0
$\begingroup$

The Robin's inequality says - If the Riemann hypothesis is true then - $$\sigma(n) < e^{\gamma}n \log(\log(n))$$ holds true for all $n \in \mathbb{N}$
Now it is proved for all $5-$ free integers .And thus for infinely many integers. So my question is does this proof also prove That there are infinitely many no - trivial zeros of zeta function zeros on real line $1/2$ ?

$\endgroup$
  • 1
    $\begingroup$ What is "The Riemann hypothesis for some specific values"? $\endgroup$ – Martín-Blas Pérez Pinilla Feb 10 '14 at 7:49
  • $\begingroup$ @Martín-BlasPérezPinilla for some particular cases? $\endgroup$ – Shivam Patel Feb 10 '14 at 7:57
  • $\begingroup$ What "particular cases" are you talking about? Do you know what the Riemann Hypothesis is? $\endgroup$ – anon Feb 10 '14 at 8:02
  • $\begingroup$ That is what I am asking $\endgroup$ – Shivam Patel Feb 10 '14 at 8:12
  • $\begingroup$ No, you were asking if Robin's inequality implies the Riemann hypothesis for "some specific values" / "some particular cases." Since there's no obvious interpretations of these phrases, it's your responsibility to explain to readers what you mean by your phrases, and you have failed to do that. I don't understand why you wouldn't want your readers to know what your questions is! If you continue to fail to have a question, then your post will eventually be closed. $\endgroup$ – anon Feb 10 '14 at 8:19
4
$\begingroup$

This result is quite nice; I should say that I am properly thought of as an amateur enthusiast about Colossally Abundant Numbers and Ramanujan's Superior Highly Composite Numbers; I do not know any of the methods of basic analytic number theory.

First things first. Robin's condition is that $\sigma(n) < e^\gamma \log \log n $ for $n \geq 5041 = 71^2 = 1 + 7! . $

Theorem 1.6 in Choie, Lichiardopol, Moree, and Sole is this: RH is true if and only if Robin's inequality is true for all natural numbers divisible by the fifth power of some prime.

There is an excellent reason that this does not change very much: The numbers that give very large values of $\sigma(n)$ relative to their size, are the Colossally Abundant Numbers, see Alaoglu and Erdos. These are almost all divisible by 32 and 243, for example, only a small finite set of them in the beginning are not so divisible. As a result, RH is true, roughly speaking, if and only if it is true for all CA numbers above 5040; note that the first counterexample is not guaranteed to be CA, only superabundant.

Oh, this is all-or-nothing. No conclusions about zeroes of the zeta function in the critical strip may be drawn.

For experimentation, I recommend the original condition of Nicolas, the adviser of Robin: RH is true if and only if, for every primorial $P,$ $$ P > \, \varphi(P) \; e^\gamma \, \log \log P, $$ where $\varphi$ is the Euler totient function. I mention this because making a computer list of consecutive CA numbers is actually a bit difficult, while computing primorials (the product of consecutive primes beginning with $2$) is relatively easy.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.