ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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For every $n \in \mathbb{N}$ there are $n$ consecutive composite integers: with $N := \prod_{j<n}(j+2)$ , each of $N+2, \dots, N+n+1$ is composite

Example

Let $n \in \mathbb{N}$ , write $\iota : \mathbb{N} \to \mathbb{Z}$ for the embedding of The naturals embed in the integers, and put

$N \;:=\; \prod_{j<n} \iota(j+2),$

the finite product of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity taken in the commutative monoid $(\mathbb{Z},\cdot,1)$ of $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ; so $N$ is the product of the $n$ integers $2, 3, \dots, n+1$ , and $N = 1$ when $n = 0$ .

Then $N + \iota(j+2)$ is composite (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) for every $j < n$ . Those $n$ integers are $N+2, N+3, \dots, N+n+1$ , consecutive because consecutive values of $j$ change the summand by $1$ . So for every $n$ there is a run of $n$ consecutive composite integers.

Facts & Assumptions

Given: $n \in \mathbb{N}$ and $N := \prod_{j<n} \iota(j+2)$ .

[L1]

$\prod_{i<0} g_i = e$ and $\prod_{i<\sigma(m)} g_i = \bigl(\prod_{i<m} g_i\bigr) g_m$ ; the value depends only on the entries named (The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , Semigroup and monoid).

[L2]

Splitting law: $\prod_{i<m+c} g_i = \bigl(\prod_{i<m} g_i\bigr)\bigl(\prod_{l<c} g_{m+l}\bigr)$ (Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either).

[L3]

$n > 1$ that is not prime is composite; $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L4]

Divisibility is reflexive and transitive and is linear: $d \mid u$ and $d \mid w$ give $d \mid u + w$ ; $d \mid u$ means $u = dc$ for some $c$ (Divisibility is reflexive and transitive on $\mathbb{Z}$ , and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$ ; also $d \mid a$ implies $d \mid ac$ , $-d \mid a$ and $d \mid -a$ , Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L6]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

[L7]

$\iota$ is injective, preserves addition, multiplication and order, and has as image the nonnegative integers, with $\iota(0) = 0$ and $\iota(1) = 1$ (The naturals embed in the integers).

[L8]

On $\mathbb{N}$ : $m + 0 = m$ and $m + \sigma(k) = \sigma(m+k)$ , so $\sigma(k) = k+1$ ; addition is commutative (Addition of natural numbers, Addition is commutative, The natural numbers $\mathbb{N}$ (von Neumann)); $a \le b$ means $a + c = b$ for some $c$ (Order on the natural numbers); $a < b$ exactly when $\sigma(a) \le b$ (Discreteness: $\sigma(n)$ is the immediate successor); and $j < \sigma(j)$ , with $i < \sigma(j)$ exactly when $i \le j$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ).

[L9]

$\mathbb{Z}$ is a commutative ring; its order is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals, The integers form a totally ordered ring, Order on the integers).

Verification

technique · direct

1.1

$0 < 1$ in $\mathbb{Z}$ , and every integer $y > 0$ satisfies $y \ge 1$ : $y = \iota(t)$ with $t \ne 0$ , so $1 = \sigma(0) \le t$ and $\iota$ preserves the order.

L7L8L9

1.2

Fix $j < n$ and put $k := \iota(j+2)$ . Then $k \mid N$ : since $j < n$ we have $\sigma(j) \le n$ , so $n = \sigma(j) + c$ for some $c \in \mathbb{N}$ , and the splitting law gives $N = \bigl(\prod_{i<\sigma(j)} \iota(i+2)\bigr)\bigl(\prod_{l<c} \iota(\sigma(j)+l+2)\bigr)$ , while $\prod_{i<\sigma(j)} \iota(i+2) = \bigl(\prod_{i<j} \iota(i+2)\bigr) k$ . Rearranging by associativity and commutativity, $N = k \cdot B$ for an integer $B$ .

L1L2L4L8L9

2.1

For every $j \in \mathbb{N}$ , $\iota(j+2) \ge 2 > 1$ , where $2 := 1+1$ . Indeed $2 = \sigma(1) = 1 + 1$ in $\mathbb{N}$ gives $\iota(2) = \iota(1) + \iota(1) = 1 + 1$ ; and $2 \le j + 2$ in $\mathbb{N}$ because $2 + j = j + 2$ , so $\iota(2) \le \iota(j+2)$ .

step 1.1L7L8L9

2.2

Hence $k \mid N + k$ , by linearity applied to $k \mid N$ and $k \mid k$ .

step 1.2L4

3.1

$N \ge 1$ . Let $T$ be the set of $m \in \mathbb{N}$ with $\prod_{i<m} \iota(i+2) \ge 1$ . Then $0 \in T$ , the empty product being $1$ . If $m \in T$ then $\prod_{i<\sigma(m)} \iota(i+2) = \bigl(\prod_{i<m} \iota(i+2)\bigr) \iota(m+2)$ has both factors $\ge 1 > 0$ , so the product is positive and hence $\ge 1$ by step 1.1. By induction $T = \mathbb{N}$ , so $N \ge 1 > 0$ .

step 1.1step 2.1L1L6L9

4.1

$N + k > k > 1$ , because $N > 0$ ; in particular $N + k > 1$ and $N + k \ne k$ .

step 2.1step 3.1L9

5.1

So $k$ is a positive divisor of $N + k$ with $k \ne 1$ and $k \ne N + k$ , and $N + k > 1$ ; therefore $N + k$ is not prime, and being greater than $1$ it is composite.

step 2.1step 2.2step 4.1L3

6.1

As $j$ runs over $0, 1, \dots, n-1$ the integers $N + \iota(j+2)$ run over $N+2, N+3, \dots, N+n+1$ , each obtained from the previous by adding $1$ , since $\iota(\sigma(j)+2) = \iota(j+2) + 1$ . All $n$ of them are composite by step 5.1.

step 5.1L7L8L9∎

Remarks

The factorial is available at this point, and is deliberately not used. The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ defines $n!$ by recursion in $\mathbb{N}$ , on a page this one may cite. The real-valued copy in For every real $x$ , $x^k/k! \to 0$ ↗ is a separate object and is already declared as a forward reference. Neither is a factorial on $\mathbb{Z}$ : naming $\iota(n!)$ as one would create a dictionary obligation this page cannot discharge. The finite product of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity in $(\mathbb{Z},\cdot,1)$ does everything needed, and $N$ is $(n+1)!$ only in the informal sense.
The case $n = 0$ is real and vacuous. The empty product is $1$ , there is no $j < 0$ , and the claim asserts nothing — correctly, since a run of $0$ consecutive composites is no claim at all.
This says the gaps are unbounded and nothing about where the primes are. It is entirely compatible with Euclid's theorem: for every $n \in \mathbb{N}$ and every list $p : n \to \mathbb{Z}$ of primes there is a prime not among $p_0, \dots, p_{n-1}$ ; consequently the set of primes is not finite: primes keep coming, and yet one can always find a stretch of any prescribed length containing none. No claim is made here about how large $N$ must be, or about the smallest run of a given length.

Depends on

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Sources

Prime gap (Wikipedia) (standard reference, not scraped)
Harvey Mudd College Math Fun Facts: Gaps in primes (standard reference, not scraped)

For every n∈Nn \in \mathbb{N}n∈N there are nnn consecutive composite integers: with N:=∏j<n(j+2)N := \prod_{j<n}(j+2)N:=∏j<n​(j+2), each of N+2,…,N+n+1N+2, \dots, N+n+1N+2,…,N+n+1 is composite