[seqfan] Re: Hilliard reference in A034387.

[Charles Greathouse]

We should really avoid linking to Google Docs, pastebin, and other such
services when we could have that information locally. (Nothing against
them, just a question of suitability for this particular purpose.)

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Sun, Mar 17, 2013 at 2:15 AM, Robert Munafo <mrob27 at gmail.com> wrote:

> On 3/17/13, L. Edson Jeffery <lejeffery2 at gmail.com> wrote:
> > The link in A034387 to a Google document by Cino Hilliard is evidently an
> > edit page only accessible by him (I think you need a password to access
> the
> > document). Could someone who knows Cino please contact him and see if he
> > can fix the problem?
>
> The same unusable link is in A006880; there are three more of these
> password-protected links in A046731, another in A131587, another in
> A133391, two more in A139390 (one of which he actually deleted, so
> it's not only unreadable, it's actually a dead link), and another one
> in A139564.
>
> I've clicked the "request access" button for all of these.
>
> The following from 2008 (and also containing one of these unreadable
> links) is probably relevant:
>
>  - - - begin quoted message - - -
>
>  RE: [primeform] Re: (sum of primes < n)
>  Posted By: hillcino368
>  Mon Jun 9, 2008 12:47 am
>
>  To: primeform at yahoogroups.com
>  From: oeis-2008 at ...
>  Date: Sun, 8 Jun 2008 14:07:37 +0000
>  Subject: [primeform] Re: (sum of primes < n)
>
> Serendipity strikes in full bloom now. Notice,
>
> SumP(10^10) = 2220822432581729238
> Pi(10^20) = 2220819602560918840
>
> This close fit prompted me to do some more heuristics.
>
> We check some other entries and do the irristable-conjecture a formula.
> Yep, this makes it it is reasonable to assume SumP(10^n) -> Pi(10^2n).
>
> Converting our input to powers of 10, we have
>
> SumP(n) ~ Pi(10^2*(log(n)/log(10))) Now 10^2*(log(n)/log(10)) = n^2.
> So there we have it . SumP(n) ~ Pi(n^2).
>
> 1. SumP(n) ~ Pi(n^2).
>
> We know that x/log(x) is asymptotic to Pi(x), so we try that first with
> x=n^2
> as the perameter in the estimator for SumP(n) ~ Pi(x). Plugging that into
> 1.,
>
> Sump(n) ~ n^2/(log(n^2) = n^2(2*log(n)). This gives us reasonable results
> but
> we also know x/(log(x)-1) gives us better results for Pi(x). So we plug
> that
> into 1.
> to get SumP(n) ~ n^2/(2log(x)-1). This justifies the formula I have been
> touting.
>
> I am not done yet. Indeed, we know Li(x) or the Logarithmetic Integral
> gives
> superior results for Pi(x) just like Gauss thought it would. Later Riemann
> came
> up with a better estimator for Pi(x) using Li(x). I scripted these Pari
> functions
> for the calculations tha follow.
>
> R(x) = local(j); (sum(j=1,400,moebius(j)*Li(x^(1/j))/j))
>
> Li(x) = local(x);-eint1(log(1/x))
>
> A tabulation of these values along with Marc Deleglise's SumP(x) up to x=
> 10^20
> can be found at
>
> http://docs.google.com/Doc?id=dgpq9w4b_26dtrq634m
>
> I post the usual suspects here using the Riemann appromation for Pi(x).
>
> SumP(10^10) = 2220822432581729238,
> R(10^20) = 2220819602556027015,
> Error = 0.0000012743142,
>
> SumP(10^15) = 14692398516908006398225702366,
> R(10^30) = 14692398897720432716641650390,
> Error = 0.0000000259190
>
> SumP(10^20) = 109778913483063648128485839045703833541,
> R(10^40) = 109778913489828302826978255361638504017,
> Error = 6.162071097138 E-11
>
> Enjoy,
> Cino Hilliard
>
>  - - - end quoted message - - -
>
> --
>   Robert Munafo  --  mrob.com
>   Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 -
> mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
>
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