What is the Visual Prime Pattern Based on Trig and Harmonics?

[JeremyEbert]

yes perfect! that helps! I'm finding e so much in my equation. the ((n-1)/(n+1))^(-n/2) ~ e part has really got me preoccupied

Just for clarification the (n-1)/(n+1) part is the x = 1-(2/d+1) part of the equation.
so its also 1-(2/d+1)^(-d/2)~e
and the rest would be
sin(acos(1-(2/d+1)) * (d+1)/2 = sqrt(d)

 

[JeremyEbert]

Just for clarification the (n-1)/(n+1) part is the x = 1-(2/d+1) part of the equation.
so its also 1-(2/d+1)^(-d/2)~e
and the rest would be
sin(acos(1-(2/d+1)) * (d+1)/2 = sqrt(d)

So I guess that means:

z = e^bt = e^(a+iw)t = (e^a) . (e^iwt) ~ (e^a) . (e^i(-1/d/2))

Or basicaly wt~(-1/d/2) right ?

Would my vertical and horizontal lines then be Gaussian integers?

 

[JeremyEbert]

also I think this means in Euler's formula e^ix = cos(x) + i sin(x)
in my equation:
x=acos(1-(2/d+1))

where 1/d = the harmonic series.

 

[dodo]

Honest, I can't make head or tails from this.

Defining the variables you're using would help a lot.

"d", last time I understood something, was an integer, going 1, 2, 3, ...
"x" .. I'm no longer sure.
Who is n, and how (n-1)/(n+1) = 1-(2/d-1) ?

And I don't really know what is this symbol ~ , or what are you trying to say with it. But begin by defining the variables that you're using. "x" is the distance from here to here. Simple things like that.

---

Here is something else that may also help you being understood. I'm noticing that you have troubles using parenthesis in formulas. Here is some advice.

If you see a formula like
$$\frac {a+b} {c+d}$$
it is incorrect to write it as a+b/c+d, because people will understand
$$a+\frac b c + d$$
You have programmed computers, so you know that the division symbol has more "precedence" than the sum. Keeping these things in mind will help everyone else understand.

 

[JeremyEbert]

Ha! Sorry, I got a little carried away. I'll start over. To be clear I'm using in radians not degrees.

n = 1,2,3,..., infinity

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

x = 1-(2/(n-1))

sqrt(n) = sin(acos(x)) / 2 * (n+1)

x^(-n/2) quickly converges to the constant e

t=acos(x)

Eulers formula e^it=cos(t) + i sin(t)

cos(t) = x = (n-1)/(n+1) = 1-(2/(n-1))

edit------e^(-1/(n/2)) quickly converges to x

edit-------

also

d=1,2,3,...(n-1)

redfine x as

x = 1-((2/(n-1)*d))

if n is prime then d can equal 1 and (n-1)

 

[dodo]

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

x = 1-(2/(n-1))

I believe you mean 1-2/(n+1), with a "+" sign.

sqrt(n) = sin(acos(x)) / 2 * (n+1)

Yes, this one was established in post#15. Assuming you correct "x" as above.

x^(-n/2) quickly converges to the constant e

Now, you have to be careful when trying things out on the computer. This may converge to e, or to something close to e but not e. At this moment I'm not entirely sure, but go on.

t=acos(x)

Eulers formula e^it=cos(t) + i sin(t)

cos(t) = x = (n-1)/(n+1) = 1-(2/(n-1))

Ok, as long as 1-2/(n-1) is changed to 1-2/(n+1).
But now you should go somewhere with these definitions; personally, I don't see where you're going to, or why the introduction of a complex variable is needed.

d=1,2,3,...(n-1)

redfine x as

x = 1-((2/(n-1)*d))

So be it, but again I don't see where you're heading to.

if n is prime then d can equal 1 and (n-1)

And now I'm lost. In which way n begin prime is a restriction to "d"? You said d was another free variable going 1,2,3,...

 

[JeremyEbert]

Sorry for the delay. I’ve been working on the next piece to this and got distracted by an interesting vector of this equation.

Sorry to jump around here but I think this is where I need some more help explaining the big picture.

I noticed that the parabolas created by this pattern, follow this equation per quadrant on the x,y grid.

Where n = 1,2,3,…infinity
Quadrant 1: y = sqrt((2xn)+n^2) = the square root of integer multiples of n

Question, if I use complex numbers, x + iy will I get both quadrants of the parabola? (1&4 for positive n and 2&3 for negative n)

The interesting vector I noticed is at 45 degrees or pi/4 radians.
The sqrt((2xn)+n^2) parabolas intersect that vector at sqrt(((1+sqrt(2))*n)^2).
I have attached an image demonstrating this.

The thing I find most interesting about these intersections is this:

q=((1+sqrt(2))*n)^2
u=((n^2)*6) – q
(q*u)^(1/4) = n

Also interesting side note:

q=((1+sqrt(2))*n)^2
u=((n^2)*12) – 2q
(2q*u)^(1/2) = 2(n^2) = maximum number of electrons an atom's nth electron shell can accommodate

 

[JeremyEbert]

I noticed something else this weekend while working with the complex exponentials demonstration here:
http://demonstrations.wolfram.com/TheComplexExponential/

if you set the equation to e^(1+3.1415 i)t you get this:

http://i98.photobucket.com/albums/l267/alienearcandy/e.png

which looks a lot like the golden ratio here:

http://i98.photobucket.com/albums/l267/alienearcandy/phi.png

in fact if you overlay them you can see they are very close:

http://i98.photobucket.com/albums/l267/alienearcandy/e-phi.png

Is there a known relationship between Phi, e and pi?

 

[JeremyEbert]

Dodo, sorry for the late response, I've been a little distracted with matricies.

I believe you mean 1-2/(n+1), with a "+" sign.

Over zealous typo. ha! 1-2/(n+1) is what I meant.

Now, you have to be careful when trying things out on the computer. This may converge to e, or to something close to e but not e. At this moment I'm not entirely sure, but go on.

Youre right. It is just an assumption bassed on a small subset of data.

But now you should go somewhere with these definitions; personally, I don't see where you're going to, or why the introduction of a complex variable is needed.

I'm not entirely sure yet either, It’s just an idea or theory that I'm trying to prove. I can visualize it; I'm just trying to depict it mathematically. I really appreciate your feedback. It helps greatly.

And now I'm lost. In which way n begin prime is a restriction to "d"? You said d was another free variable going 1,2,3,...

I'm going to come at this from a different perspective... give me a few.

 

[Raphie]

Is there a known relationship between Phi, e and pi?

5 arccos (Golden Ratio/2) = pi
2 cos (pi/5) = Golden Ratio

Or, alternatively, to better express the symmetrical relationship in terms that requires nought but a single sign change and an exchange of variables...

(7+3)/2 cos^-1 (phi/((7-3)/2)) = pi
(7- 3)/2 cos^1 (pi/((7+3)/2)) = phi

Ceiling [(sqrt e)^(n-2)] is a Fibonacci Forgery for n = 1-->10 (stemming from the fact that phi^2 = 2.618033988... and e = 2.718281828...)

e is associated with compound growth, while the Golden Ratio is associated with optimal reception and transmission of information. More another time, particularly in relation to all 3 mathematical constants and the binomial theorem. (e.g. Both the Eulerian and Stirling Triangles give row sums equal to n!. As such, if you sum the inverses of both these triangles by row sum, you will eventually, at the limit = infinity, converge to e)

Also, Jeremy, note the following relationship:

zeta (n) / 2^(n-1) gives lower bounds on the density of lattice sphere packings in n-dimensions
Kissing Number
http://mathworld.wolfram.com/KissingNumber.html

2^n, of course, is the row sum of Pascal's Triangle [SUM C (n,k) for n = row number and -1 < k < n+1 ]. If, instead, you take the sum of squares of row entries you get the Central Binomial Coefficients, which can be related (in tandem with the powers of 2) to both pi and, by extension, the summed volume of 2n-dimensional spheres.

- RF

P.S. One interesting note in relationship to pi vs. phi. pi is transcendental, but not phi, which is "only" irrational (but according to some, the most irrational of the irrationals).

 

[Raphie]

Also consider, Jeremy, the following relationship:

Let x = Divisors of 12 = 1,2,3,4,6,12

x + 1 = 2, 3, 4, 5, 7, 13 --> {4 U Unique Prime Divisors of the Leech Lattice} --> {4 U First Five Mersenne Prime Exponents}

Let y = Divisors of Divisors of 12 = 1, 2, 2, 3, 4, 6

y - 1 = 0, 1, 1, 2, 3, 5 --> {Fibonacci_n (modulo 6)}

The point being that Dodo may be on the right track in cautioning you not to over-extrapolate regarding the apparent Golden Ratio relationship you have come across.

Which is not to say you are wrong or that, even if you are, you might not be on to something important:

1, 2, 3, 4, 6 are the only positive integer solutions to 2*cos (2*pi/n) is in N. This is the formula that is used in the short proof of the Crystallographic Restriction Theorem.

- RF

P.S. Also note that for K_n a Maximal Lattice Packing for Dimension n...

1*2 = x_1 * x_2 = 2 = K_1
2*3 = x_2 * x_3 = 6 = K_2
3*4 = x_3 * x_4 = 12 = K_3
4*6 = x_4 * x_5 = 24 = K_4
6*12 = x_5 * x_6 = 72 = K_6

 

[JeremyEbert]

5 arccos (Golden Ratio/2) = pi
2 cos (pi/5) = Golden Ratio

Or, alternatively, to better express the symmetrical relationship in terms that requires nought but a single sign change and an exchange of variables...

(7+3)/2 cos^-1 (phi/((7-3)/2)) = pi
(7- 3)/2 cos^1 (pi/((7+3)/2)) = phi

I posted arcos(Phi/2) = 36 deg earlier so I guess I knew this although I really like the symmetry you have shown.

Ceiling [(sqrt e)^(n-2)] is a Fibonacci Forgery for n = 1-->10 (stemming from the fact that phi^2 = 2.618033988... and e = 2.718281828...)

Hence the similarity of the images I assume.

e is associated with compound growth, while the Golden Ratio is associated with optimal reception and transmission of information. More another time, particularly in relation to all 3 mathematical constants and the binomial theorem. (e.g. Both the Eulerian and Stirling Triangles give row sums equal to n!. As such, if you sum the inverses of both these triangles by row sum, you will eventually, at the limit = infinity, converge to e)

Also, Jeremy, note the following relationship:

zeta (n) / 2^(n-1) gives lower bounds on the density of lattice sphere packings in n-dimensions
Kissing Number
http://mathworld.wolfram.com/KissingNumber.html

2^n, of course, is the row sum of Pascal's Triangle [SUM C (n,k) for n = row number and -1 < k < n+1 ]. If, instead, you take the sum of squares of row entries you get the Central Binomial Coefficients, which can be related (in tandem with the powers of 2) to both pi and, by extension, the summed volume of 2n-dimensional spheres.

- RF

P.S. One interesting note in relationship to pi vs. phi. pi is transcendental, but not phi, which is "only" irrational (but according to some, the most irrational of the irrationals).

I’m reading your post here:
https://www.physicsforums.com/showthread.php?p=3212052#post3212052
fascinating to say the least.

The interesting vector I noticed is at 45 degrees or pi/4 radians.
The sqrt((2xn)+n^2) parabolas intersect that vector at sqrt(((1+sqrt(2))*n)^2).
I have attached an image demonstrating this.

The thing I find most interesting about these intersections is this:

q=((1+sqrt(2))*n)^2
u=((n^2)*6) – q
(q*u)^(1/4) = n

Evidently the 1+SQRT(2) part is known as the Silver Ratio:
http://en.wikipedia.org/wiki/Silver_ratio
which is the also limiting ratio of consecutive Pell numbers

 

[Raphie]

Evidently the 1+SQRT(2) part is known as the Silver Ratio:
http://en.wikipedia.org/wiki/Silver_ratio
which is the also limiting ratio of consecutive Pell numbers

Indeed. Which, as we (now) know, is also not unrelated to the Sophie Germain Triangular Numbers. I believe there to be some manner of as yet undiscovered linkage between those and Sophie Germain Primes, the first three of which are 2, 3, 5 = (5-1)/2, (7-1)/2, (11-1)/2, for 5, 7 & 11 the first 3 safe primes associated with the Ramanujan Congruences (aka "0-Dimensional Ono Primes")

Pell Numbers are resursively constructed thusly:

1A + 2B = C

Whereas Fibonacci Numbers are recursively constructed thusly:

1A + 1B = C

So, in spite of my caution, I would not, if I were you, assume as a given that the (provisional) Golden Ratio relationship you came across is just a chimera.

- RF

 

[Raphie]

I posted arcos(Phi/2) = 36 deg earlier so I guess I knew this although I really like the symmetry you have shown.

One of the benefits of geting the two formulas "talking together in the same language" is that it then becomes a simple matter to construct a single generating formula uniting both constants, phi and pi:

For...
(7- 3)/2 cos^1 (pi/((7+3)/2)) = phi
(7+3)/2 cos^-1 (phi/((7-3)/2)) = pi

Let..
b = (n) (mod 2)
a = (n + 1) (mod 2)

Then...
(7- 3(-1)^b))/2 cos^((-1)^b) ((phi^a*pi^b)/((7+3(-1)^a))/2)) = phi^b*pi^a

And then you can generate other formulas from those same variables...

(7- 3(-1)^b))/2 cos^((-1)^b) ((phi^a*pi^b)/((7+3(-1)^a))/2))
*
(7- 3(-1)^a))/2 cos^((-1)^a) ((phi^b*pi^a)/((7+3(-1)^b))/2))
= phi*pi

The phi*pi product is known as the "Biwabik Sum" and it relates to the set of all odd numbers, of which all primes, excluding 2, are a subset. See...

Pi, Phi and Fibonacci Numbers
http://goldennumber.net/pi-phi-fibonacci.htm

- RF

 

[JeremyEbert]

Thank you Raphie, you have given me some amazing morsels to digest.

 

[JeremyEbert]

Also, just found that e ^ asinh(.5) = phi

 

[JeremyEbert]

I wonder if I should look at my y = sqrt((2xn)+n^2) parabolas rotated pi/2 radians?
Now y =( (x/ sqrt(2n) )^2) - n/2.
Can this be treated as a sort of parabolic function of n? Essentially the pattern I’m talking about comes from a conic section of a cone in complex exponential space. Where the growing amplitude and the circular motion create the cone. Make any sence?
http://en.wikipedia.org/wiki/Conic_section
where the cone looks similar to this:
http://i98.photobucket.com/albums/l267/alienearcandy/econic.png

 

[JeremyEbert]

something like this:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90.png

 

[JeremyEbert]

better yet:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90-3d-1.png

 

[JeremyEbert]

ball in cone:
http://www.vic.com/~syost/utk/BallInCone.html

 

[JeremyEbert]

a kind of modified shell theorem right?:
http://en.wikipedia.org/wiki/Shell_theorem
and
http://en.wikipedia.org/wiki/Inverse-square_law

 

[Raphie]

I wonder if I should look at my y = sqrt((2xn)+n^2) parabolas rotated pi/2 radians?
Now y =( (x/ sqrt(2n) )^2) - n/2.
Can this be treated as a sort of parabolic function of n? Essentially the pattern I’m talking about comes from a conic section of a cone in complex exponential space. Where the growing amplitude and the circular motion create the cone. Make any sence?

Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.


Raphie

 

[dimension10]

I noticed something else this weekend while working with the complex exponentials demonstration here:
http://demonstrations.wolfram.com/TheComplexExponential/

if you set the equation to e^(1+3.1415 i)t you get this:

http://i98.photobucket.com/albums/l267/alienearcandy/e.png

which looks a lot like the golden ratio here:

http://i98.photobucket.com/albums/l267/alienearcandy/phi.png

in fact if you overlay them you can see they are very close:

http://i98.photobucket.com/albums/l267/alienearcandy/e-phi.png

Is there a known relationship between Phi, e and pi?

Yes. 1-1/4(Pi)r2 is related to e by the science of fluctuations.

 

[JeremyEbert]

Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.


Raphie

Raphie
Thanks again! I can see where my animation highlights the divisor function and its summation by counting the lattice points that intersect.

http://www.tubeglow.com/test/Fourier.swf

I'lll have the equation here shortly.

 

[FlexGunship]

Is there a known relationship between Phi, e and pi?

Yes! When you add them together you get 7.47790847! Amazing!

 

[Raphie]

Yes! When you add them together you get 7.47790847! Amazing!

Thank you for this extremely, uhh... "constructive" input Flexgunship. That 1 + 1 = 2 is also quite "amazing," but hardly (IMHO) worthy of an exclamation point, derisively intended or otherwise...

 

[JeremyEbert]

Raphie
Thanks again! I can see where my animation highlights the divisor function and its summation by counting the lattice points that intersect.

http://www.tubeglow.com/test/Fourier.swf

I'lll have the equation here shortly.

so if we go back to the y = sqrt((2xn)+n^2) parabolas. Another way to generate them would be:
d=(1,2,3,...n)
as y increases by sqrt(n)
x = (n-d^2)/(2d)
when x = 0, d = sqrt(n)

primes in mod(0.5) x = 0 when d = 1 or n

make sense?

 

[JeremyEbert]

a little more detail:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90-3d-2.png
could this be applied to spherical harmonics?

 

[JeremyEbert]

a little more detail:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90-3d-2.png
could this be applied to spherical harmonics?

Albrecht Durer's cone: http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=2591&bodyId=3058

 

[JeremyEbert]

so if we go back to the y = sqrt((2xn)+n^2) parabolas.

the parabola sqrt((2xn)+n^2) is the same as sqrt(4 * (n/2) * (x + (n/2)) which I guess fits the more common definition of "y^2 = 4px".

Also Raphie, is this part clear?

d=(1,2,3,...n)
as y increases by sqrt(n)
x = (n-d^2)/(2d)
when x = 0, d = sqrt(n)

primes in mod(0.5) x = 0 when d = 1 or n

I've made reference to it before with little success.

 

[Raphie]

a little more detail:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90-3d-2.png
could this be applied to spherical harmonics?

Factorials, double factorials (product of odd numbers) and powers of 2 come into 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...)RF

As for this...

Also Raphie, is this part clear?.

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

 

[Raphie]

Incidentally, Jeremy, if the train of thought I outlined above is on the right track, then relationships/mappings such as the following are most likely not coincidence...

For...

K --> Kissing Number (Lattice)
p --> Prime Number
B --> Bernoulli Number
5,7,11 --> Primes of the Ramanujan Partition Number Congruences (aka "0-D Ono Primes")

===================================

K_1 = 002 = 5^0 + 1 = 5^((05-5)/2) + 1 = 2^2 -2 = 2^(05-1)/2 - 2 = totient (p_2)
K_2 = 006 = 5^1 + 1 = 5^((07-5)/2) + 1 = 2^3 -2 = 2^(07-1)/2 - 2 = totient (p_4)
K_7 = 126 = 5^3 + 1 = 5^((11-5)/2) + 1 = 2^7 -2 = 2^(11-1)/2 - 2 = totient (p_31)

1 Division of 4-space --> 2 Regions
2 Divisions of 4-space --> 4 Regions
5 Divisions of 4-space --> 31 Regions
------------------------------------------
2*4*31 = 248 --> 240 + 8 = Dimensions of E_8

p_001 = p_(2^(05-1)/2 - 3) = 2
p_005 = p_(2^(07-1)/2 - 3) = 11
p_125 = p_(2^(11-1)/2 - 3) = 691
------------------------------------------
Partition Numbers: 1,1,2,3,5,7,11...

B_(05+1) = -1/30; 30 = 2*3*5
B_(07+1) = 1/42; 42 = 2*3*7
B_(11+1) = -691/2730 = 2*3*5*7*13
------------------------------------------
2,3,5 = ({5,7,11} - 1)/2 --> First 3 Sophie Germain Primes
pi (2,3,5,7,13) = 1,2,3,4,6 --> Allowable n-fold Rotational Symmetries under the Crystallographic Restriction Theorem

===================================

(p_(2^((5-1)/2) - 3) * p_(2^((7-1)/2) - 3) * p_(2^((11-1)/2) - 3) / ((5 * 7 * 11) * (B_(5+1) * (B_(7+1) * (B_(11+1))
(p_(5^((5-5)/2) - 0) * p_(5^((7-5)/2) - 0) * p_(5^((11-5)/2) - 0) / ((5 * 7 * 11) * (B_(5+1) * (B_(7+1) * (B_(11+1))
= (2*11*691)/(5*7*11*(-1/30 * 1/42 * - 691/2730))
= 196560

196560 --> Vertices of the Leech Lattice
= K_24

Bernoulli Numbers: Ramanujan's congruences
(Note: This set of Congruences is different from Ramanujan's Partition Number Congruences)
http://en.wikipedia.org/wiki/Bernoulli_number#Ramanujan.27s_congruences

Bernoulli Numbers: Restatement of the Riemann Hypothesis
http://en.wikipedia.org/wiki/Bernoulli_number#A_restatement_of_the_Riemann_hypothesis

196560/24 (Average Spheres/Dimension of the Leech Lattice), btw, = 8190, which is the 12th Ore's Harmonic Number and the totient of the 1028-th (=2^10 + 2^2) prime number. Also: sqrt ((196560*(-1/30 * 1/42 * - 691/2730))) gives a not very accurate (pi is 99.9907% of this...), but interesting in a contextual sense, approximation of 2*pi.

And sqrt ((196560*(-1/30 * 1/42 * - 691/2730))), by extension, gives a not very accurate approximation of 4*pi^2, which, of course, prominently figures into Kepler's Third Law, perhaps of interest to you in relation to the Shell Theorem.RF

 

[Raphie]

Just a thought for you Jeremy, but if you're looking to relate the primes to gravity in the exploratory vein you seem to be doing, then try substituting (Zeta_2)^-1 = 6/pi^2 into the formula for a pendulum (replacing "L/g"); Zeta_2, of course, being the probability that two randomly selected numbers will be relatively prime. When you square the period, you get the result: 24 seconds^2.

All primes > 3 when squared, as you may already know, are == 1 mod (24), while ALL odd primes are congruent to either 1 mod (24), or 1 mod (totient 24) = 1 mod (8) .

Also, (p^2 - 1)/24 is pentagonal for n > 3. Thus, lim n --> infinity p^2/T^2 is a Pentagonal "Number," as well as 1/3 a Triangular "Number," at least conceptually speaking (since infinity is not a "number"). Add in a distance parameter to the numerator (p^2 meters^2 - 1^2 meter^2)/T^2 and you end up with velocity^2 unit-wise.

If you plug in Zeta_4, you also get some interesting results related to the steradian, 1/90-th of a steradian, to be specific (totient 90 = 24); the steradian also being known as a "solid angle." 4*pi Steradian (= 4*pi*radians^2) traces the surface area of a sphere.RF

RELATED LINK
John Baez
The Rankin Lectures, University of Glasgow
September 15-19, 2008
My Favorite Numbers (5, 8 & 24)
http://math.ucr.edu/home/baez/numbers/

Also see the discussion on the below referenced thread. There's a great link there, posted by (former poster) Goongyae, to a Euler paper that relates generalized pentagonal numbers to the primes...
Ken Ono and Factoring?
https://www.physicsforums.com/showthread.php?t=472486

 

[JeremyEbert]

Raphie,
Thanks for the wealth of information. Your posts really help me to see things I've never thought about before. Sorry for the delay but I've been trying to formalize my ideas before I get to lost exploring other areas. I have attached a rough draft of what I've been working on. I'd love any feedback you can offer.
Jeremy

 

[JeremyEbert]

I fixed a few mistakes.
http://www.tubeglow.com/test/Pythagorean lattice.pdf