Visual Prime Pattern identified

  • #101
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Factorials, double factorials (product of odd numbers) and powers of 2 come in to play in regards to Volumes of n-balls...

n-ball
http://en.wikipedia.org/wiki/N-sphere#n-ball

And so too Spherical Harmonics...

Hyperspherical volume element
http://en.wikipedia.org/wiki/N-sphere#Hyperspherical_volume_element

So... when summing volumes for a unit sphere, then e will naturally also be involved.

e.g.
pi^e/n! = SUM[V_2n]

Insofar as e relates to the prime number distribution [pi (x) ~ x/ln(x)] specifically, and compound growth generally, that there is some manner of relationship, discovered (and I am unaware of it) or undiscovered, seems evident. The precise nature of this relationship, however, is far less clear.

Keep in mind, however, that the number of conjugacy classes in the Symmetric Group S_n is a partition number:

Conjugacy class
http://en.wikipedia.org/wiki/Conjugacy_class

Since we now know, by the work of Ono et al, that partitions of prime numbers evidence fractal-like behavior, we can also logically surmise that the growth sequences of n-dimensional spaces of dimension p and/or p-1 (and/or p+1) will also be found to exhibit fractal-like behaviors. Think of it this way, and then the root system of a lattice such as E8 (241 is prime, and so too 239...) can, in some manner at least, be thought of as if it were a freeze-framed cross-section of a fractal iterating through multi-dimensional space.

And, insofar as all of this is the case, then Periodicity (e.g. The Crystal Restriction Theorem) and Quasi-periodicity (e.g. Penrose Tilings, related to the Golden Ratio) should also make an appearance is some form. (And so too, for that matter, the Shell theorem that you posted, which has everything to do with theoretical physics...)

Best,
RF

As for this...

I need to look more closely at what you've been doing before I can answer.

Raphie, I have a direct link form my equation into Apollonian sphere packing. http://oeis.org/A045506

ex:

5 + 2^2 = 9
7 + 3^2 = 16
11 + 5^2 = 36
13 + 6^2 = 49
17 + 8^2 = 81
19 + 9^2 = 100
23 + 11^2 = 144
......
of course this is linked to the fact that (2^(p-1)-1)/p is congruent to 0 (mod 3), for all primes p greater than 3
....
 
  • #102
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Subtract any two + and - Pentagonal Pyramid numbers of equal index and you get a square. Add them together and you get a cube.

e.g.
40 - 24 = 4^2
40 + 24 = 4^3

One can use this mathematical fact to easily obtain integer solutions to the following:

Period^2 = 4*pi^2/GM * Distance^3 (Kepler's 3rd Law)

e.g.
(40 + 24)^2 = (40 - 24)^3 = 4^6 = 4096
(6 + 2)^2 = (6 - 2)^3 = 2^6 = 64 (= sqrt 4096)

The Pentagonal Pyramid numbers, of course, are the summation of the Pentagonal numbers, which are already well-known to be related to the "timing" and/or "tuning" of the primes.

p^2 - 1 == 1 mod (24) for all p > 3

(p^2 - 1)/24 is Pentagonal for all p > 3.

And, also, as I mentioned previously, 24 s^2 is the Period^2 one obtains if one replaces L/g in the formula for a pendulum with zeta(2)^-2 = (pi^2/6)^-2, where (the reciprocal of) zeta(2) gives the probability of two randomly selected integers being relatively prime.

- RF

Note: Pentagonal Pyramid Numbers have a very easy to remember formula n*T_n = (+) Pentagonal Pyramid # and n*T_-n = (-) Pentagonal Pyramid #, for T_n a Triangular Number.

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)
1/2 bh = a (triangle area)

2(a^2) = Pentagonal pyramidal number
 
  • #103
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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......
 
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  • #104
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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
 
  • #105
204
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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
 
  • #106
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well, since raphie seems to be restricted at the moment, i have to ask, is anyone else following this?
 
  • #108
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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
 
  • #109
PAllen
Science Advisor
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  • #110
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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
 
  • #111
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a spherical version of my equation:
http://dl.dropbox.com/u/13155084/PL3D2SPHERE/P_Lattice_3D_Sphere.html [Broken]
 
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  • #112
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))

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
 
  • #113
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update:
http://dl.dropbox.com/u/13155084/prime.png [Broken]
 
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