I want a number that has a prime factorization that contains all prime numbers less than that number (besides $2$), anyone with an answer please show a proof.
I have made a little progress, if this number exists, then it is one less than a prime number.
proof:
make a list of all the primes less than $n$ $\{P_1,\space P_2,\space P_3,\cdots\}$
$n+1$ is not a factor of any of these primes so it ether is a prime or has a prime that our list missed and since the list contains all primes less than $n$, $n+1$ must be prime.
and more evidence that suggests this exists is the proof that any arbitrary prime gap exists. if $n!+1$ is prime then the prime gap between $n!+1$ and the next prime is at least $n-1$ because:
$n!+2$ is a multiple of $2$ and
$n!+3$ is a multiple of $3$ and so-on until
$n!+n$ which is a multiple of $n$
any more help would be appreciated
Best Answer
This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.
Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.