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>