Loading ...
Sorry, an error occurred while loading the content.

Re: primes between n^2 and (n+1)^2. Primes between n^3 and (n+1)^3.

Expand Messages
  • Werner D. Sand
    In the same way one proves that the number of primes between cubes amounts to approximately: N = pi(n+1)^3-pi(n^3) ~ (n/ln(n))*(n+1) ~ pi(n)*(n+1), better ~
    Message 1 of 4 , Apr 12, 2007
    • 0 Attachment
      In the same way one proves that the number of primes between
      cubes amounts to approximately:

      N = pi(n+1)^3-pi(n^3) ~
      (n/ln(n))*(n+1) ~
      pi(n)*(n+1),
      better ~ n*pi(n).

      WDS

      --- In primenumbers@yahoogroups.com, "Werner D. Sand"
      <Theo.3.1415@...> wrote:
      >
      > Proof:
      >
      > pi[(n+1)^2]-pi(n^2) ~ (PNT)
      > [(n+1)^2]/ln[(n+1)^2] - (n^2)/ln(n^2) =
      > [(n+1)^2]/2ln(n+1) - (n^2)/(2ln n) =
      > [(n+1)^2]/2[ln n + ln(1+1/n)] - (n^2)/(2ln n) -> (n->inf)
      > [(n+1)^2]/(2ln n) - (n^2)/(2ln n) =
      > (2n+1)/(2ln n) =
      > n/ln n + 1/(2ln n) -> (n->inf)
      > n/ln n ~
      > pi(n)
      >
      > qed
      >
      > Werner
      >
      >
      > --- In primenumbers@yahoogroups.com, "Mark Underwood"
      > <mark.underwood@> wrote:
      > >
      > > --- In primenumbers@yahoogroups.com, "Mark Underwood"
      > > <mark.underwood@> wrote:
      > > >
      > > >
      > > > As we know, the number of primes up to n is about n/log(n).
      Given
      > > > this, it is easy to show that the number of primes between n^2
      and
      > > > (n+1)^2 is also about n/log(n).
      > > >
      > > > How much does the actual count of primes between n^2 and (n+1)^2
      > > > differ from n/log(n) ? On a very cursory inspection, it seems
      the
      > > > prime count is no more than sqrt(n) removed from n/log(n).
      > > >
      > > > Mark
      > > >
      > >
      > >
      > > Just did some prime counting and so far it holds that
      > > the number of primes between n^2 and (n+1)^2 is within the range
      > > n/log(n) +/- sqrt(n). At least for n up to about 47,000.
      > >
      > > There have been some close calls though. Here are the cases where
      > the
      > > difference between the actual prime count and n/log(n) was at
      least
      > 75
      > > percent of the square root of n.
      > >
      > > Format: (n, primecount between n^2 and (n+1)^2, percentage)
      > >
      > > (696,81)
      > > (696,85,81)
      > > (1760,204,75)
      > > (2456,268,94)
      > > (2761,390,79)
      > > (3516,486,93)
      > > (3788,508,78)
      > > (5266,675,83)
      > > (9980,1168,84)
      > > (10706,1250,93)
      > > (15646,1718,78)
      > > (23515,2458,79)
      > > (23924,2503,84)
      > > (28678,2923,76)
      > > (28678,2923,76)
      > > (32460,2986,77)
      > > (39590,3904,83)
      > >
      > > Mark
      > >
      >
    Your message has been successfully submitted and would be delivered to recipients shortly.