# Finding Primes: A Divisor summatory function

1. Jul 18, 2011

### JeremyEbert

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

Last edited by a moderator: May 5, 2017
2. Jul 19, 2011

### JeremyEbert

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. Jul 23, 2011

### Anti-Crackpot

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

Last edited: Jul 23, 2011
4. Jul 23, 2011

### JeremyEbert

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

Last edited by a moderator: May 5, 2017
5. Jul 24, 2011

### Anti-Crackpot

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. Jul 24, 2011

### JeremyEbert

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. Jul 24, 2011

### Anti-Crackpot

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. Jul 25, 2011

### JeremyEbert

Now that would be truly glorious. :)

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]

Last edited by a moderator: May 5, 2017
9. Jul 26, 2011

### JeremyEbert

10. Jul 26, 2011

### JeremyEbert

Last edited by a moderator: May 5, 2017
11. Sep 23, 2011

### JeremyEbert

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

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. Sep 24, 2011

### JeremyEbert

Borrowing from some insight on Harmonic Numbers…

(-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. Sep 24, 2011

### JeremyEbert

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)

Last edited: Sep 24, 2011
14. Sep 27, 2011

### JeremyEbert

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. Oct 16, 2011

### JeremyEbert

A nice little chart:
http://dl.dropbox.com/u/13155084/chart.pdf [Broken]

Last edited by a moderator: May 5, 2017
16. Oct 21, 2011

### JeremyEbert

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?

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.

Last edited by a moderator: May 5, 2017
17. Nov 10, 2011

### JeremyEbert

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.

Last edited by a moderator: May 5, 2017
18. Nov 14, 2011

### JeremyEbert

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}]

Last edited by a moderator: May 5, 2017
19. Nov 15, 2011

### JeremyEbert

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]

Last edited by a moderator: May 5, 2017