Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Let $L = \text{lcm}(2,\ldots, n+1)$. Then $p = Lm -1$ satisfies your conjecture for $n$ if $Lm/(k+1)-1$ for $k = 0 \ldots n$ are all primes. Dixon's Dickson's conjecture implies that this is the case for infinitely many $m$.
Let $L = \text{lcm}(2,\ldots, n+1)$. Then $p = Lm -1$ satisfies your conjecture for $n$ if $Lm/(k+1)-1$ for $k = 0 \ldots n$ are all primes. Dixon's conjecture implies that this is the case for infinitely many $m$.
Let $L = \text{lcm}(2,\ldots, n+1)$. Then $p = Lm -1$ satisfies your conjecture for $n$ if $Lm/(k+1)-1$ for $k = 0 \ldots n$ are all primes. Dickson's conjecture implies that this is the case for infinitely many $m$.
Let $L = \text{lcm}(2,\ldots, n+1)$. Then $p = Lm -1$ satisfies your conjecture for $n$ if $Lm/(k+1)-1$ for $k = 0 \ldots n$ are all primes. Dixon's conjecture implies that this is the case for infinitely many $m$.
