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Finding Primes: A Divisor summatory function

  1. Jul 18, 2011 #1
    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
     
  2. jcsd
  3. Jul 19, 2011 #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
     
  4. Jul 23, 2011 #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
     
    Last edited: Jul 23, 2011
  5. Jul 23, 2011 #4
    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
     
  6. Jul 24, 2011 #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
     
  7. Jul 24, 2011 #6
    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.
     
  8. Jul 24, 2011 #7
    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
     
  9. Jul 25, 2011 #8
    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
     
  10. Jul 26, 2011 #9
  11. Jul 26, 2011 #10
  12. Sep 23, 2011 #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...
     
  13. Sep 24, 2011 #12
    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
     
  14. Sep 24, 2011 #13

    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
  15. Sep 27, 2011 #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
     
  16. Oct 16, 2011 #15
    A nice little chart:
    http://dl.dropbox.com/u/13155084/chart.pdf
     
  17. Oct 21, 2011 #16
    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.
     
  18. Nov 10, 2011 #17
    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.
     
  19. Nov 14, 2011 #18
    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}]
     
  20. Nov 15, 2011 #19
    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
     
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