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

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  • ianredwood
    Hi, What s the highest known n for which, for all 7 pi(n) - pi(n/2)? Thankyou.
    Message 1 of 6 , Feb 10, 2010
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      Hi,

      What's the highest known n for which, for all 7 <= k <= n, pi(n/2) > pi(n) - pi(n/2)?

      Thankyou.
    • 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 2 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 3 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 4 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 5 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 6 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.


                [Non-text portions of this message have been removed]
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