# Prime Numbers in Linear Patterns

Prime numbers in a table with 180 columns
Field: Number Theory
Image Created By: Iris Yoon

Prime numbers in a table with 180 columns

Create a table with 180 columns and write down positive integers from 1 in increasing order from left to right, top to bottom. When we mark the prime numbers on this table, we obtain the linear pattern as shown in the figure.

# Basic Description

Arranging natural numbers in a particular way and marking the prime numbers can lead to interesting patterns. For example, consider a table with 180 columns and infinitely many rows. Write positive integers in increasing order from left to right, and top to bottom. If we mark all the prime numbers, we get a pattern shown in the figure. We can see that prime numbers show patterns of vertical line segments, which implies that the prime numbers only appear on certain columns.

# A More Mathematical Explanation

Instead of studying a table with 180 columns, we will study a table with 30 columns, as shown in [[#1 [...]

Instead of studying a table with 180 columns, we will study a table with 30 columns, as shown in Image 1.

Image 1

## Construction

First, create a table with 30 columns and sufficiently many rows. Write all positive integers starting from 1 as one moves from left to right, and top to bottom. Then, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . If we mark the prime numbers in this table we get Image 2.

Image 2

## Properties

Theorem 1. All prime numbers appear on columns that have a $1$ or a prime number on its top row. In other words, for every prime number $p$, either $p \equiv 1\pmod {30}$, or there exists a prime number $q$ less than $30$ such that $p \equiv q \pmod {30}$.

Proof.

Given any prime number $p$, assume that $p$ is neither congruent to $1 \pmod {30}$ nor $q \pmod {30}$ for every prime $q$ less than $30$. Then,

$p \equiv x \pmod {30}$,

where $x$ is some integer less than $30$ that is not $1$ and not a prime. Prime factorization of $x$ must contain one of $2, 3,$ and $5$. (If the prime factorization of $x$ did not contain any of $2, 3,$ or $5$, then the smallest possible value of $x$ will be $7 \cdot 7 =49$, which is greater than $30$). Thus,

$x=2^a3^b5^c$,

where $a,b,c \ge 0$, and at least one of $a,b,c$ is greater than $0$.

Since $p$ is congruent to $x$, we can write $p$ as $p=30n+x=30n+2^a3^b5^c$, where $n$ is an integer greater than or equal to 1. Then,

$p=30n+2^a3^b5^c=(2 \cdot 3 \cdot 5)n+2^a3^b5^c$.

$p$ is then equal to one of $2(3 \cdot 5 \cdot n+2^{a-1}3^b5^c)$, $3(2 \cdot 5 \cdot n+2^a3^{b-1}5^c)$, and $5(2 \cdot 3 \cdot n+2^a3^b5^{c-1})$. This contradicts that $p$ is a prime number. Thus, $p \equiv 1 \pmod {30}$ or $p \equiv q \pmod {30}$, for some prime number $q$ less than $30$.$\Box$

However, the statement does not generalize to other integer modulo groups. For instance, consider a table with $60$ columns. The number $49$ appears on the first row, and $49$ is not a prime number. However, the column containing $49$ will contain other prime numbers, such as $109$.

Moreover, not all integers that are congruent to $1 \pmod {30}$ or $q \pmod {30}$, where $q$ is a prime number less than $30$, are prime numbers. For instance, $49$, which is congruent to $19 \pmod {30}$, is not a prime number, but $49$ still appears on the same column as$19$. Let's call the columns that have a $1$ or a prime number greater than $5$ on its top row as prime-concentrated columns. One can observe that for all composite numbers that appear on these prime-concentrated columns, say $49,77,91,119,121,133,143,161,169,...,$ all prime factors must be greater than or equal to $7$. In other words, these composite numbers do not have $2, 3,$or $5$ as a prime factor.

Theorem 2. Composite numbers that appear on prime-concentrated columns do not have $2,3,$ or $5$ as a prime factor.

Proof.

Let $x$ be a composite number that appears on a prime-concentrated column, and assume that $x$ has at least one of $2, 3,$ or $5$ as a prime factor. Since $x$ appears on the prime-concentrated column, $x$ can be written as

$x=30n+k$,

where $n$ is a positive integer, and $k =1$or $k$ is a prime number greater than $5$ and smaller than $30.$ If $x$ had $2$ as a prime factor, $k$ must also have $2$ as a factor because $30$ has $2$ as a factor. This contradicts the fact that $k$is equal to $1$ or is a prime number between $7$ and $30$. The same argument works for the case when $x$ has $3$ or $5$ as prime factors.$\Box$

Another pattern to notice is that the prime-concentrated columns seem symmetric about the column that contains $15$, which leads to the following observation.

Theorem 3. If $p$ is a prime number less than$30$ and if $p$ is not equal to $2,3,$ or $5,$ then $30-p$ is a prime number.

Proof.

Let $p$ be a prime number less than $30$ that is not equal to $2, 3,$or $5$. Let $q=30-p$. If $q$ were not a prime, then $q$ must have $2, 3,$ or $5$ as a prime factor. Since $p=30-q$, $p$ will also be divisible by $2, 3,$ or $5$, contradicting our condition that $p$ is a prime number. $\Box$

One can also observe that each prime-concentrated column seems to contain infinitely many prime numbers. In fact, such observation is consistent with Dirichlet's Theorem in Arithmetic Progressions.

Dirichlet's Theorem On Primes In Arithmetic Progressions

Let $a,N$ be relatively prime integers. Then there are infinitely many prime numbers $p$ such that $p \equiv a \pmod {N}$.

The proof of Dirichlet's Theorem is not written in this page. One can easily note that Dirichlet's Theorem implies that each prime-concentrated column contains infinitely many prime numbers.
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