# 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...@ThankS-NO-

SperM.hotmail.com> wrote:

bill <b92...@xxxxxxxxx> wrote in message

53f02011-eca4-41b2-8059-a1e7630e3...@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

.

**Follow-Ups**:**Re: The nth Prime***From:*Gerry Myerson

**References**:**The nth Prime***From:*bill

**Re: The nth Prime***From:*Dirk Van de moortel

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