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An elementary problem equivalent to the Riemann hypothesis

  1. Jul 18, 2006 #1


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    Let [tex]H_{n}=\sum_{k=1}^{n}\frac{1}{k}[/tex] be the nth harmonic number, then the Riemann hypothesis is equivalent to proving that for each [tex]n\geq 1[/tex],

    [tex]\sum_{d|n}d\leq H_{n}+\mbox{exp}(H_{n})\log H_{n}[/tex]​

    where equality holds iff n=1. The paper that this came from is here: An Elementary Problem Equivalent to the Riemann Hypothesis by Jeffrey C. Lagarias.

    No questions, just thought it would be appreciated.
    Last edited: Jul 19, 2006
  2. jcsd
  3. Jul 18, 2006 #2
    I dont understand... explain! This thread made my haed hurt!
  4. Jul 18, 2006 #3
    This is based on [itex]\sigma(n) = \sum_{d|n}d < e^{\gamma} n \log \log n[/itex] for all [itex]n >= 5041[/itex] being logically equivalent to the Riemann Hypothesis as shown in
    Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypoth`ese de Riemann,J. Math. Pures Appl. (9) 63 (1984), 187–213.
    See also: http://arxiv.org/abs/math.NT/0604314
    Last edited: Jul 19, 2006
  5. Jul 19, 2006 #4


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    should be
    [tex]H_n= \sum_{k=1}^{n}\frac{1}{k}[/tex]
  6. Jul 19, 2006 #5
    -Beatiful problem..but completely useless...:frown: :frown: the problem itself is even more problematic than RH..since you can calculate every Harmonic Number (approximately) the problem is to calculate:

    [tex] \sum_{d|n} d [/tex] for every n even for n big.
    Last edited by a moderator: Jul 21, 2006
  7. Jul 21, 2006 #6


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    There are other equivalences that can be considered 'elementary', the error term in prime number theorem, bounds for the sum of the moebius function..

    This one is nice quite nice though. I wonder if there's any connection between the height of the first zero off the line and the smallest counterexample to this criterea (assuming false of course).

    It's hard to know where to start on such an open question. There are lots of threads on the Riemann Hypothesis, try doing a search for them. What specific things are bothering you?
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