## Visual Prime Pattern identified

 Quote by JeremyEbert Also found a direct link to Pentagonal Pyramid numbers while looking for the area or quadrature of the parabolas in my equation. (n-1)/2 = h (height) 2*sqrt(n)= b (base) (b*h)/2 = a (triangle area) 2*(a^2) = Pentagonal pyramidal number
also

(n-1)/2 = h
2*sqrt(n)= b
(b*h)/2 = a

2*(a^2) = Pentagonal pyramidal number

if (a*(4/3))^2 is an integer then n is a number having a digital root of 1, 4, 7 or 9.

1, 4, 7, 9, 10, 13, 16, 18, 19, 22, 25, 27, 28, 31, 34, 36, 37, 40, 43, 45, 46, 49, 52, 54......

 also... The area of a rectangle A = sqrt(n) * ((n-1)/2) * 4 (this area directly relates to my equation) When “n” is a square then A/12 = Tetrahedral (or triangular pyramidal) number Or reducing… (4n(((n^2)-1)/2)) /12 = Tetrahedral (or triangular pyramidal) number (n(((n^2)-1)/2)) /3 = Tetrahedral (or triangular pyramidal) number

 Quote by JeremyEbert also... The area of a rectangle A = sqrt(n) * ((n-1)/2) * 4 (this area directly relates to my equation) When “n” is a square then A/12 = Tetrahedral (or triangular pyramidal) number Or reducing… (4n(((n^2)-1)/2)) /12 = Tetrahedral (or triangular pyramidal) number (n(((n^2)-1)/2)) /3 = Tetrahedral (or triangular pyramidal) number
which directly relates to the Close-packing of spheres:
http://en.wikipedia.org/wiki/Close_packing

 well, since raphie seems to be restricted at the moment, i have to ask, is anyone else following this?

 Quote by JeremyEbert Here is a visual prime pattern: http://plus.maths.org/content/catching-primes I have developed one of my own based upon trig, square roots and the harmonic sequence. Here is an animation/application that shows the formula visually: http://tubeglow.com/test/Fourier.html Thoughts? Questions?
Wow, but it is rather a hard method. Is it, by any chance, related to the sieve of erasothones?

 Quote by dimension10 Wow, but it is rather a hard method. Is it, by any chance, related to the sieve of erasothones?
dimension10 ,
It is a sieve and all prime sieves seem to smack of Eratosthenes to me. I'm approaching it in my head from a different angle though. My method relates to the fact that a square number added to a prime number only equals another square number when the square added to the prime is equal to ((n-1)/2)^2. Or basically:

n+((n-1)/2)^2 = ((n+1)/2)^2

Its true that all integers share this property despite their primality but composite numbers will have other square congruence, less than the ((n-1)/2)^2 ratio, according to their integer divisors.

These ratios form a lattice when you deal with integers at their square root the way I have. This lattice creates a parabolic coordinate system. This coordinate system is what I'm using to exploit the sieve.

jeremy
* http://en.wikipedia.org/wiki/Congruence_of_squares
* http://en.wikipedia.org/wiki/Parabolic_coordinates

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 Quote by JeremyEbert Here is a visual prime pattern: http://plus.maths.org/content/catching-primes I have developed one of my own based upon trig, square roots and the harmonic sequence. Here is an animation/application that shows the formula visually: http://tubeglow.com/test/Fourier.html Thoughts? Questions?
Ok, maybe I'm the first that doesn't see it. In the first link, I see the primes. In the second link I don't see what identifies the primes. Clue me in.

 Quote by PAllen Ok, maybe I'm the first that doesn't see it. In the first link, I see the primes. In the second link I don't see what identifies the primes. Clue me in.
PAllen,
As an intger n increases, the first blue horizontal line north (north/south = y axis) of the green line (east/west = x axis) increases by the square root of n. The intersections of the vertical lines and the concentric circles at the square root of n (blue horizontal line) equate to the divisors d of n by (n-d^2)/2d = 0 mod(.5). Does that help?

Jeremy

 a spherical version of my equation: http://dl.dropbox.com/u/13155084/PL3...3D_Sphere.html

Raphie (quoted below),

I'm probably late on this but when saying things like:
(11+13/((1+1)+(1+3)) = 4

You should see what mod9 (*notated by %9) gives you...

I.e.
((x+y)/(x%9+y%9))

It matches most of your numbers...
since mod9 is the infinite digital sum.. (digital sum taken as many times as possible until a single digit is reached)

I.e.
(11+13/((1+1)+(1+3)) == ((11+13)/(11%9+13%9))

 Quote by Raphie A POSSIBLY RELATED SEQUENCE Suppose the sum of the digits of prime(n) and prime(n+1) divides prime(n) + prime(n+1). Sequence gives prime(n). http://oeis.org/A127272 2, 3, 5, 7, 11, 17, 29, 41, 43, 71, 79, 97, 101, 107... e.g. (2 + 3)/(2+3) = 1 (3+5)/(3+5) = 1 (5+7)/(5+7) = 1 (7+11)/(7+(1+1)) = 2 (11+13/((1+1)+(1+3)) = 4 (17+19/((1+7)+(1+9)) = 2 (29+31/((2+9)+(3+1)) = 4 (41+43/((4+1) + (4+3)) = 7 (43+47/((4+3)+(4+7)) = 5 (71+73)/((7+1)+(7+3)) = 8 (79+83)/((7+9)+(9+7)) = 5 (97+101)/((9+7)+(1+0+1)) = 11 (101+103)/((1+0+1) + (1+0+3) = 34 (107+109)/((1+0+7)+(1+0+9) = 12 ALSO... Numbers n such that 1 plus the sum of the first n primes is divisible by n+1. http://oeis.org/A158682 2, 6, 224, 486, 734, 50046, 142834, 170208, 249654, 316585342, 374788042, 2460457826, 2803329304, 6860334656, 65397031524, 78658228038 002 - 002 = 000 = K_00 012 - 006 = 006 = K_02 (Max) 600 - 224 = 336 = K_10 (Lattice Max known) 924 - 486 = 438 = K_11 (Lattice Max known) 6/(5+1) = 1 42/(6+1) = 6 143100/(224+1) = 636 775304/(486+1) = 1592 Like I said, especially given that these two progressions are ones I came across in the process of writing that last post to you, "hmmmm..." RELATED PROGRESSIONS Integer averages of first n noncomposites for some n. http://oeis.org/A179860 1, 2, 6, 636, 1592, 2574, 292656, 917042, 1108972, 1678508, 3334890730, 3981285760, 28567166356, 32739591796, 83332116034 a(n) is the sum of the first A179859(n) noncomposites. http://oeis.org/A179861 1, 6, 42, 143100, 775304, 1891890, 14646554832, 130985694070, 188757015148, 419047914740, 1055777525624570390, 1492138298614167680, 70288308055831268412, 91779857115464381780, 571686203669195590338 Numbers n that divide the sum of the first n noncomposites. http://oeis.org/A179859 1, 3, 7, 225, 487, 735, 50047, 142835, 170209, 249655, 316585343, 374788043, 2460457827, 2803329305, 6860334657 This number, in particular, I find interesting... 142835 = 5*7^2*11*53 = (142857 - par_8) = (142857 - 22) vs. 1/7 = .142857 (repeating) Indexing from 0, 142857 is the 24th Kaprekar Number 1, 3, 7 and 225, the 1st 4 terms in that last sequence above == (2^1 - 1)^1, (2^2 - 1)^1, (2^3 - 1)^1, (2^4 - 1)^2. - RF