Existence of Prime-Free Sequence of Natural Numbers

Theorem

Let $n$ be a natural number.

Then there exists a sequence of consecutive natural numbers of length $n$ which are all composite.

Proof

Consider the number:

$N := \paren {n + 1}!$

where $!$ denotes the factorial.

Let $i \in I$ where:

$I = \set {i \in \N: 2 \le i \le n + 1}$

We have that $\size I = n$.

Then for all $i \in I$:

\(\ds i\)

\(\divides\)

\(\ds N\)

where $\divides$ denotes divisibility

\(\ds \leadsto \ \ \)

\(\ds i\)

\(\divides\)

\(\ds N + i\)

and so $N + i$ is composite.

That is:

$N + 2, N + 3, \ldots, N + n, N + n + 1$

are all composite.

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.16$: The Sequence of Primes: Footnote $1$