[math-fun] Riemann hypothesis equivalent to bounding the count of divisors?
If d(N) is the number of divisors of N, I can prove -- except for a gap -- the following: The Riemann Hypothesis is equivalent to the claim that there exists a constant J such that d(N) < a certain known function of N and J for all N>0. Specifically: The Riemann hypothesis is equivalent to the existence of a constant J such that log2( d(N) ) <= Li(lnN) + K*Li((lnN)^K) - (lnN)^K / lnlnN + J*(lnN)^(1/2) / (lnlnN)^2 for all integers N>1. Here K=log2(3/2)=0.5849625... Actually I could probably compute a suitable J with more work, in which case the bound would become a fully specified function of N only. Most of the work behind "my" proof was done by Ramanujan 1915 and I am taking some of his claims on trust, I mostly have not checked his work: S.Ramanujan: Highly composite numbers, Proc. London Math. Soc. (2) 14 (1915) 347-409. http://ramanujan.sirinudi.org/Volumes/published/ram15.pdf THE MISSING LEMMA (which I need to complete the proof): Letting z be the non-real zeros of the Riemann zeta function, I need that S2(x) = SUM x^z / z^2 must obey S2(x) < -x^P for some fixed P>1/2, for an infinite number of integers x>0, if the Riemann hypothesis is false. This "missing lemma" is presumably true and presumably has nothing particularly to do with the Riemann hypothesis -- all you really (presumably) need is the fact that if the RH is false the set of z obey: 1. If z exists, then ComplexConjugate(z) also exists. [Which causes S2(x) to be real-valued.] 2. The z all obey 0<Re(z)<1 and with at least one z having Re(z)>1/2. 3. There are O(T*logT) values of z having |Im(z)|<T, for T>20. 4. Indeed, there are O(logT) values of z having T<Im(T)<T+7, for T>20. 5. There are a countably infinite number of z. 6. All the z are non-real, indeed |Im(z)|>14.. I.e. any set of z with these 6 properties would presumably work. The lemma is easy to see it is true if the set of z were finite (with, say, a million elements) because then the finite subset of z with maximum real part, alone, would suffice to prove it with P=max(Re(z))>1/2 because the non-maximal Re(z) have effects on S2(x) which become relatively negligible for large x. The difficulty for me is when the set of z is infinite. It is possible this lemma is already known or standard (to those who know it). STRONG EXPLICIT BOUNDS ON THE d(N) FUNCTION: Empirically, my computer says that log2( d(N) ) < Li(lnN) + K*Li((lnN)^K) - (lnN)^K / lnlnN + 1.19113*(lnN)^(1/2) / (lnlnN)^2 is true where K=log2(3/2)=0.5849625... for all integers with 1<N<exp(8140) thus "confirming" the Riemann hypothesis. Also, for all N with 1<N<exp(8140) that are "highly composite" [have greater d(N) than all lesser numbers] my computer says log2( d(N) ) > Li(lnN) + K*Li((lnN)^K) - (lnN)^K / lnlnN - 3.71217*(lnN)^(1/2) / (lnlnN)^2 Also, for all N with 1<N<exp(8140) that are "highly composite" [have greater d(N) than all lesser numbers] my computer says Li(lnN) + ((lnN)^K)/(K*(lnlnN)^2) - 0.96824 *(lnN)^(1/2) / (lnlnN)^2 < log2( d(N) ) < Li(lnN) + ((lnN)^K)/(K*(lnlnN)^2) - 0.94223 *(lnN)^(1/2) / (lnlnN)^2 which seems pretty damn impressively tight, although I do not think a bound window of this form can be valid for all N>1. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Warren D Smith