New Mathematical Conjecture? - Data Science Central 2020-05-30T06:39:55Z https://www.datasciencecentral.com/forum/topics/new-mathematical-conjecture?feed=yes&xn_auth=no A reader posted the following… tag:www.datasciencecentral.com,2018-11-02:6448529:Comment:774487 2018-11-02T14:12:00.238Z Vincent Granville https://www.datasciencecentral.com/profile/VincentGranville <p>A reader posted the following, see below. So this is not new, it is actually related to the Pólya conjecture.</p> <p><em>I believe this fact follows from some upper bounds on the Mertens function.</em></p> <p><em>Assuming the conjecture is referring to the limiting natural density, then it suffices to show that the <a href="https://en.wikipedia.org/wiki/M%C3%B6bius_function" rel="nofollow">Mobius function</a> is -1 and 1 with limiting proportion 1; this entails the desired result over…</em></p> <p>A reader posted the following, see below. So this is not new, it is actually related to the Pólya conjecture.</p> <p><em>I believe this fact follows from some upper bounds on the Mertens function.</em></p> <p><em>Assuming the conjecture is referring to the limiting natural density, then it suffices to show that the <a href="https://en.wikipedia.org/wiki/M%C3%B6bius_function" rel="nofollow">Mobius function</a> is -1 and 1 with limiting proportion 1; this entails the desired result over squarefree integers (which are of positive density in the naturals). Then if we go up to N large enough, we can just apply this limit recursively on N/4 (for squarefree integers times 4), N/9, N/16, etc. and get a ratio within epsilon 1/2 for arbitrarily small epsilon.</em></p> <p><em>But showing that the Mobius function is equally often 1 and -1 just amounts to saying the Mertens function (defined to be the partial sums of the Mobius function) divided by the number of squarefree integers less than n goes to 0, which is equivalent to saying |M(x)| = o(x). But we have the <a href="https://en.wikipedia.org/wiki/Mertens_function#cite_note-3" rel="nofollow">bound</a> |M(x)|&lt;0.58782x/log^(11/9)(x) for x&gt;685 due to Kotnik. (Note that assuming the Riemann Hypothesis, we can get down to x^(1/2 + e), but we don't need anything that strong.)</em></p> <p><em>Edit: Forgot my relevant number theory functions for a second, I was trying to describe the Liouville function.</em></p> <p><em>Related trivia: The smallest integer at which there are more integers with an even number of prime factors than an odd number is 906150257, which is a counterexample to the <a href="https://en.wikipedia.org/wiki/P%C3%B3lya_conjecture" rel="nofollow">Pólya conjecture</a>.</em></p>