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Finding Primes: A Divisor summatory functionby JeremyEbert
Tags: divisor function 
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#1
Jul1811, 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 


#2
Jul1911, 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(n1) = 2 then n is prime 


#3
Jul2311, 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 


#4
Jul2311, 10:15 PM

P: 205

Finding Primes: A Divisor summatory function
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 


#5
Jul2411, 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 


#6
Jul2411, 09:35 PM

P: 205




#7
Jul2411, 09:54 PM

P: 75

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
Jul2511, 06:40 AM

P: 205

http://dl.dropbox.com/u/13155084/Pyt...%20lattice.pdf 


#9
Jul2611, 09:00 PM

P: 205

a related thread.
http://www.physforum.com/index.php?showtopic=29372&st=0 


#10
Jul2611, 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 "Inversesquare law"? http://en.wikipedia.org/wiki/Shell_theorem http://en.wikipedia.org/wiki/Inversesquare_law http://dl.dropbox.com/u/13155084/PL3...3D_Sphere.html 


#11
Sep2311, 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 ( ((nk^2)/(2k))  (((nk^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
Sep2411, 10:07 AM

P: 205

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


#13
Sep2411, 10:18 PM

P: 205

This might be easier to understand for some: c(n) = sum((nk^2)/(2k)) * 2 where k = 1 to n H(n) = nth Harmonic Number T(n) = nth Triangular Numner j(n) = sum((nk^2)/(2k) mod .5) * 2 where k = 1 to n nondivisor 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) 


#14
Sep2711, 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*(((((nu)k^2)/(2k)) mod .5)  (((nk^2)/(2k)) mod .5 ))) ] where k=1 to n, u = 0.000001/n sigma(x,n) = ceiling [ SUM(((2*k*((((nu)k^2)/(2k)) mod .5))  (n mod k))^x) ] where k=1 to n, u = 0.0000001/n, x>0 


#15
Oct1611, 05:28 PM

P: 205



#16
Oct2111, 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 halfinteger 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 hypotenusebase difference of k and a height of sqrt(n). when the base and hypotenuse are both either integers or halfintegers then k is a divisor of n. 


#17
Nov1011, 07:45 PM

P: 205

complex number equation for my lattice points: amplitude = a = (n+1)/2 angular frequency = w = acos((n1)/(n+1)) time (moments) = t = acos(1k(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. 


#18
Nov1411, 02:14 PM

P: 205

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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