       Re: Prime count question

• To: mathgroup at smc.vnet.net
• Subject: [mg126491] Re: Prime count question
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Mon, 14 May 2012 01:36:52 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201205130702.DAA16830@smc.vnet.net>

```Since the number of primes greater than or equal to any real number is infinite, I assume that you mean less than or equal to. In that case use, use the built-in function PrimePi.

Your parentheses are unbalanced so the following may not be your intended equation. And once again you have failed to use proper Mathematica syntax in posing your question.

Maximize returns a local maxima in this case so I applied brute force.

t[x_] = PrimePi[x]/((x Log[x] (1 + 2.51/(Log[x]^2))));

ymin = 599; tax = 355991;

data = ({#, t[#]}& /@ Range[ymin, ymax];

dataMax = Max[data[[All, 2]]];

Select[data, #[] == dataMax&]

{{59797, 1.00019}}

t

1.00019

Bob Hanlon

On May 13, 2012, at 3:02 AM, "J.Jack.J." <jack.j.jepper at googlemail.com> wrote:

> Let pi(x) be the number of primes greater than or equal to x.
>
> Then how do I find, through Mathematica, x such that
>
> t(x) = pi(x) / ((x/ln(x))*(1+1/ln(x) + 2.51/(ln^2(x))))
>
> is the highest t(y) such that  599 <= y <= 355991?
>
> Many thanks in advance -- thanks also to those who helped with my
> previous question.
>

```

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