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Old 2011-02-19, 19:07   #1
Don Blazys
 
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Default Standard crank division by zero thread

Here:

httр://donblazys.com/on_рolygonal_numbers_3.рdf

you will find my "Special Polygonal Number Counting Function" that approximates (to a very high degree of accuracy)
how many "polygonal numbers of order greater than 2" there are under some given number x
in much the same way that the function: Li(x) approximates how many primes there are under some given number x.

The reason that I am posting it here is because this forum seems to have many experienced "coders"
who have access to some very powerful computers.

Here is my question...

Would it be possible to calculate \varpi(x) to say, x=10^{18} or so?
Given that \pi(x) (the number of primes under x) has been calculated to x=10^{24},
I should think that this would be "easy", or at least, "possible in a reasonable amount of time",
but as it turns out, the coders who determined the present "world record" \varpi(1,100,000,000,000)=704,398,597,754
informed me that determining \varpi(10^{13}) would probably take about a year.

There is an intrepid young coder who is letting his computer run constantly
and will verify that "world record" in a few days from now. He estimates that he will be able to determine \varpi(10^{13})
in about 6 or 7 months, so I doubt that his machine is powerful enough to determine \varpi(10^{18}) within our lifetimes!

I don't own a computer (I use my grand daughters laptop to post) nor do I know anything about "coding" or "programing".
Thus, I would greatly appreciate any help or advice whatsoever as to how a determination of \varpi(10^{18}) might be possible.

Thanks in advance,

Don.
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Old 2011-02-19, 23:18   #2
science_man_88
 
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Quote:
Originally Posted by Don Blazys View Post
Here:

httр://donblazys.com/on_рolygonal_numbers_3.рdf

you will find my "Special Polygonal Number Counting Function" that approximates (to a very high degree of accuracy)
how many "polygonal numbers of order greater than 2" there are under some given number x
in much the same way that the function: Li(x) approximates how many primes there are under some given number x.

The reason that I am posting it here is because this forum seems to have many experienced "coders"
who have access to some very powerful computers.

Here is my question...

Would it be possible to calculate \varpi(x) to say, x=10^{18} or so?
Given that \pi(x) (the number of primes under x) has been calculated to x=10^{24},
I should think that this would be "easy", or at least, "possible in a reasonable amount of time",
but as it turns out, the coders who determined the present "world record" \varpi(1,100,000,000,000)=704,398,597,754
informed me that determining \varpi(10^{13}) would probably take about a year.

There is an intrepid young coder who is letting his computer run constantly
and will verify that "world record" in a few days from now. He estimates that he will be able to determine \varpi(10^{13})
in about 6 or 7 months, so I doubt that his machine is powerful enough to determine \varpi(10^{18}) within our lifetimes!

I don't own a computer (I use my grand daughters laptop to post) nor do I know anything about "coding" or "programing".
Thus, I would greatly appreciate any help or advice whatsoever as to how a determination of \varpi(10^{18}) might be possible.

Thanks in advance,

Don.

I know my values may be off so first I''l check them up (no I'm not a mathematician), mass of proton according to my calculator (which if I'm not mistaken is needed for the mass ratio you talk of) is about 1.67 *10^-27 , mass of electron is about 9.109 * 10^-31, the ratio my calculator gives me with the ratio of the 2 constants is 1836.15 but I'll shorten that to 1.83*10^3
assuming the e talked of that's not defined is 2.71828............ ? , and you give the fine structure constant. I find that B(x)\times(1-\frac{\alpha}{\mu-2\times {e}}) is about 9.99**10^-1 which if my math is correct means the figures in the table are wrong. never mind I don't see why the first equation then but like I said I'm not mathematical

Last fiddled with by science_man_88 on 2011-02-20 at 00:01
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Old 2011-02-21, 00:44   #3
Don Blazys
 
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The counting function works.

In fact, it approximates the number of polygonal numbers of order greater than 2
to a much higher degree of accuracy than Li(x) approximates the prime numbers.

The challenge now is to determine \varpi(x) to about x=10^{18}.

That will tell us a lot.

Don.
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Old 2011-02-21, 01:30   #4
CRGreathouse
 
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Quote:
Originally Posted by Don Blazys View Post
In fact, it approximates the number of polygonal numbers of order greater than 2
to a much higher degree of accuracy than Li(x) approximates the prime numbers.
Of course -- these are easy to count by virtue of being well-behaved polynomials. There's no need to account for the irregularities of the zeta function.

Your approximation is constant * n + lower order terms and your paper shows you modifying the coefficient of n based on counts done so far. That you use constants from physics is of no particular significance here; higher counts will show that the constant term is still off and needs to be modified.

Now I'm curious about this "young coder" you have working on the task. What algorithm is he using? Based on the speed it seems like direct enumeration, which would seem to be a very slow method for the task. I would think that the theory of Diophantine equations could be used to construct an inclusion-exclusion method that would be orders of magnitude faster.

Last fiddled with by CRGreathouse on 2011-02-21 at 01:31
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Old 2011-02-21, 15:09   #5
Don Blazys
 
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Quoting CRGreathouse:
Quote:
Of course --
these are easy to count by virtue of being well-behaved polynomials.
There's no need to account for the irregularities of the zeta function.
The sequence of "polygonal numbers of order greater than2",
which can be found here:

http://oeis.org/A090466

is not a "well behaved" sequence.

As with the sequence of primes, it can only be described as
erratic, irregular, patternless, random and unpredictable.

The random fluctuations in \varpi(x) may not be as pronounced as
the random fluctuations in \pi(x) (the number of primes under x)
but they are nevertheless quite random, and the sequence is
extraordinarily difficult to count.

At least a dozen coders crashed their computers trying to break the
present "world record" \varpi(1,100,000,000,000)=704,398,597,754.

Quoting CRGreathouse
Quote:
Your approximation is constant * n + lower order terms and your paper
shows you modifying the coefficient of n based on counts done so far.
Well, my counting function can be "simplified" as follows:

B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)= \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)

=.64036274309582*x-.40011254372008*\sqrt{x},

and since I am the first mathematician to develop a "counting function" for
"polygonal numbers of order greater than 2", I suppose that I could name
the coefficients .6403627... and .4001125... after myself and call them
"Blazys constants", but I am much too humble and modest to do that!

Quoting CRGreathouse:
Quote:
That you use constants from physics is of no particular significance here;
higher counts will show that the constant term is still off and needs to be modified.
When I first began work on this counting function,
there was very little data to work with. However,
as the coders provided me with higher counts of \varpi(x).
the "physical" constants \alpha and \mu emerged naturally.

That's why I am quite certain that the function is correct,
and will hold up regardless of how high the counts are.

Quoting CRGreathouse:
Quote:
Now I'm curious about this "young coder" you have working on the task.
What algorithm is he using? Based on the speed it seems like direct enumeration,
which would seem to be a very slow method for the task.

I would think that the theory of Diophantine equations could be used to
construct an inclusion-exclusion method that would be orders of magnitude faster.
Well, the young coder is working with me... not for me.

Since I don't know anything about computers or coding,
I really can't comment on his methods, but he recently
informed me that \varpi(1,200,000,000,000) will be determined
by this coming Friday, and that will be the new world record.

If you are interested, I will post the result here as it comes in.

Don.
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Old 2011-02-21, 15:26   #6
R.D. Silverman
 
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Quote:
Originally Posted by Don Blazys View Post
Quoting CRGreathouse:


The sequence of "polygonal numbers of order greater than2",
which can be found here:

http://oeis.org/A090466

is not a "well behaved" sequence.

As with the sequence of primes, it can only be described as
erratic, irregular, patternless, random and unpredictable.
He did not say that it was. You need to learn to read.
He said that the polynomials were well behaved.


Quote:
The random fluctuations in \varpi(x) may not be as pronounced as
the random fluctuations in \pi(x) (the number of primes under x)
but they are nevertheless quite random,
Oh? What is the underlying density function? Please specify.


Quote:
and since I am the first mathematician to develop a "counting function" for
"polygonal numbers of order greater than 2",
May we ask: Where did you get your math degree?

Quote:
I suppose that I could name
the coefficients .6403627... and .4001125... after myself and call them
"Blazys constants", but I am much too humble and modest to do that!
You have an approximation that works for the range(s) for which you
have computed values. Please show us the derivation of your
counting function. Or is merely an emprical result from fitting curves?
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Old 2011-02-21, 20:07   #7
CRGreathouse
 
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Quote:
Originally Posted by Don Blazys View Post
I am the first mathematician to develop a "counting function" for
"polygonal numbers of order greater than 2"
How do you know that you're the first?

Also, you seem to have only an empirical fit. What have you actually proved? I can prove bounds on the distribution function; can you?

Quote:
Originally Posted by Don Blazys View Post
That's why I am quite certain that the function is correct,
and will hold up regardless of how high the counts are.
I don't suppose you'd care to back that up with a friendly bet?


I'm not prepared at the moment to commit the time or processor power, but at some point I may want to compete with the team of you and your coder to reach 10^14.
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Old 2011-02-21, 21:10   #8
science_man_88
 
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Quote:
Originally Posted by Don Blazys View Post
Quoting CRGreathouse:


The sequence of "polygonal numbers of order greater than2",
which can be found here:

http://oeis.org/A090466

is not a "well behaved" sequence.

As with the sequence of primes, it can only be described as
erratic, irregular, patternless, random and unpredictable.

The random fluctuations in \varpi(x) may not be as pronounced as
the random fluctuations in \pi(x) (the number of primes under x)
but they are nevertheless quite random, and the sequence is
extraordinarily difficult to count.

At least a dozen coders crashed their computers trying to break the
present "world record" \varpi(1,100,000,000,000)=704,398,597,754.

Quoting CRGreathouse


Well, my counting function can be "simplified" as follows:

B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)= \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)

=.64036274309582*x-.40011254372008*\sqrt{x},

and since I am the first mathematician to develop a "counting function" for
"polygonal numbers of order greater than 2", I suppose that I could name
the coefficients .6403627... and .4001125... after myself and call them
"Blazys constants", but I am much too humble and modest to do that!

Quoting CRGreathouse:


When I first began work on this counting function,
there was very little data to work with. However,
as the coders provided me with higher counts of \varpi(x).
the "physical" constants \alpha and \mu emerged naturally.

That's why I am quite certain that the function is correct,
and will hold up regardless of how high the counts are.

Quoting CRGreathouse:


Well, the young coder is working with me... not for me.

Since I don't know anything about computers or coding,
I really can't comment on his methods, but he recently
informed me that \varpi(1,200,000,000,000) will be determined
by this coming Friday, and that will be the new world record.

If you are interested, I will post the result here as it comes in.

Don.
Assigning all the values you give and the final equation I can give you that result for x = 1.2*10^12 but you need PARI for me to give you the script I wrote of the equation.

according to my math:

Code:
alpha = 137.035999084^-1
Code:
micro = 1836.15267247
and:

Code:
blazy(x)=(x-(x/(alpha*micro*Pi*exp(1)+exp(1)))-(.5*(sqrt(x-(x/(alpha*Pi*exp(1)+exp(1)))))))*(1-(alpha/(micro-(2*exp(1)))))
try it out for 1.2*10^12 and I get:

Quote:
1189750897554.266557436437709
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Old 2011-02-21, 21:42   #9
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never mind coding error.. added a term in.
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Old 2011-02-21, 22:02   #10
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Default result after correction

the only error I found was a micro where it didn't belong, with this corrected I redid the test (took 0 ms according to my timer) and I got the new result of:

Quote:
768434853413.6797063854667429

Last fiddled with by science_man_88 on 2011-02-21 at 22:22
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Old 2011-02-21, 22:39   #11
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also with the corrected code and maximum memory allocation my machine can handle in the PARI console I calculated this one:

Code:
(18:24)>blazy(1.2*10^100000000)
%34 = 7.684352917150111789071860349 E99999999
(18:30)>##
  ***   last result computed in 41,469 ms.
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