Let's find natural numbers that are followed by 7, and also those that are followed by 15 consecutive composite numbers!
Let the sought number be $n+1$. Then each of the 7 consecutive numbers: $n+2, n+3, n+4, n+5, n+6, n+7, n+8$ is certainly a composite number if $n$ is divisible by all prime numbers not greater than 8. Thus, $n=k \cdot 2 \cdot 3 \cdot 5 \cdot 7$. The sought number is therefore $k \cdot 2 \cdot 3 \cdot 5 \cdot 7+1=210 \cdot k+1$.

With the same reasoning, we can determine the numbers for which the next 15 numbers are all composite. This is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 k+1=30030 k+1$.

Note: Among these numbers, $8!+1$ and $16!+1$ also occur.