Howard 7871 [illegible] 2 b's, n's & m's are always equal to 3/4 (where is 5?), and y's = [illegible] [illegible] It's worth a guess that the answer is no. If not, it may be true for a specific case. The question asks if there exists an infinite set of primes such that each prime divides at least one number in the sequence defined by $a_n = n! + 1$. The answer is yes. For example, consider the sequence $b_k = k! + 1$. By Wilson's theorem, $(p-1)! \equiv -1 \mod p$, so if we choose $k=p-1$, then $b_{p-1} = (p-1)! + 1$ is divisible by $p$. Since there are infinitely many primes, there are infinitely many such numbers. However, the question specifies that each prime must divide at least one number in the sequence. This means we need to find an infinite set of primes where each prime divides some term in the sequence. Let's consider the sequence $c_k = k! + 1$. We know that for any prime $p$, there exists a $k$ such that $p | c_k$. Specifically, if we choose $k=p-1$, then $c_{p-1} = (p-1)! + 1$ is divisible by $p$. Therefore, the set of all primes is an infinite set of primes where each prime divides at least one number in the sequence $c_k = k! + 1$.