[QUOTE=akruppa;253362]Allowing larger values would require partitioning the sieve which would take several more minutes, which I don't think worth the time.[/QUOTE]

I wonder if what you have in mind was what I had in mind.

I don't claim having had anything in mind at all when I wrote the code. It simply looped through n,r combinations and lit up bits in an array. I can't imagine what OP's haxor friends are doing if their code is two orders of magnitude slower than that.

[QUOTE=akruppa;253376]I don't claim having had anything in mind at all when I wrote the code. It simply looped through n,r combinations and lit up bits in an array. I can't imagine what OP's haxor friends are doing if their code is two orders of magnitude slower than that.[/QUOTE]

I mean what you mentioned when you suggested partitioning the sieve.

To: R.D. Silverman,

Quoting R.D. Silverman:
[QUOTE]

Your "counting function" is a function (as you gave it) of the form

C1 x + C2 sqrt(x)

You are so totally clueless that you don't even realize that this is NOT
a polynomial.
[/QUOTE]
Are you hallucinating?

Where oh where did I ever say that my [B]counting function[/B] is a polynomial?

It was CRGreathouse who casually and informally introduced
the idea of "counting polynomials" in this discussion.

But you see, unlike you, I [B][I]know[/I][/B] which expression he was referring to.

[B]Unlike you[/B], I know that he was referring to the [B][I]polynomial[/I][/B]:

[TEX]\left(\frac{n}{2}-1\right)*r^2-\left(\frac{n}{2}-2\right)*r[/TEX]

and [I][B]not[/B][/I] to the "counting function"!

Quoting R.D. Silverman:
[QUOTE]
You are a classic crank.
You are an idiot studying to become an imbecile.
You have made my ignore list faster than anyone
else ever has.
[/QUOTE]

When you keep making remarks like that, I can only [I][B]wish[/B][/I] that you would
[B][U]hurry up and start ignoring me instead of continuing to be obsessed with me![/U][/B]

Don.

Quoting CRGreathouse
[QUOTE]
They're quite close (by construction) at the end of your calculated range,
so that would be unwise.
[/QUOTE]
The word "close" is somewhat arbitrary.

[TEX]\varpi(10^{12})[/TEX] and [TEX]\varpi(10^{13})[/TEX] are apart by a factor of 10.

That's roughly the difference between traveling across the U.S.A.
and traveling all the way around the Earth!

[TEX]\varpi(10^{12})[/TEX] and [TEX]\varpi(10^{14})[/TEX] are apart by a factor of 100.

That's roughly the difference between traveling across the U.S.A.
and traveling all the way around the Earth ten times!

Now, I'm confident enough in the accuracy of my function to take that bet,
but I'm not all that confident that anyone here can determine [TEX]\varpi(10^{14})[/TEX]
in a reasonable amount of time.

Don.

[QUOTE=Don Blazys;253564]The word "close" is somewhat arbitrary.

[TEX]\varpi(10^{12})[/TEX] and [TEX]\varpi(10^{13})[/TEX] are apart by a factor of 10.[/QUOTE]

I'm not talking about how close those are, but how close the true value and your estimate are at 10^13. Simply because they're close, move at approximately the same rate, and jump around somewhat unpredictably (in the case of the true value) we'd expect many more crossings on that basis alone, even if they were not asymptotically equivalent.

I could do some Browning modelling to see how many crossings we'd expect between 10^13 and 10^14 if they're not of the same order, though I'd have to estimate how far apart they are and also how much 'noise' there is in the true function. Right now I'm not inclined to do that since it would take a few hours to do it right, but you can imagine that the number would be dozens if not hundreds or more.

[QUOTE=Don Blazys;253564]Now, I'm confident enough in the accuracy of my function to take that bet,
but I'm not all that confident that anyone here can determine [TEX]\varpi(10^{14})[/TEX]
in a reasonable amount of time.[/QUOTE]

I'll leave the offer open until the end of this month. Of course if the count to 10^14, ..., 5 * 10^14 aren't completed then neither will have to pay out, so that's not a concern. I'm more likely to invest time and effort into the calculation if I have some 'skin' in the game -- of course that's much more about my ego being on the line than the money.

[P.S. If you'd like we could also do it for charities of the winner's choice.]

To: CRGreathouse,
Quoting CRGreathouse:
[QUOTE]I'm more likely to invest time and effort into the calculation if I have some 'skin' in the game --
of course that's much more about my ego being on the line than the money.

[/QUOTE]It's not about the money for me either, (although it [I][B]should[/B][/I] be,
since my son lost his job in this screwed up economy and I am now
helping support him, my grand daughter, and my wife on one salary!)

How about this idea... regardless of whether or not the function:

[TEX]B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)=[/TEX] [TEX]\left(x-\frac{x}{\alpha*\pi*e+e}-\frac{1}{2}*\sqrt{x-\frac{x}{\alpha*\pi*e+e}}\right)*\left(1-\frac{\alpha}{\mu-2*e}\right)[/TEX]

is "absolutely correct", it is [B][I]still[/I][/B] good enough to give [B][U]exellent[/U][/B] estimates of [TEX]\varpi(x)[/TEX]
to well past [TEX]x=1,000,000,000,000[/TEX] which means that the coefficients of:

[TEX]=.64036274309582*x-.40011254372008*\sqrt{x}[/TEX]

are correct to about 14 decimal places.

There is [B][I]still[/I][/B] a lot of reasearch to be done here,
and this [B][I]is[/I][/B] to polygonal numbers as [TEX]Li(x)[/TEX] is to [TEX]\pi(x)[/TEX],
so why not consider co-authoring a paper on this topic with me.

You can have "top billing" and we can call .6403627... and .4001125...
the Greathouse-Blazys constants regardless of whether or not we will
ultimately have to jettison the function involving [TEX]\pi[/TEX],[TEX]e[/TEX],[TEX]\alpha[/TEX] and [TEX]\mu[/TEX].:smile:

I don't know enough about computers to use them for doing math,
(but I am learning how to use a pocket calculator) nor do I know how to
write "publication ready" papers in LaTex, so I [B][U]need[/U][/B] to find a professional
who can do all that, (and probably a lot more that I am unaware of).

You see, I don't hold it against you that you called me a "crank".

Here's why...

I have a [B][I]friend[/I][/B] who is a well known professor at a prestigious university
who told me point blank that although he knows that my proof of BC is good,
he can't say it publicly without the risk of being ostracised by his colleagues
who don't understand it and who would probably damage his career for
[B][I]associating[/I][/B] with a "crank". (His initials are B.B., in case he ever reads this.):wink:

I am well aware of the possibility that your situation may be similar to his,
and that it is therefore all but obligatory that you "distance" yourself from me.

I, on the other hand, have no "math career" to protect,
and couldn't care less what "real" mathematicians call me!

I'm [B][I]not[/I][/B] a "real" mathematician, and the only math I know is what my grandfather
(who [B][I]was[/I][/B] a mathematician and a bridge engineer) stuffed down my throat when I was a kid,
and from the books that I read to alleviate the boredom of having to wait between fares
while driving a taxi to support my then young family.

Thus, I am [B][I]free[/I][/B] to splatter and scatter my findings all over cyberspace,
and that is [B][I]exactly[/I][/B] what I will continue to do until one of you professionals
musters up the courage and risks your reputation to give me a helping hand!

I am now over 60 years old, and I want to [B][I]forget[/I][/B] about math and finish up my life
playing music (which is what I used to do to supplement my income as a taxi driver).

But I will [B][I][U]not[/U][/I][/B] let polygonal numbers flounder about without their very own "counting function".

Heck, even the relatively obscure "practical numbers" have a "counting function"
and considering how extraordinarily difficult polygonals of order > 2 are to count,
a counting function for them has been long, long overdue!

Moreover, and perhaps, just perhaps most importantly, it has recently been discovered that
quantum chaotic Hamiltonians may be responsible for the non-trivial zeroes of the Riemann Zeta Function
which would mean that the prime counting function is intimately connected to quantum mechanics...
so it may not be so "far fetched" to suppose that the counting function for that [B][U][I]other[/I][/U][/B] erratic sequence,
the [B]polygonal numbers of order greater than 2[/B], might give rise to the fine structure constant.

Anyway, I find you to be intellectually honest, and that's what's most important to me,
so please consider the above ramblings and let me know what you think.

Don.

if you download PARI and walk through some of the PARI commands thread, that might help the math on a computer part, as for LaTex you can possibly use :

[url]http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf[/url] it's a symbol list not all of them work everywhere at last check but it's a good help.

for example I've learned enough from and playing around with symbols without posting them from quotes from others that I can now do :

[TEX]\pi = 4\sum_{k=1}^{\infty} \frac{-1^k}{2k+1}[/TEX]

or

[TEX]\pi = \sqrt{12}\sum_{k=1}^{\infty}\frac{-3^{-k}}{2k+1}[/TEX]

[QUOTE=Don Blazys;253579]is "absolutely correct", it is [B][I]still[/I][/B] good enough to give [B][U]exellent[/U][/B] estimates of [TEX]\varpi(x)[/TEX]
to well past [TEX]x=1,000,000,000,000[/TEX] which means that the coefficients of:

[TEX]=.64036274309582*x-.40011254372008*\sqrt{x}[/TEX]

are correct to about 14 decimal places.[/QUOTE]

I agree that it's a great approximation. I reserve judgment on how many decimal places are right -- I'd have to do some analysis on my own to distinguish the primary from secondary terms and I wouldn't do the problem justice by guessing right now.

[QUOTE=Don Blazys;253579]I have a [B][I]friend[/I][/B] who is a well known professor at a prestigious university
who told me point blank that although he knows that my proof of BC is good,
he can't say it publicly without the risk of being ostracised by his colleagues
who don't understand it and who would probably damage his career for
[B][I]associating[/I][/B] with a "crank". (His initials are B.B., in case he ever reads this.)[/QUOTE]

Interesting. I've looked at your T-Z (Beal) conjecture proofs and they were utter rubbish. That your friend wouldn't see that straightaway makes me question his ability, if not credentials.

I was pleased to see that for this problem you took a more reasonable position. The count is clearly Theta(n) and you have that; the constant is almost surely close to the value you give for it; you provide evidence for your claim. That we disagree on the 'numerology' of the constant is little in comparison.

[QUOTE=Don Blazys;253579]Anyway, I find you to be intellectually honest, and that's what's most important to me,
so please consider the above ramblings and let me know what you think.[/QUOTE]

I think that the current choice of constants is overfit, so that the closeness of the match is not as great as it appears by looking at the count up to 1.1 * 10^12. In particular, I suspect the prediction for the interval
[TEX]\left(1-\frac{\alpha}{\mu-2e}\right)\left(\mathcal{B}(1.2\cdot10^{12})-\mathcal{B}(1.1\cdot10^{12})\right)[/TEX]
will have a relative error substantially greater than the 10[SUP]-11[/SUP] up to 1.1 * 10^12.

I think that I'd need to look at the residuals more closely before commenting on the Ansatz you've chosen, a*x + b*sqrt(x).

I think that the leading term is quite close but, as I wrote above, I can't say how close without looking at it more closely.

--

I have an idea for counting solutions more quickly. Depending on the details, it may also be a good way to estimate the number of solutions to high precision. To avoid embarrassment if I'm wrong I'm not going to give out detail until I have a chance to work with the idea and see if it pans out. There are very specific criteria the problem will need to have to work for efficient counting and for precise estimation (and actually they're different, so conceivably this could give a very good approximation but be inefficient if used to count exactly).

Quoting CRGreathouse:
[QUOTE]
Interesting. I've looked at your T-Z (Beal) conjecture proofs and they were utter rubbish.
That your friend wouldn't see that straightaway makes me question his ability, if not credentials.
[/QUOTE]
Well, my friend has impressive and impeccable credentials. (Phd's in both mathematics [I][B]and[/B][/I] physics.)

Moreover, he is not the only one who thinks that my proof of BC is both true and correct.
If you look at the "articles and letters" on my website, then you will find a lot more evidence that I'm right!
Even the [B]Journal of the London Mathematical Society gave my proof some support[/B] while declining to
publish it due only to a lack of available journal space, and back then my proof was a handwritten manuscript!

And that's not all! My proof [B][I][U]is[/U][/I][/B] [B][I][U]published[/U][/I][/B] in the online journal: "Unsolved Problems" where
it can be refereed not just by one or two referees, but by the [B][I]entire math community[/I][/B],
and no one has ever found a "fatal flaw" in it. Google searching "Beal's Conjecture Proof"
shows that it is consistently in the top 5, so clearly, it is both well known and popular.

By contrast, there is absolutely [B]no evidence whatsoever[/B] that my proof of BC is wrong.
You calling it "utter rubbish", and kids with names like "Punky Munky" calling me a "crank"
in no way constitutes evidence, and I can be assure you that if a fatal flaw was ever found,
then I would drop my proof like a hot potato! I would [B][U]never[/U][/B] waste my [B][I]precious[/I][/B] time on a lie.

Thus, the controversy continues, and I think that a [B][I]formal[/I][/B] online debate between
a [B]recognized panel of experts[/B] and [B]myself[/B] is in order. Maybe you can help arrange that!

Quoting CRGreathouse:

[QUOTE]
I think that the current choice of constants is overfit.
[/QUOTE]
The function [TEX]B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)[/TEX] starts out as an overestimate,

but then becomes mostly an underestimate as the "random excursions" of [TEX]\varpi(x)[/TEX]

periodically overtake [TEX]B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)[/TEX] and then drop back down below it.

Interestingly, [TEX]B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)[/TEX] never seems to overestimate [TEX]\varpi(x)[/TEX] by much.

Here's an interesting article on the possible connection between prime numbers and quantum physics:

[URL]http://www.americanscientist.org/issues/id.3349,y.0,no.,content.true,page.1,css.print/issue.aspx[/URL]