Re: Prime count question
- To: mathgroup at smc.vnet.net
- Subject: [mg126485] Re: Prime count question
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Mon, 14 May 2012 01:34:47 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201205130702.DAA16830@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
Here are the k largest t@x, where k = 5:
Clear[t]
t[x_] := Evaluate@
Rationalize[PrimePi@x / ((x/Log@x)*(1 + 1/Log@x + 2.51/(Log@x)^2))]
k = 5;
pairs = Table[{N[t@x - 1] + 1, x}, {x, 599, 355991}];
pairs[[Ordering[pairs, -k]]]
{{1.00018, 59754}, {1.00018, 59798}, {1.00018, 59809}, {1.00019,
59753}, {1.00019, 59797}}
Bobby
On Sun, 13 May 2012 02:02:40 -0500, 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.
>
--
DrMajorBob at yahoo.com
- References:
- Prime count question
- From: "J.Jack.J." <jack.j.jepper@googlemail.com>
- Prime count question