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_summatory_function#Definition related: http://mathworld.wolfram.com/GausssCircleProblem.html
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
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 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
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.
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
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 lattice.pdf
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/PL3D2SPHERE/P_Lattice_3D_Sphere.html
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
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)
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
More detail... http://dl.dropbox.com/u/13155084/divisor semmetry.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.
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/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}]
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 recursion 36 http://dl.dropbox.com/u/13155084/Recusion36.png recursion 48 http://dl.dropbox.com/u/13155084/Recusion48.png 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