Finding Primes: A Divisor summatory function


by JeremyEbert
Tags: divisor function
JeremyEbert
JeremyEbert is offline
#1
Jul18-11, 07:19 AM
P: 205
Divisor summatory function is a function that is a sum over the divisor function. It can be visualized as the count of the number of lattice points fenced off by a hyperbolic surface in k dimensions. My visualization is of a different conic , one of a parabola. In fact my lattice points are not arranged in a square either, they are arranged in parabolic coordinates. My lattice point counting algorithm is simple enough though.
for k = 0 --> floor [sqrt n]
SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)
my visualization:
http://dl.dropbox.com/u/13155084/prime.png

reference:
http://en.wikipedia.org/wiki/Divisor...ion#Definition
related:
http://mathworld.wolfram.com/GausssCircleProblem.html
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JeremyEbert
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#2
Jul19-11, 06:33 AM
P: 205
To find a prime with DSUM(n):

for k = 0 --> floor [sqrt n]
SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)

if DSUM(n) - DSUM(n-1) = 2 then n is prime
Anti-Crackpot
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#3
Jul23-11, 08:18 PM
P: 75
In other words, Jeremy, may posters to this forum infer that you are suggesting that you have rediscovered the Dirichlet Divisor Sum [D(n)]?

If so, I suggest you post some data, say up to 10,000, to better convince the skeptical, although it is an elementary insight to note that your formula matches, in slightly different manner, the definition.

Divisor Summatory Function: Definition
http://en.wikipedia.org/wiki/Divisor...ion#Definition

- AC

JeremyEbert
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#4
Jul23-11, 10:15 PM
P: 205

Finding Primes: A Divisor summatory function


Quote Quote by Anti-Crackpot View Post
In other words, Jeremy, may posters to this forum infer that you are suggesting that you have rediscovered the Dirichlet Divisor Sum [D(n)]?

If so, I suggest you post some data, say up to 10,000, to better convince the skeptical, although it is an elementary insight to note that your formula matches, in slightly different manner, the definition.

Divisor Summatory Function: Definition
http://en.wikipedia.org/wiki/Divisor...ion#Definition

- AC
Yes I am suggesting that and yes it does look similar.
here is 10,000 with javascript:
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Also I would like to point out that the terms in this version give the count of the number of ways that integers <=n can be written as a product of k
Anti-Crackpot
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#5
Jul24-11, 09:25 PM
P: 75
It doesn't just "look similar," Jeremy. Your formula is, algebraically, in terms of the sums, an exact match for the Dirichlet Divisor Sum. In other words, the formula is not in question. What is in question is the Geometry.

Can you -- and I know this might seem silly -- but can you prove that the formula accurately describes the Geometrical construction from which you derived the formula?

- AC
JeremyEbert
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#6
Jul24-11, 09:35 PM
P: 205
Quote Quote by Anti-Crackpot View Post
It doesn't just "look similar," Jeremy. Your formula is, algebraically, in terms of the sums, an exact match for the Dirichlet Divisor Sum. In other words, the formula is not in question. What is in question is the Geometry.

Can you -- and I know this might seem silly -- but can you prove that the formula accurately describes the Geometrical construction from which you derived the formula?

- AC
I'm not sure....What does the hyperbolic proof look like? In essence I’m just deforming the” hyperbola across a square lattice” into a straight line across a parabolic coordinate lattice at the square root.
Anti-Crackpot
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#7
Jul24-11, 09:54 PM
P: 75
Quote Quote by JeremyEbert View Post
In essence I’m just deforming the” hyperbola across a square lattice” into a straight line across a parabolic coordinate lattice at the square root.
Well, then, that's what you need to show. You need a formula to describe the geometric transformation. Maybe one of the many Mathematics or Physics PhD's upon this forum will take an interest in your problem and help you do that.

Personally, I don't know enough about the specific maths involved to be of much help. I am only trying to show you where your model seems to fall short.

- AC
JeremyEbert
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#8
Jul25-11, 06:40 AM
P: 205
Quote Quote by Anti-Crackpot View Post
Maybe one of the many Mathematics or Physics PhD's upon this forum will take an interest in your problem and help you do that.
Now that would be truly glorious. :)

Quote Quote by Anti-Crackpot View Post
I am only trying to show you where your model seems to fall short.
Here is some of the process I went through to come up with this model. It is a VERY rough, unfinished draft but at least it descibes some of the math involved.

http://dl.dropbox.com/u/13155084/Pyt...%20lattice.pdf
JeremyEbert
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#9
Jul26-11, 09:00 PM
P: 205
a related thread.
http://www.physforum.com/index.php?showtopic=29372&st=0
JeremyEbert
JeremyEbert is offline
#10
Jul26-11, 09:09 PM
P: 205
After you click ok;

press "1"

press "space bar"
wait 5 sec...

press "2"
wait 5 sec....

press "space bar"


now play around with the views, 3D and zoom.

Can you see how I'm trying to relate it to the "shell theorem" and the "Inverse-square law"?

http://en.wikipedia.org/wiki/Shell_theorem


http://en.wikipedia.org/wiki/Inverse-square_law


http://dl.dropbox.com/u/13155084/PL3...3D_Sphere.html
JeremyEbert
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#11
Sep23-11, 11:52 AM
P: 205
My model shows:

Dirichlet Divisor function

sum ((2*floor[(n - k^2)/k]) + 1) where k=1 to floor[sqrt(n)]

= A006218 http://oeis.org/A006218


Cicada function

sum ( ((n-k^2)/(2k)) - (((n-k^2)/(2k)) mod .5) ) * -2 where k=1 to n


= A161664 http://oeis.org/A161664



I would think this would be of great interest to professional mathematicians, maybe not...
JeremyEbert
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#12
Sep24-11, 10:07 AM
P: 205
Quote Quote by JeremyEbert View Post
My model shows:

Dirichlet Divisor function

sum ((2*floor[(n - k^2)/k]) + 1) where k=1 to floor[sqrt(n)]

= A006218 http://oeis.org/A006218


Cicada function

sum ( ((n-k^2)/(2k)) - (((n-k^2)/(2k)) mod .5) ) * -2 where k=1 to n


= A161664 http://oeis.org/A161664



I would think this would be of great interest to professional mathematicians, maybe not...
Borrowing from some insight on Harmonic Numbers…

Cicada function =
(-1/2 (2 n H(n)-n^2-n) ) + (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

Dirichlet Divisor function =
n H(n) - (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

So I guess I’m open to suggestions on what to call the function:

f(n) = sum ( (((n-k^2)/(2k)) mod .5) ) * 2 where k=1 to n
JeremyEbert
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#13
Sep24-11, 10:18 PM
P: 205
Quote Quote by JeremyEbert View Post
Borrowing from some insight on Harmonic Numbers…

Cicada function =
(-1/2 (2 n H(n)-n^2-n) ) + (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

Dirichlet Divisor function =
n H(n) - (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

So I guess I’m open to suggestions on what to call the function:

f(n) = sum ( (((n-k^2)/(2k)) mod .5) ) * 2 where k=1 to n

This might be easier to understand for some:

c(n) = sum((n-k^2)/(2k)) * -2 where k = 1 to n

H(n) = nth Harmonic Number

T(n) = nth Triangular Numner

j(n) = sum((n-k^2)/(2k) mod .5) * 2 where k = 1 to n


non-divisor base = c(n)

divisor base = n*H(n)

c(n) + j(n) = Cicada function

n*H(n) - j(n) = sum(tau(n))

n*H(n) + c(n) = T(n)
JeremyEbert
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#14
Sep27-11, 10:22 AM
P: 205
I need some help guys.... I'm looking for a way to get a single formula for the sigma function. So far I have got it down to 2.

sigma(0,n) = ceiling [ SUM(2*(((((n-u)-k^2)/(2k)) mod .5) - (((n-k^2)/(2k)) mod .5 ))) ] where k=1 to n, u = 0.000001/n


sigma(x,n) = ceiling [ SUM(((2*k*((((n-u)-k^2)/(2k)) mod .5)) - (n mod k))^x) ] where k=1 to n, u = 0.0000001/n, x>0
JeremyEbert
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#15
Oct16-11, 05:28 PM
P: 205
Quote Quote by JeremyEbert View Post
This might be easier to understand for some:

c(n) = sum((n-k^2)/(2k)) * -2 where k = 1 to n

H(n) = nth Harmonic Number

T(n) = nth Triangular Numner

j(n) = sum((n-k^2)/(2k) mod .5) * 2 where k = 1 to n


non-divisor base = c(n)

divisor base = n*H(n)

c(n) + j(n) = Cicada function

n*H(n) - j(n) = sum(tau(n))

n*H(n) + c(n) = T(n)
A nice little chart:
http://dl.dropbox.com/u/13155084/chart.pdf
JeremyEbert
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#16
Oct21-11, 09:07 AM
P: 205
More detail...

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png

how many unique right triangles with side lengths less than n,,can be formed with the height equal to the sqrt(n) and the base and hypotenuse are both either integer or half-integer solutions?


Answer:

sigma(0,n) / 2 where n is not a perfect square
(sigma(0,n)-1) / 2 where n is a perfect square


the parabolas are formed by tracing the right triangles with a hypotenuse-base difference of k and a height of sqrt(n).

when the base and hypotenuse are both either integers or half-integers then k is a divisor of n.
JeremyEbert
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#17
Nov10-11, 07:45 PM
P: 205
Quote Quote by JeremyEbert View Post
Divisor summatory function is a function that is a sum over the divisor function. It can be visualized as the count of the number of lattice points fenced off by a hyperbolic surface in k dimensions. My visualization is of a different conic , one of a parabola. In fact my lattice points are not arranged in a square either, they are arranged in parabolic coordinates. My lattice point counting algorithm is simple enough though.
for k = 0 --> floor [sqrt n]
SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)
my visualization:
http://dl.dropbox.com/u/13155084/prime.png

reference:
http://en.wikipedia.org/wiki/Divisor...ion#Definition
related:
http://mathworld.wolfram.com/GausssCircleProblem.html
http://dl.dropbox.com/u/13155084/prime.png

complex number equation for my lattice points:

amplitude = a = (n+1)/2
angular frequency = w = acos((n-1)/(n+1))
time (moments) = t = acos(1-k(2/(n+1)))/w

a * e^iwt

http://en.wikipedia.org/wiki/Trigono...omplex_numbers


Interesting wolfram view...

((n+1)/2) * e^iwt where w = acos(1-(2/(n+1)))

Re:

http://www.wolframalpha.com/input/?i...36%2C+36%7D%5D

Im:
http://www.wolframalpha.com/input/?i...36%2C+36%7D%5D

I wish wolfram had a scatter plot function.
JeremyEbert
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#18
Nov14-11, 02:14 PM
P: 205
Quote Quote by JeremyEbert View Post
http://dl.dropbox.com/u/13155084/prime.png

complex number equation for my lattice points:

amplitude = a = (n+1)/2
angular frequency = w = acos((n-1)/(n+1))
time (moments) = t = acos(1-k(2/(n+1)))/w

a * e^iwt

http://en.wikipedia.org/wiki/Trigono...omplex_numbers


Interesting wolfram view...

((n+1)/2) * e^iwt where w = acos(1-(2/(n+1)))

Re:

http://www.wolframalpha.com/input/?i...36%2C+36%7D%5D

Im:
http://www.wolframalpha.com/input/?i...36%2C+36%7D%5D

I wish wolfram had a scatter plot function.
Here is a nice table view of my ((n+1)/2) * e^(i acos(1-k(2/(n+1)))) function:

http://www.wolframalpha.com/input/?i...7Bk%2C+n%7D%5D

Squaring the imaginary part gives us the multiplication table:

http://www.wolframalpha.com/input/?i...7Bk%2C+n%7D%5D

A little deeper view of how the function works:

http://www.wolframalpha.com/input/?i...%7Bk%2Cn%7D%5D


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