Finding Primes: A Divisor summatory function

In summary, the conversation discusses the concept of a divisor summatory function, which is a function that is a sum over the divisor function. It is described as the count of the number of lattice points fenced off by a hyperbolic surface in k dimensions. The visualization of this function is of a different conic, such as a parabola, and the lattice points are arranged in parabolic coordinates instead of a square. The conversation also includes a lattice point counting algorithm and references to related topics. The conversation concludes with the presenter seeking suggestions on what to call the function and providing alternative representations of it.
  • #1
JeremyEbert
204
0
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 [Broken]

reference:
http://en.wikipedia.org/wiki/Divisor_summatory_function#Definition
related:
http://mathworld.wolfram.com/GausssCircleProblem.html
 
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  • #2
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
 
  • #3
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_summatory_function#Definition

- AC
 
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  • #4
Anti-Crackpot said:
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_summatory_function#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 [Broken]

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
 
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  • #5
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
 
  • #6
Anti-Crackpot said:
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.
 
  • #7
JeremyEbert said:
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
 
  • #8
Anti-Crackpot said:
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. :)

Anti-Crackpot said:
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/Pythagorean%20lattice.pdf [Broken]
 
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  • #10
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  • #11
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...
 
  • #12
JeremyEbert said:
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
 
  • #13
JeremyEbert said:
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)
 
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  • #14
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
 
  • #15
JeremyEbert said:
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 [Broken]
 
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  • #16
More detail...

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

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.
 
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  • #17
JeremyEbert said:
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 [Broken]

reference:
http://en.wikipedia.org/wiki/Divisor_summatory_function#Definition
related:
http://mathworld.wolfram.com/GausssCircleProblem.html

http://dl.dropbox.com/u/13155084/prime.png [Broken]

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/Trigon...p_to_exponential_function_and_complex_numbers


Interesting wolfram view...

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

Re:

http://www.wolframalpha.com/input/?...+++n)])+(1+++n)]/2,+{n,+-1,+9},+{t,+-36,+36}]

Im:
http://www.wolframalpha.com/input/?...+++n)])+(1+++n)]/2,+{n,+-1,+9},+{t,+-36,+36}]

I wish wolfram had a scatter plot function.
 
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  • #18
JeremyEbert said:
http://dl.dropbox.com/u/13155084/prime.png [Broken]

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/Trigon...p_to_exponential_function_and_complex_numbers


Interesting wolfram view...

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

Re:

http://www.wolframalpha.com/input/?...+++n)])+(1+++n)]/2,+{n,+-1,+9},+{t,+-36,+36}]

Im:
http://www.wolframalpha.com/input/?...+++n)])+(1+++n)]/2,+{n,+-1,+9},+{t,+-36,+36}]

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/?...rcCos[1+-+k+(2/(n+++1))])]},+{n,+25},+{k,+n}]

Squaring the imaginary part gives us the multiplication table:

http://www.wolframalpha.com/input/?...Cos[1+-+k+(2/(n+++1))])]^2},+{n,+25},+{k,+n}]

A little deeper view of how the function works:

http://www.wolframalpha.com/input/?...+*+e^(+i+acos(1-k(2/(n+1)))++)},{n,25},{k,n}]
 
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  • #19
JeremyEbert said:
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/?...rcCos[1+-+k+(2/(n+++1))])]},+{n,+25},+{k,+n}]

Squaring the imaginary part gives us the multiplication table:

http://www.wolframalpha.com/input/?...Cos[1+-+k+(2/(n+++1))])]^2},+{n,+25},+{k,+n}]

A little deeper view of how the function works:

http://www.wolframalpha.com/input/?...+*+e^(+i+acos(1-k(2/(n+1)))++)},{n,25},{k,n}]

Recursion via a variable t:


(n+1)/t * e^(i cos^(-1)(1-k(2/(n+1))))

1<=k<=n
1<=n<=t
1<=t<=25

Gives us our fimiliar pattern:

http://dl.dropbox.com/u/13155084/recursion.png [Broken]

recursion 36
http://dl.dropbox.com/u/13155084/Recusion36.png [Broken]

recursion 48
http://dl.dropbox.com/u/13155084/Recusion48.png [Broken]

To see the recursion in action:
Click "OK"
Press "3" = Recursion Pattern
Press "Space" = Start
Press "v" = Changes view from the standard "square" multiplication table to the "square root" multiplication table via a matrix transform.

**pressing "m" brings up the menu for other options.

http://dl.dropbox.com/u/13155084/PL3D2/P_Lattice_3D_2.html [Broken]
 
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1. What is a Divisor Summatory Function?

A Divisor Summatory Function is a mathematical function that calculates the sum of all the divisors of a given number. In other words, it adds up all the numbers that can evenly divide the given number.

2. Why is Finding Primes important?

Finding Primes is important in many areas of mathematics and computer science. Prime numbers are used in cryptography and data encryption, as well as in algorithms for efficient data storage and retrieval. They also have applications in number theory and other branches of pure mathematics.

3. How do you use a Divisor Summatory Function to find primes?

The Divisor Summatory Function can be used to identify prime numbers by evaluating the sum of divisors for a given number. If the sum is equal to 1 plus the number itself, then the number is prime. Otherwise, it is a composite number.

4. Is there a limit to how large of a prime number can be found using this method?

Technically, there is no limit to the size of a prime number that can be found using the Divisor Summatory Function method. However, as the numbers get larger, it becomes increasingly difficult and time-consuming to find primes using this method.

5. Can this method be used to find all prime numbers?

No, the Divisor Summatory Function method cannot be used to find all prime numbers. It is only able to identify prime numbers that can be expressed as the sum of their divisors. There are infinitely many prime numbers that cannot be expressed in this way.

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