Clarification regarding Prime theorem This is one theorem which I came across the book:

For every positive integer $n$, there is a sequence of $n$ consecutive positive integers containing no primes.

Is this theorem valid ?
Because if $n = 2$, then its next consecutive integer $3$ itself is prime. 
 A: It's easy to develop such a sequence. However, it may not be the first such sequence. If you want to find a sequence of $n$ consecutive composite numbers, the classical way is to first calculate $(n+1)!$. Then the sequence:
$$
\{(n+1)!+2,(n+1)!+3,...,(n+1)!+(n+1)\}
$$
Is a sequence of n consecutive composite numbers. The first is divisible by 2 as both $(n+1)!$ and $2$ will be divisible, and so on for all these $n$ numbers.
A: Notice that it does not say where the span of consecutive numbers has to start, nor does it say that it is a span of exactly $n$ composite numbers.
So, for $n = 2$, we want to find two consecutive composite numbers. 8 and 9 fit the bill, so that works for us. Now, 10 is also composite, so that helps us out for $n = 3$. The span of consecutive numbers from 24 to 28 helps us for $n = 4$ or 5. In fact, take a look at Sloane's A030296. (Sloane's OEIS is an indispensable resource for many number theory inquiries).
Now, if you don't care about finding the first span of at least $n$ consecutive composite numbers, you can just do $(n + 1)!$ Then every number from $(n + 1)! + 2$ to $(n + 1)! + n$ is guaranteed to be composite. In some cases, you'll also get $(n + 1)! + 1$ is composite, too (see Wilson's theorem), and if $n$ is odd, then $(n + 1)! + n + 1$ is composite as well.
