to a specific theta function, $G(q) = f_2^2 f_3 f_{12} / (f_1 f_4 f_6)$ , which is a holomorphic modular form of weight 1/2. The author establishes a congruence modulo 2 between the generating function of $a(n)$ and the product of the generating function for the partition function $p(n)$ and the theta function $G(q)$ .
The coefficients of $G(q)$ , denoted $g(n)$ , are non-zero only when $n$ takes the form $3k^2 + 2k$ . The proof then reduces to a number-theoretic argument: it demonstrates that the arithmetic progressions $4N+2$ and $4N+3$ cannot be represented in the form $3k^2 + 2k$ . This is shown by analyzing the quadratic equation $3n+1 = (3k+1)^2$ and observing that for $n \in \{4N+2, 4N+3\}$ , the resulting value $3n+1$ is congruent to 3 or 2 modulo 4, respectively. Since no perfect square is congruent to 2 or 3 modulo 4, the coefficients $g(4N+2)$ and $g(4N+3)$ are identically zero.

Key Contributions and Results

Elementary Proofs: The paper provides two "truly elementary" proofs for the congruences $a(4n+2) \equiv 0 \pmod 2$ and $a(4n+3) \equiv 0 \pmod 2$ , directly addressing Merca's request for classical derivations.
Simplified Identities: The author derives new, much simpler generating function identities for the subsequences $a(4n+2)$ and $a(4n+3)$ (Theorems 1.4 and 1.5), which are far less complex than the identities previously established by Merca (Theorems 1.1 and 1.2).
Structural Insight: The second proof method sheds light on why these specific arithmetic progressions arise. By linking the problem to the non-existence of squares in certain arithmetic progressions (specifically modulo 4), the paper reveals the number-theoretic mechanism driving the divisibility by 2.

Significance
The paper claims significance primarily in its pedagogical and explanatory value. The author notes that while the automated proofs by Radu and Smoot are correct, they are "unnecessary" in terms of power and provide "no insights into the results themselves." By contrast, these elementary proofs:

Fulfill the specific request made by Merca for a classical approach.
Demonstrate that simpler generating function identities exist for these congruences.
Offer a clear, combinatorial, and number-theoretic explanation for the divisibility properties, specifically highlighting the role of quadratic residues modulo 4.

The work does not propose new applications or future experimental directions but rather serves to clarify and simplify existing results in the theory of partitions and overpartitions.

The Mystery

Method 1: The "Magic Filter" (Generating Functions)

Method 2: The "Square Hunt" (Theta Functions)

Summary

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