I have already asked this question here. No answers despite the bounty (which has now ended)

Let $p$ be a prime number, $p > 3$.

Does there always exist $k \in \mathbb N_{\ge 1}$ such that the prime factors of $2^kp - 1$ are all less than $p$?

Thoughts

Well, we can easily see that if $2p - 1$ is not prime, then there are no primes bigger than $p$ which divide it (hence $k=1$ would work). But $2p-1$ being prime is pretty common when $p$ is prime; it happens with $p= 7,19, 37$ etc.

For those last values 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?

Note: It seems like an interesting question, but if it's not up to the standards of mathoverflow tell me and I'll remove it :-)

I have already asked this question here. No answers despite the bounty (which has now ended)

Let $p$ be a prime number, $p > 3$.

Does there always exist $k \in \mathbb N_{\ge 1}$ such that the prime factors of $2^kp - 1$ are all less than $p$?

Thoughts

Well, we can easily see that if $2p - 1$ is not prime, then there are no primes bigger than $p$ which divide it (hence $k=1$ would work). But $2p-1$ being prime is pretty common when $p$ is prime; it happens with $p= 7,19, 37$ etc.

For those last values 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?

Note: It seems like an interesting question, but if it's not up to the standards of mathoverflow tell me and I'll remove it :-)

I have already asked this question here. No answers despite the bounty (which has now ended)

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 if $2p - 1$ is not prime then there are no primes bigger than $p$ which divide it (hence $k=1$ would work). 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?

Note It seems like an interesting question, but if it's not up to the standards of mathoverflow tell me and I'll remove it :-)

I have already asked this question here. No answers despite the bounty (which has now ended)

Let $p$ be a prime number, $p > 3$.

Does there always exist $k \in \mathbb N_{\ge 1}$ such that the prime factors of $2^kp - 1$ are all less than $p$?

Thoughts

Well, we can easily see that if $2p - 1$ is not prime, then there are no primes bigger than $p$ which divide it (hence $k=1$ would work). But $2p-1$ being prime is pretty common when $p$ is prime; it happens with $p= 7,19, 37$ etc.

For those last values 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?

Note: It seems like an interesting question, but if it's not up to the standards of mathoverflow tell me and I'll remove it :-)