[seqfan] Re: Hilliard reference in A034387.

[Robert Munafo]

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 - - -

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