Re: The nth Prime
 From: bill <b92057@xxxxxxxxx>
 Date: Fri, 16 May 2008 11:43:37 0700 (PDT)
On May 16, 10:59 am, "Dirk Van de moortel" <dirkvandemoor...@ThankSNO
SperM.hotmail.com> wrote:
bill <b92...@xxxxxxxxx> wrote in message
53f02011eca441b28059a1e7630e3...@xxxxxxxxxxxxxxxxxxxxxxxxxxx
P(n) is the nth prime. There are n primes less than or equal to
P(n).
Therefore, n = pi [ P(n) ] ~ P(n) / ln[ P(n) ] Rearranging;
P(n) ~ n* ln[ P(n) ]
If X = n * ln(n), then
P(n) ~ n * ln(X)
For n's up to 1,000,121,668,853 this formula gives consistently
better estimates for P(n) than "n * ln(n)". For the same range; it
also
gives better estimates than "n * ln[ P(n)]".
It is possible that as 'n' increases; that "n * ln(n)" or "n *
ln[ P(n)]" may
give a more accurate estimate of P(n).
If we write out the expression we get;
P(n) ~ n * ln[ n* ln(n)] Is it possible to simplify the right hand
term?
Bill J
Of course n ln( n ln(n) ) = n ln(n) + n ln( ln(n) ).
But then you should worry about simplifying the next better approximation:
http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_...
;)
Dirk Vdm
Thanks  for nothing. I thought that I had discovered
something new.
Bill J
.
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