0
$\begingroup$

Sylvester and Schur Theorem says that for $x > n$, there is always at least one integer with a prime factor $> n$.

Bertrand's Postulate shows there is at least one and provides a method for determining a minimum number that increases as $n$ increases.

Bertrand's Postulate's proof depends on calculus to show that a function is increasing.

If my logic is sound, I have found a very elementary argument that relies on nothing more advanced than the Arithmetic Mean/Geometric Mean and argues that this is true for $n\ge 11$ (if I did my analysis correctly).

If this is already well known, could someone post a citation? If this is easily shown to be incorrect, could you provide the reasoning or counter example.

I was able to find this very interesting paper by Shorey and Tijdeman. (It takes me a long time to digest these papers so I apologize if the answer to my question is obvious from reading this paper.)

Is there any other such paper that would be interesting? I am trying to see how my method compares to what is well known.

$\endgroup$
7
  • 3
    $\begingroup$ If an integer $x$ is between $n$ and $2n$, and it has a prime factor larger than $n$, then $x$ is prime. $\endgroup$ – Arthur Mar 25 '17 at 20:02
  • $\begingroup$ Thanks. I updated my question. $\endgroup$ – Larry Freeman Mar 25 '17 at 20:26
  • 2
    $\begingroup$ ramanujan.sirinudi.org/Volumes/published/ram24.pdf check the result at the bottom of the 2nd page. $\endgroup$ – rtybase Mar 25 '17 at 23:54
  • $\begingroup$ That's right. Ramanujan showed this! Thanks. I am using a different method than he did. I had read through Ramanujan's paper a few years ago. A classic paper! :-) $\endgroup$ – Larry Freeman Mar 26 '17 at 0:06
  • 1
    $\begingroup$ @daniel, unfortunately, since posting this question, I found an error in my reasoning. I am not sure if it is a fatal error or not. If I can save the argument, I am very glad to post it. $\endgroup$ – Larry Freeman Mar 26 '17 at 7:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.