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For every $n \in N$ 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 $n \in N$ , write $ι : N \to Z$ for the embedding of The naturals embed in the integers, and put

$N : = \prod_{j < n} ι (j + 2),$

the finite product of The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity taken in the commutative monoid $(Z, \cdot, 1)$ of $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ 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 $N + ι (j + 2)$ 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: $n \in N$ and $N : = \prod_{j < n} ι (j + 2)$ .

[L1]

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

[L2]

Splitting law: $\prod_{i < m + c} g_{i} = (\prod_{i < m} g_{i}) (\prod_{l < c} g_{m + l})$ (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: $d ∣ u$ and $d ∣ w$ give $d ∣ u + w$ ; $d ∣ u$ means $u = d c$ for some $c$ (Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L6]

Induction on $N$ (The principle of mathematical induction).

[L7]

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

[L8]

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

[L9]

$Z$ 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 $Z$ , and every integer $y > 0$ satisfies $y \geq 1$ : $y = ι (t)$ with $t \neq 0$ , so $1 = σ (0) \leq t$ and $ι$ preserves the order.

L7L8L9

1.2

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

L1L2L4L8L9

2.1

For every $j \in N$ , $ι (j + 2) \geq 2 > 1$ , where $2 : = 1 + 1$ . Indeed $2 = σ (1) = 1 + 1$ in $N$ gives $ι (2) = ι (1) + ι (1) = 1 + 1$ ; and $2 \leq j + 2$ in $N$ because $2 + j = j + 2$ , so $ι (2) \leq ι (j + 2)$ .

step 1.1L7L8L9

2.2

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

step 1.2L4

3.1

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

step 1.1step 2.1L1L6L9

4.1

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

step 2.1step 3.1L9

5.1

So $k$ is a positive divisor of $N + k$ with $k \neq 1$ and $k \neq N + k$ , 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 $N + ι (j + 2)$ run over $N + 2, N + 3, \dots, N + n + 1$ , each obtained from the previous by adding $1$ , since $ι (σ (j) + 2) = ι (j + 2) + 1$ . 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 $N$ defines $n!$ by recursion in $N$ , 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 $Z$ : naming $ι (n!)$ as one would create a dictionary obligation this page cannot discharge. The finite product of The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity in $(Z, \cdot, 1)$ 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 $n \in N$ and every list $p : n \to Z$ 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

Used by

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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∈N there are nn consecutive composite integers: with N:=∏j<n(j+2)N:=∏j<n​(j+2), each of N+2,…,N+n+1N+2,…,N+n+1 is composite