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

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  • 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 1 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 2 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 3 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 4 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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