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Re: Highest n, prime count pi(n), pi(n/2)

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  • ianredwood
    Whoops - I ve got to say that again. What s the highest known n for which, for any 7 pi(k) - pi(k/2)? Is it bigger than 10^(60)?
    Message 1 of 6 , Feb 10, 2010
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      Whoops - I've got to say that again.
      What's the highest known n for which, for any 7 <= k <= n, pi(k/2) > pi(k) - pi(k/2)?
      Is it bigger than 10^(60)?


      --- In primenumbers@yahoogroups.com, "ianredwood" <ianredwood@...> wrote:
      >
      > Hi,
      >
      > What's the highest known n for which, for all 7 <= k <= n, pi(n/2) > pi(n) - pi(n/2)?
      >
      > Thankyou.
      >
    • maximilian_hasler
      ... pi(k/2) pi(k) - pi(k/2) pi(k/2)*2 pi(k) The function pi(k/2)*2 - pi(k) tends to infinity in view of the asymptotic behavious of the pi() function
      Message 2 of 6 , Feb 11, 2010
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        --- In primenumbers@yahoogroups.com, "ianredwood" <ianredwood@...> wrote:
        >
        > Whoops - I've got to say that again.
        > What's the highest known n for which, for any 7 <= k <= n, pi(k/2) > pi(k) - pi(k/2)?
        > Is it bigger than 10^(60)?
        >

        pi(k/2) > pi(k) - pi(k/2)

        <=> pi(k/2)*2 > pi(k)

        The function pi(k/2)*2 - pi(k) tends to infinity
        in view of the asymptotic behavious of the pi() function
        so for all k>5 it is nonnegative and for all k>21 it is strictly positive.

        vector(20,i,(primepi((1+i)\2)*2-primepi(1+i))) /* i+1 to avoid pi(0) */
        = [-1, -2, 0, -1, 1, 0, 0, 0, 2, 1, 1, 0, 2, 2, 2, 1, 1, 0, 0, 0]

        vector(99,i,(primepi((1+i)\2)*2-primepi(1+i)))
        = [-1, -2, 0, -1, 1, 0, 0, 0, 2, 1, 1, 0, 2, 2, 2, 1, 1, 0, 0, 0, 2, 1, 1, 1, 3, 3, 3, 2, 2, 1, 1, 1, 3, 3, 3, 2, 4, 4, 4, 3, 3,
        2, 2, 2, 4, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 4, 3, 3, 2, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 3, 3, 3, 3, 3, 2, 2, 2, 4, 3, 3, 3, 5,
        5, 5, 4, 4, 4, 4, 4, 6, 6, 6, 5, 5, 5, 5]

        Maximilian
      • ianredwood
        ... Many thanks but the vector part is beyond me. How do you know that the bounds are the ones, 5 and 21, that you cite? Can you supply any literary
        Message 3 of 6 , Feb 11, 2010
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          --- In primenumbers@yahoogroups.com, "maximilian_hasler" <maximilian.hasler@...> wrote:
          >
          >
          >
          > --- In primenumbers@yahoogroups.com, "ianredwood" <ianredwood@> wrote:
          > >
          > > Whoops - I've got to say that again.
          > > What's the highest known n for which, for any 7 <= k <= n, pi(k/2) > pi(k) - pi(k/2)?
          > > Is it bigger than 10^(60)?
          > >
          >
          > pi(k/2) > pi(k) - pi(k/2)
          >
          > <=> pi(k/2)*2 > pi(k)
          >
          > The function pi(k/2)*2 - pi(k) tends to infinity
          > in view of the asymptotic behavious of the pi() function
          > so for all k>5 it is nonnegative and for all k>21 it is strictly positive.
          >
          > vector(20,i,(primepi((1+i)\2)*2-primepi(1+i))) /* i+1 to avoid pi(0) */
          > = [-1, -2, 0, -1, 1, 0, 0, 0, 2, 1, 1, 0, 2, 2, 2, 1, 1, 0, 0, 0]
          >
          > vector(99,i,(primepi((1+i)\2)*2-primepi(1+i)))
          > = [-1, -2, 0, -1, 1, 0, 0, 0, 2, 1, 1, 0, 2, 2, 2, 1, 1, 0, 0, 0, 2, 1, 1, 1, 3, 3, 3, 2, 2, 1, 1, 1, 3, 3, 3, 2, 4, 4, 4, 3, 3,
          > 2, 2, 2, 4, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 4, 3, 3, 2, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 3, 3, 3, 3, 3, 2, 2, 2, 4, 3, 3, 3, 5,
          > 5, 5, 4, 4, 4, 4, 4, 6, 6, 6, 5, 5, 5, 5]
          >

          Many thanks but the 'vector' part is beyond me. How do you know that the bounds are the ones, 5 and 21, that you cite? Can you supply any literary reference?

          Thankyou.
        • maximilian_hasler
          ... (...) ... It was mainly meant as illustration... among others, of the fact that I don t understand your k = 7 lower limit... Could you first explain that
          Message 4 of 6 , Feb 11, 2010
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            > > --- In primenumbers@yahoogroups.com, "ianredwood" wrote:
            > > >
            > > > Whoops - I've got to say that again.
            > > > What's the highest known n for which,
            > > > for any 7 <= k <= n, pi(k/2) > pi(k) - pi(k/2)?
            > > > Is it bigger than 10^(60)?
            (...)
            > > vector(99,i,(primepi((1+i)\2)*2-primepi(1+i)))
            > > = [-1, -2, 0, -1, 1, 0, 0, 0, 2, 1, 1, 0, 2, 2, 2, 1, 1, 0, 0, 0, 2, 1, 1, 1, 3, 3, 3, 2, 2, 1, 1, 1, 3, 3, 3, 2, 4, 4, 4, 3, 3,
            > > 2, 2, 2, 4, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 4, 3, 3, 2, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 3, 3, 3, 3, 3, 2, 2, 2, 4, 3, 3, 3, 5,
            > > 5, 5, 4, 4, 4, 4, 4, 6, 6, 6, 5, 5, 5, 5]
            > >
            > Many thanks but the 'vector' part is beyond me.

            It was mainly meant as illustration...
            among others, of the fact that I don't understand your k >= 7 lower limit... Could you first explain that one ?
            (From the above, one sees that for example for k=21 (position of the last "0" in the list), one has equality

            pi(k/2) = pi(k) - pi(k/2)

            i.e. for this k your initial inequality is not satisfied,
            and one sees that k=5 is the largest value for which (in the domain calculated above), we have strict inequality in the "wrong" sense,

            pi(k/2) < pi(k) - pi(k/2)

            So I don't know what the k >= 7 limit comes from.)

            > How do you know that the bounds are the ones, 5 and 21,
            > that you cite? Can you supply any literary reference?

            no, sorry...
            Hm, at a second look, it seems to follow from the equations in
            http://en.wikipedia.org/wiki/Prime_number_theorem#Bounds_on_the_prime-counting_function

            According to that, the function

            f:= x -> x/ln(x)*(1+1/ln(x));

            is a lower bound, and

            g:= x -> x/ln(x)*(1+1/ln(x)+251/100/ln(x)^2);

            is an upper bound of pi(x), for x large enough, thus:

            D(k) = pi(k/2)*2 - pi(k) > f(k/2)*2 - g(k)

            which yields, if my Maple is correct, something equivalent to
            D(k) ~ ln(2) * k * ln(k)^2 / ln(k/2)^4.

            Can you confirm this calculation ?

            Maximilian
          • Maximilian Hasler
            ... x ≥ 355991 according to the WP page. And I plotted the difference up to 10^6, there s really absolutely no doubt for smaller values ! M. [Non-text
            Message 5 of 6 , Feb 11, 2010
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              > It's the term 'for x large enough' that undoes the claim. How large is large enough?

              x ≥ 355991 according to the WP page.

              And I plotted the difference up to 10^6, there's really absolutely no
              doubt for smaller values !

              M.


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