\[(n+1)!+2, \; (n+1)!+3, \; (n+1)!+4,..., \; (n+1)!+(n+1)\]
Each of these numbers is composite, divisible by 2, 3, 4,...,n+1 respectively and as
\[n \rightarrow \infty\]
so does the leg nth of consecutive numbers. Consecutive Composite Numbers
There are only 2 consecutive prime numbers - 2 and 3. There are infinitely many twin primes, defined as consecutive odd prime numbers e.g. 101 and 103. As the numbers get bigger, the gap between consecutive prime numbers can get arbitrarily large, notwithstanding the fact that there are infinitely many twin primes, and between consecutive prime numbers there are of course only composite numbers. In fact the number of consecutive composite numbers can be arbitrarily large. Consider the sequence
Each of these numbers is composite, divisible by 2, 3, 4,...,n+1 respectively and as
Each of these numbers is composite, divisible by 2, 3, 4,...,n+1 respectively and as