Let $p$ be a prime number, $p > 3$.
Does it always exists a $k \in \mathbb N, k \ge 1$ such that the prime factors of $2^kp - 1$ are all less then $p$?
Thoughts
Well we can easily see that either $2p - 1$ is prime or there no primes bigger than $p$ which divide him. But being prime is pretty common, it happens with $p= 7,19, 37$ etc.
With those I looked at $k=2$ and they all work, but there is a prime less than $100$ (I don't remember which one) for which you have to use $k=3$.
Anyhow it seems like a good bet, but is it actually true?