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Re: [PrimeNumbers] proving the Riemann hypothesis

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  • Phil Carmody
    ... Correct and Correct. However, Bertrand s postulate indicates that at prime P1 the gap must be less than P1. RH itself does not bound the gaps, they are
    Message 1 of 40 , Feb 1, 2007
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      --- Nathan Russell <windrunner@...> wrote:
      > On 2/1/07, Phil Carmody <thefatphil@...> wrote:
      > >
      > > --- Shi Huang <shuangtheman@... <shuangtheman%40yahoo.com>> wrote:
      > > > If the RH is false, the gap between prime P1 and the
      > > > next prime P2 could be anywhere from 1 to infinitely
      > > > large.
      > >
      > > False. Your attempt at a proof ends here.
      >
      > To further clarify for the original poster's benefit, of course there exist
      > arbitrarily large (greater than any chosen integer) gaps following primes.
      > This follows from the prime number theorem. There can be no infinite gaps,
      > however, or the prime preceding the gap would be the largest (finite) prime,
      > and we know that the number of primes is infinite.

      Correct and Correct.
      However, Bertrand's postulate indicates that at prime P1 the gap must be less
      than P1.
      RH itself does not bound the gaps, they are already bounded. RH simply reduces
      that bound (to something almost as pathetically weak as Bertrand, compared with
      the empirical evidence).

      This means that my comment about 3 days ago (about the RH) was in fact
      incorrect.

      Phil

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    • Werner D. Sand
      For example 2 adjacent gaps cannot be equal if they aren t multiple of 6. For example the gap between 2 pairs of twins is at least 4. For example each prime
      Message 40 of 40 , Feb 7, 2007
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        For example 2 adjacent gaps cannot be equal if they aren't multiple of
        6. For example the gap between 2 pairs of twins is at least 4. For
        example each prime number has the form 2n+/-1, 3n+/-1, 4n+/-1, 6n+/-1.
        Each pair of twins has the form 12n+-1, there are approximate formulas
        for the nth prime and the number of primes < x and so on. You cannot
        call all this random ore unpredictable. Of course the prime numbers are
        distributed as regularly as possible, that's a tautology. In
        mathematics everything is as regular as possible. Is pi random? Build
        P=2,357111317192329…, and you have the same case as pi. Consider the
        primes to be an irrational number, and there are no problems. If you
        mean there is no formula f(n) which produces primes for each n, then
        you are right. In this sense primes are random. (I am not quite sure –
        there is a formula p=[k^n^3] (H.W.Mills) which is said to produce only
        prime numbers). If you define "formula" as an algorithm, as a
        calculation instruction such as the sieve of Eratosthenes, then the
        primes are not random but simply what they are. Perhaps the compound
        numbers are random? Or are they only non-transparently complicated?

        Werner
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