What is the Klein Four-Group Geometry and Its Applications?

[JeremyEbert]

I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

 

[JeremyEbert]

I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

I'm working on the explination.
http://dl.dropbox.com/u/13155084/Given%20a%20divisor%20k.pdf

 

[JeremyEbert]

I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

A better visual.
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

 

[JeremyEbert]

Maybe its more than the Klein four group.

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."
I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = nZ/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}

Z/3Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,2}

Z/4Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,3} Z/6Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,5} Z/8Z
(s+k)^2 = {1,4,1,0}
s^2 = {0,1,4,1}
n = {1,3,5,7} Z/12Z
(s+k)^2 = {1,9,4,0}
s^2 = {0,4,9,1}
n = {1,5,7,11} Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23} (s+k)^2 + s^2 = 1 (mod 12) ** only modulo 12. ? additive inversion of the identity element to its multiplication group Z/12Z*?
((s+k)^2 - s^2)^2 = 1 (mod 12)

In the Klein Four-Group the quadratic residues of (s+k)^2 and s^2 are permutations of the symmetric group of order 4 (S4).

 

[DonAntonio]

Maybe its more than the Klein four group.

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."



I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = n


Z/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}

Z/3Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,2}

Z/4Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,3}


Z/6Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,5}


Z/8Z
(s+k)^2 = {1,4,1,0}
s^2 = {0,1,4,1}
n = {1,3,5,7}


Z/12Z
(s+k)^2 = {1,9,4,0}
s^2 = {0,4,9,1}
n = {1,5,7,11}


Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23}


(s+k)^2 + s^2 = 1 (mod 12) ** only modulo 12. ? additive inversion of the identity element to its multiplication group Z/12Z*?
((s+k)^2 - s^2)^2 = 1 (mod 12)

In the Klein Four-Group the quadratic residues of (s+k)^2 and s^2 are permutations of the symmetric group of order 4 (S4).

It seems to be that only, or perhaps mainly, you know what you're talking about: "Klein four group geometry"? "A group geometry"?

If time passes by and you get no answer, perhaps it's time to give more background, books, papers for people to know what you mean. It may be that

what you call "some group geometry" is known by another name to someone else...

DonAntonio

 

[JeremyEbert]

It seems to be that only, or perhaps mainly, you know what you're talking about: "Klein four group geometry"? "A group geometry"?

If time passes by and you get no answer, perhaps it's time to give more background, books, papers for people to know what you mean. It may be that

what you call "some group geometry" is known by another name to someone else...

DonAntonio

Thank you for your reply DonAntonio.

Let me preface my responses with the fact that I am self taught so I may be describing things incorrectly or may be over looking something obvious, but that is why I am here, to learn. Also, I quote wiki a lot. Sorry for that.

wiki:
Geometric group theory

"the study of finitely generated groups via exploring the connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act (that is, when the groups in question are realized as geometric symmetries or continuous transformations of some spaces)."

The geometry I speak of relates to the points in Euclidean space defined by (x,y,z) where
x=s=(n-k^2)/(2k)
y=sqrt(n)
z=(s+k)

I had hoped that this would provide more detail:
http://dl.dropbox.com/u/13155084/Given%20a%20divisor%20k.pdf

The reason I say "Klein four group geometry" is because that group seems to fit best or "mainly" as you stated.

 

[DonAntonio]

Thank you for your reply DonAntonio.

Let me preface my responses with the fact that I am self taught so I may be describing things incorrectly or may be over looking something obvious, but that is why I am here, to learn. Also, I quote wiki a lot. Sorry for that.

wiki:
Geometric group theory

"the study of finitely generated groups via exploring the connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act (that is, when the groups in question are realized as geometric symmetries or continuous transformations of some spaces)."

The geometry I speak of relates to the points in Euclidean space defined by (x,y,z) where
x=s=(n-k^2)/(2k)
y=sqrt(n)
z=(s+k)

I had hoped that this would provide more detail:
http://dl.dropbox.com/u/13155084/Given%20a%20divisor%20k.pdf

The reason I say "Klein four group geometry" is because that group seems to fit best or "mainly" as you stated.

Well, it still looks a little odd: geometric group theory is a well-known, pretty advanced subject within group theory. I studied

a little of it no less than with Prof. Iliyah Ripps while in graduate school. Undoubtedly this could be a rather tough subject for

a non-mathematician.

What you're talking about, though, seems to be something else, related, as above, to group theory, geometry, number theory, etc., but in

a different way, apparently...

The link you wrote looks interesting but if you can I'd like to see books, papers, etc. about that in order to decide whether it is

something I can mess with (interesting, level, etc.) or not.

DonAntonio

 

[JeremyEbert]

Well, it still looks a little odd: geometric group theory is a well-known, pretty advanced subject within group theory. I studied

a little of it no less than with Prof. Iliyah Ripps while in graduate school. Undoubtedly this could be a rather tough subject for

a non-mathematician.

What you're talking about, though, seems to be something else, related, as above, to group theory, geometry, number theory, etc., but in

a different way, apparently...

The link you wrote looks interesting but if you can I'd like to see books, papers, etc. about that in order to decide whether it is

something I can mess with (interesting, level, etc.) or not.

DonAntonio

I'll see if I can find any real reference material for you.

It seems to me to be related to Pythagorean Triples. Here is something I wrote quite a while ago when I started down this path. It's very armature I know but it was my first attempt to make since of the relations I was noticing. It might help give a little background I guess.

http://dl.dropbox.com/u/13155084/Pythagorean%20lattice.pdf
Another visual:
http://dl.dropbox.com/u/13155084/CircleRecusion.png

 

[JeremyEbert]

As far as number theory goes, one relation is the Divisor Summatory Function equivalence.

D(n)=\sum_{k=1}^{\lfloor\sqrt{n}\rfloor}\left(2 \cdot \left\lfloor \frac{n-k^2}{k}\right\rfloor+1\right)


http://en.wikipedia.org/wiki/Divisor_summatory_function

also the Coupon collector's problem


http://en.wikipedia.org/wiki/Coupon_collector's_problem





related:

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png

http://dl.dropbox.com/u/13155084/DSUMv2.htm

http://dl.dropbox.com/u/13155084/prime.png

http://math.stackexchange.com/quest...ntation-for-the-divisor-summatory-function-dx

 

[Anti-Crackpot]

For whatever it's worth Don Antonio, I used Jeremy Ebert as an example of potentially undiscovered genius in a paper submitted last fall to two professors at Columbia University (Sociology Department).

He has already rediscovered the Dirichlet Divisor Sum geometrically, and there is no telling how far he could go with proper support from experts. It's a tough trick to pull to go backwards from simply "getting it" to understanding how it is that you "got it."

 

[DonAntonio]

For whatever it's worth Don Antonio, I used Jeremy Ebert as an example of potentially undiscovered genius in a paper submitted last fall to two professors at Columbia University (Sociology Department).

He has already rediscovered the Dirichlet Divisor Sum geometrically, and there is no telling how far he could go with proper support from experts. It's a tough trick to pull to go backwards from simply "getting it" to understanding how it is that you "got it."

Uneducated intelligence can easily be wasted in vain, in particular in realms such as mathematics where so often a hefty base is

needed to build upon it.

If he wants support then he'd rather go to some Maths Depts. in some university and approach some people there.

DonAntonio

 

[JeremyEbert]

Uneducated intelligence can easily be wasted in vain, in particular in realms such as mathematics where so often a hefty base is

needed to build upon it.

If he wants support then he'd rather go to some Maths Depts. in some university and approach some people there.

DonAntonio

I've shown this to one of my employees who is also a Math teacher at a local College. I'm really interested in what your take on this is Don Antonio.

 

[JeremyEbert]

As far as number theory goes, one relation is the Divisor Summatory Function equivalence.

D(n)=\sum_{k=1}^{\lfloor\sqrt{n}\rfloor}\left(2 \cdot \left\lfloor \frac{n-k^2}{k}\right\rfloor+1\right)


http://en.wikipedia.org/wiki/Divisor_summatory_function

also the Coupon collector's problem


http://en.wikipedia.org/wiki/Coupon_collector's_problem





related:

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png

http://dl.dropbox.com/u/13155084/DSUMv2.htm

http://dl.dropbox.com/u/13155084/prime.png

http://math.stackexchange.com/quest...ntation-for-the-divisor-summatory-function-dx

A related number theory paper:

Note the geometry of Gauss circle problem.

http://www.homepages.ucl.ac.uk/~ucahipe/Lfunctions.pdf

 

[Anti-Crackpot]

I've shown this to one of my employees who is also a Math teacher at a local College. I'm really interested in what your take on this is Don Antonio.

A natural question to ask:

"What was the response of the math teacher?"

Hopefully he or she even knew what it was, but, based on anecdotal experience, I would not bank on it, tho', in principle, the basic formulation for D(n) is junior high school, if not elementary school, level. It's just a quotient table when you get down to it, albeit a quotient table that currently confounds all modern techniques available with regards to our ability to fully understand it...

- AC

 

[chiro]

I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

Hey JeremyEbert.

I'm a little lost as to what you are trying to ask. Do you want specific kinds of comments on your diagrams and your PDF or do you have some specific question you had in mind?

You have put in a lot of effort and I don't want to see it go to waste by asking undirected questions. Are you trying to solve something in particular? Are you trying to investigate something for a particular reason? Are you looking for comments on a particular issue?

If the above is the case then this would help us give a more directed answer and know exactly what to focus our attention on to initiate the conversation.

 

[JeremyEbert]

A natural question to ask:

"What was the response of the math teacher?"

Hopefully he or she even knew what it was, but, based on anecdotal experience, I would not bank on it, tho', in principle, the basic formulation for D(n) is junior high school, if not elementary school, level. It's just a quotient table when you get down to it, albeit a quotient table that currently confounds all modern techniques available with regards to our ability to fully understand it...

- AC

Radio silence thus far.

 

[JeremyEbert]

Hey JeremyEbert.

I'm a little lost as to what you are trying to ask. Do you want specific kinds of comments on your diagrams and your PDF or do you have some specific question you had in mind?

You have put in a lot of effort and I don't want to see it go to waste by asking undirected questions. Are you trying to solve something in particular? Are you trying to investigate something for a particular reason? Are you looking for comments on a particular issue?

If the above is the case then this would help us give a more directed answer and know exactly what to focus our attention on to initiate the conversation.

Chiro,
Thanks for your interest. I see you've had some post that deal with quantum. I've noticed that the geometry of this equation produces a type of parabolic coordinate system.

http://dl.dropbox.com/u/13155084/CircleRecusion.png

http://mathworld.wolfram.com/ParabolicCoordinates.html

This coordinate system seems to be of importance in the quantum world when it comes to angular momentum.
http://www.ejournal.unam.mx/rmf/no546/RMF005400609.pdf

With the similarities between the Riemann zeros and the quantum energy levels of classically chaotic systems and the parabolic coordinates created by this geometry, it seems like this might be another connection between primes and the quantum word. What do you think?

 

[chiro]

Chiro,
Thanks for your interest. I see you've had some post that deal with quantum. I've noticed that the geometry of this equation produces a type of parabolic coordinate system.

http://dl.dropbox.com/u/13155084/CircleRecusion.png

http://mathworld.wolfram.com/ParabolicCoordinates.html

This coordinate system seems to be of importance in the quantum world when it comes to angular momentum.
http://www.ejournal.unam.mx/rmf/no546/RMF005400609.pdf

With the similarities between the Riemann zeros and the quantum energy levels of classically chaotic systems and the parabolic coordinates created by this geometry, it seems like this might be another connection between primes and the quantum word. What do you think?

Well oddly enough, there has been shown to be a link between the Riemann Zeta function and Quantum Field Theory.

I remember reading about this ages ago from work associated with Alain Connes, but I don't have any deep knowledge, only superficial awareness of the fact. I would continue your work if this is what you are working on, because if you want ways to solve the Zeta problems and connect it with general quantum phenomena (discrete structures, diophantine systems and anything involving some kind of discrete system or finite-field) then that would be extroadinarily useful.

Here are two results from a google search involving the Zeta function and quantum field theory:

http://www.ricam.oeaw.ac.at/specsem/srs/groeb/download/Moch.pdf

http://arxiv.org/pdf/hep--th/0701284.pdf

 

[Anti-Crackpot]

if you want ways to solve the Zeta problems and connect it with general quantum phenomena (discrete structures, diophantine systems and anything involving some kind of discrete system or finite-field) then that would be extroadinarily useful.

Chiro,

Give me any 6 Lucas Sequences with P's and Q's constructed from 4 integer variables (r_0, r_1, r_2, r_3) in toto and if the ratio between successive terms of those six sequences at the limit as n approaches infinity is an integer, then those same P's and Q's can be used in a very simple, regular and consistent manner to construct a quartic and its resolvent cubic that will have integer roots.

P = (r_x + r_y)
Q = (r_x * r_y)
D = P^2 - 4Q (Discriminant)

Minimal Start Terms:
U-Type Lucas Sequence = 0, 1
V-Type Lucas Sequence = 2, P

Recursive Rule:
-Q*a(n) + P*a(n+1) = a(n+2)

In other words, seems to me not only that there is a very clear linkage between the mathematics of the quartic (and thereby V4, the Klein Four-Group...) and Diophantine systems, but that this linkage may help explain how Jeremy is uncovering what he is uncovering.

If you are unclear as to what I am referencing, I can show you how to map those P's and Q's to the quartic. It took me all of about a day to work out the "translation" once Jeremy got me thinking about it.

- AC

 

[chiro]

Chiro,

Give me any 6 Lucas Sequences with P's and Q's constructed from 4 integer variables (r_0, r_1, r_2, r_3) in toto and if the ratio between successive terms of those six sequences at the limit as n approaches infinity is an integer, then those same P's and Q's can be used in a very simple, regular and consistent manner to construct a quartic and its resolvent cubic that will have integer roots.

P = (r_0 + r_1)
Q = (r_0 * r_1)
D = (r_0 - r_1)^2 = P^2 - 4Q

Minimal Start Terms:
U-Type Lucas Sequence = 0, 1
V-Type Lucas Sequence = 2, P

Recursive Rule:
-Q*a(n) + P*a(n+1) = a(n+2)

In other words, seems to me not only that there is a very clear linkage between the mathematics of the quartic (and thereby V4, the Klein Four-Group...) and Diophantine systems, but that this linkage may help explain how Jeremy is uncovering what he is uncovering.

If you are unclear as to what I am referencing, I can show you how to map those P's and Q's to the quartic. It took me all of about a day to work out the "translation" once Jeremy got me thinking about it.

- AC

I'm not sure what a Lucas sequence is. I only know the basics of number theory, but I think if you show me something like an external web-page or something that clarifies your ideas I can take a look at it.

 

[JeremyEbert]

Well, it still looks a little odd: geometric group theory is a well-known, pretty advanced subject within group theory. I studied

a little of it no less than with Prof. Iliyah Ripps while in graduate school. Undoubtedly this could be a rather tough subject for

a non-mathematician.

What you're talking about, though, seems to be something else, related, as above, to group theory, geometry, number theory, etc., but in

a different way, apparently...

The link you wrote looks interesting but if you can I'd like to see books, papers, etc. about that in order to decide whether it is

something I can mess with (interesting, level, etc.) or not.

DonAntonio

Don Antonio,
Here is a link with some useful information regarding this although this site seems to be political in nature.
http://wlym.com/~animations/ceres/InterimII/Arithmetic/Reciprocity/Reciprocity.html

 

[JeremyEbert]

Chiro,

Give me any 6 Lucas Sequences with P's and Q's constructed from 4 integer variables (r_0, r_1, r_2, r_3) in toto and if the ratio between successive terms of those six sequences at the limit as n approaches infinity is an integer, then those same P's and Q's can be used in a very simple, regular and consistent manner to construct a quartic and its resolvent cubic that will have integer roots.

P = (r_x + r_y)
Q = (r_x * r_y)
D = P^2 - 4Q (Discriminant)

Minimal Start Terms:
U-Type Lucas Sequence = 0, 1
V-Type Lucas Sequence = 2, P

Recursive Rule:
-Q*a(n) + P*a(n+1) = a(n+2)

In other words, seems to me not only that there is a very clear linkage between the mathematics of the quartic (and thereby V4, the Klein Four-Group...) and Diophantine systems, but that this linkage may help explain how Jeremy is uncovering what he is uncovering.

If you are unclear as to what I am referencing, I can show you how to map those P's and Q's to the quartic. It took me all of about a day to work out the "translation" once Jeremy got me thinking about it.

- AC

Nice AC! I'll have to play with this!

 

[JeremyEbert]

Maybe its more than the Klein four group.

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."
I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = nZ/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}

Z/3Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,2}

Z/4Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,3} Z/6Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,5} Z/8Z
(s+k)^2 = {1,4,1,0}
s^2 = {0,1,4,1}
n = {1,3,5,7} Z/12Z
(s+k)^2 = {1,9,4,0}
s^2 = {0,4,9,1}
n = {1,5,7,11} Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23} (s+k)^2 + s^2 = 1 (mod 12) ** only modulo 12. ? additive inversion of the identity element to its multiplication group Z/12Z*?
((s+k)^2 - s^2)^2 = 1 (mod 12)

In the Klein Four-Group the quadratic residues of (s+k)^2 and s^2 are permutations of the symmetric group of order 4 (S4).

with the equivalence relation
s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s+k=cosine

given the divisors of 12 have a different property than the divisors of 24, is there a link to equal temperament?

http://en.wikipedia.org/wiki/Equal_temperament

http://upload.wikimedia.org/wikipedia/commons/a/a5/ComplexSinInATimeAxe.gif

 

[JeremyEbert]

with the equivalence relation
s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s+k=cosine

given the divisors of 12 have a different property than the divisors of 24, is there a link to equal temperament?

http://en.wikipedia.org/wiki/Equal_temperament

http://upload.wikimedia.org/wikipedia/commons/a/a5/ComplexSinInATimeAxe.gif

mind typo, s+k=radius:

s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s=cosine

 

[Anti-Crackpot]

I'm not sure what a Lucas sequence is. I only know the basics of number theory, but I think if you show me something like an external web-page or something that clarifies your ideas I can take a look at it.

Consider the form:

C(a(n+2), j) (Binomial Coefficient)
a(n + 2) = x*a(n) + y*a(n+1) + z

The Lucas Sequences are just a special case of the above where:
j = 1
z = 0
x = r_0 * r_1 = Q
y = r_0 + r_1 = P

Take, for example, r_0 = phi and r_1 = -1/phi
r_0 * r_1 = -1 = Q
r_0 + r_1 = 1 = P

Start terms: 0, 1 (U-Type Lucas Sequence) gives the Fibonacci Series
Start terms: 2, 1 (V-Type Lucas Sequence) gives the Lucas Series

The recursive rule is -(-1)*a(n) + 1*a(n+1) = a(n + 2)

(sqrt (P^2 - 4Q) +/- 1)/2 = (sqrt(1^2 - 4(-1)) +/- 1)/2 = phi, 1/phi

You can read more about Lucas Sequences on Wolfram MathWorld or on Wikipedia. All kinds of identities follow from the maths and it's pretty simple stuff actually. http://mathworld.wolfram.com/LucasSequence.html

Lucas Sequences are also related to Carmichael's Theorem which involves the introduction of new prime factors into integers associated with recursively based sequences. Thus, for example, Mersenne Numbers (2^x - 1) [which are a Lucas Sequence] can only be prime where x is prime.

- AC

 

[Anti-Crackpot]

You've got a clock that goes to 12. In step increments you can go 1, 2, 3, 4, 6 or 12 at a time to end up back at 12. These are the subgroups of 12. Now take the totient of 1, 2, 3, 4, 6, 12 equal to 1, 1, 2, 2, 2, 4. These are the number of elements in each subgroup. Add them together and what do you get? 1 + 1 + 2 + 2 + 2 + 4 = 12. This holds for any integer, not just 12.

But what if you want to make sure you hit every number on your clock rather than skipping over, for instance, 1, 3, 5, 7, 9 and 11?

There are 4 ways to do that, which is the essence of the term automorphism aka "self-mapping." You can step forward by 1 or 7 and back by 1 or 7, which is the same as stepping back by 5 or 11 or forward by 5, or 11. Thus, 1, 5, 7, 11.

This fact underlies the maths of the Circle of Fifths of the Western Musical System which is based on the Perfect 5th. 12 steps, 7 notes at a time and there you are, from C through G, D, A, E, B, F# etc. and back to C after 84 notes.

I'll take a look at the link you posted, but, in short, the mathematics of Automorphism Group Z(12) underlies all of Western Music. Think about that in relation to what you posted:

Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23}

Restate n...
n = 1, 5, 7, 11, 24 - 11, 24 - 7, 24 - 5, 24 - 1

Two times, not 1 time, around the clock for a 720 degree rotation comprised of 168 notes, "coincidentally" the number of symmetries associated with the Fano plane. Whoever thought to divide the day and week the way it's divided, I might add, seems to have been a natural group theorist.

- AC

 

[JeremyEbert]

You've got a clock that goes to 12. In step increments you can go 1, 2, 3, 4, 6 or 12 at a time to end up back at 12. These are the subgroups of 12. Now take the totient of 1, 2, 3, 4, 6, 12 equal to 1, 1, 2, 2, 2, 4. These are the number of elements in each subgroup. Add them together and what do you get? 1 + 1 + 2 + 2 + 2 + 4 = 12. This holds for any integer, not just 12.

But what if you want to make sure you hit every number on your clock rather than skipping over, for instance, 1, 3, 5, 7, 9 and 11?

There are 4 ways to do that, which is the essence of the term automorphism aka "self-mapping." You can step forward by 1 or 7 and back by 1 or 7, which is the same as stepping back by 5 or 11 or forward by 5, or 11. Thus, 1, 5, 7, 11.

This fact underlies the maths of the Circle of Fifths of the Western Musical System which is based on the Perfect 5th. 12 steps, 7 notes at a time and there you are, from C through G, D, A, E, B, F# etc. and back to C after 84 notes.

I'll take a look at the link you posted, but, in short, the mathematics of Automorphism Group Z(12) underlies all of Western Music. Think about that in relation to what you posted:

Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23}

Restate n...
n = 1, 5, 7, 11, 24 - 11, 24 - 7, 24 - 5, 24 - 1

Two times, not 1 time, around the clock for a 720 degree rotation comprised of 168 notes, "coincidentally" the number of symmetries associated with the Fano plane. Whoever thought to divide the day and week the way it's divided, I might add, seems to have been a natural group theorist.

- AC

Nice! The connection to the Fano plane opens up all kinds of wonderful things.

as far as the equivalence relation
s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s=cosine

this can be simplified to:
SIN(theta) = sqrt(n)/(s+k)
COS(theta) = s/(s+k)

which reduces to a very simple:

\frac{-k + i \sqrt{n}}{k + i \sqrt{n} }

 

[JeremyEbert]

Nice! The connection to the Fano plane opens up all kinds of wonderful things.

as far as the equivalence relation
s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s=cosine

this can be simplified to:
SIN(theta) = sqrt(n)/(s+k)
COS(theta) = s/(s+k)

which reduces to a very simple:

\frac{-k + i \sqrt{n}}{k + i \sqrt{n} }

so s+k is the scalar from the unit circle. s+k = (n+k^2)/(2k)

\frac{n+k^2}{2k} \cdot \frac{-k + i \sqrt{n}}{k + i \sqrt{n} }

alternative form:

\frac{(\sqrt{n} + i k)^2}{2k }

an interesting result of this equation is the additive inverse of the divisor k is -k produces the quotient and vice versa.

12/3=4
where n=12, k=3 the result is 0.5 + i sqrt(12)
where n=12, k=-3 the result is -0.5 + i sqrt(12)
where n=12, k=4 the result is -0.5 + i sqrt(12)
where n=12, k=-4 the result is 0.5 + i sqrt(12)

 

[Anti-Crackpot]

Nice! The connection to the Fano plane opens up all kinds of wonderful things.

Glad to have been of some help. If you'd like to see how the Fano plane fits into Minkowski SpaceTime, John Baez discusses this in week 219. Here's an excerpt...

"So, there's an interesting but complicated relation between hyperbolic geometry over Z/7 and the Fano plane. How does the Klein quartic curve fit in? There's more to this side of the story than I've managed to absorb, so I'll just say a few words - probably more than you want to hear. For more detail, try my Klein quartic curve webpage.

There are 48 nonzero lightlike vectors in 3d Minkowksi spacetime, but if you take one of them and apply elements of PSL(2,Z/7) to it, you get an orbit consisting of only 24. These 24 guys correspond to the 24 heptagons in the heptagonal tiling of the Klein quartic curve! In other words, PSL(2,Z/7) acts in precisely the same way."
http://math.ucr.edu/home/baez/week219.html

Bear in mind that if John Baez hasn't been able to fully absorb it, no one's expecting you to either. But good to know that the relationships are there.

- AC

 

[JeremyEbert]

AC,

I'm working on some notes about this.

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]" https://dl.dropbox.com/u/13155084/Barycentric%20coordinates%20on%20triangles.txt

 

[JeremyEbert]

AC,
You have made reference to my models and Minkowski space before. I think you were spot on.

[Click ok then key sequence = 4,d,space]

https://dl.dropbox.com/u/13155084/PL3D2/P_Lattice_3D_2.html

http://en.wikipedia.org/wiki/Minkowski_space

 

[Anti-Crackpot]

Funny, Jeremy, but you are quite likely the only person in the world other than myself who recognizes the potential significance of what you just posted, which, unfortunately, cannot be said aloud. Enough said. Many thanks for this.

- AC

 

[Anti-Crackpot]

AC,
You have made reference to my models and Minkowski space before. I think you were spot on.

[Click ok then key sequence = 4,d,space]

https://dl.dropbox.com/u/13155084/PL3D2/P_Lattice_3D_2.html

http://en.wikipedia.org/wiki/Minkowski_space

This is a work of art Jeremy. Quite literally. And to think it's got the Dirichlet Divisor Sum embedded within. Very cool.

Btw, if you're going to play around with roots of the quartic, a natural place to begin would be with either 1, 5, 7, 11 or (24 - 1, 5, 7, 11). Two other sets of roots that are rather interesting are:

12, 19, 29, 59 & 11, 4, 40, 10. Perform any binary operation excepting division upon any two of those terms and you'll have an integer that divides the Monster Group. In fact, the first 4 roots (12, 19, 29, 59) gets you 14 of the 15 supersingular primes as factors if you combine in all possible ways. The only supersingular prime missing is 23, which is where the second set of roots comes in. Add those terms to the first set of roots and you get a multiple of 23.

Should you ever take the next step (to the quintic), then 11, 12, 19, 29, 59 is pretty "nifty," since if you add, subtract and multiply out all possible combinations and then multiply all the terms together, you get a subgroup of the Monster.

Combine 12, 19, 29, 59, 11, 4, 40, 10 in all possible ways and multiply out and here's what you get:

2^53 * 3^24 * 5^17 * 7^10 * 11^8 * 13^3 * 17^2 * 19^3 * 23^4 * 29^2 * 31 * 41 * 47 * 59^2 * 71

That's a pretty big number, about 17780 times the estimated number of atoms in the observable universe, and the Monster just happens to be one of it's subgroups.

- AC

 

[Anti-Crackpot]

AC,

I'm working on some notes about this.

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]"

https://dl.dropbox.com/u/13155084/Barycentric%20coordinates%20on%20triangles.txt

As a perhaps overly simplistic observation since we're just talking about the basic multiplication table here...
5 + 8 + 9 + 8 + 5 = (9 - (-2))^2 + (9 - (-1))^2 + (9 - (0))^2 + (9 - (1))^2 + (9 - (2))^2

This symmetry property holds for all even rows. Odd row entries have 1st differences from the central terms equal to a Pronic Number.

I would tell you how to relate this table to the divisors of the p^k*q^(n - k) triangle, but I know you already know that.

Interesting though, how the very structure of the basic multiplication table embeds the tetrahedron, one of the most important structural forms in nature (i.e. what do water and diamonds have in common?).

- AC

 

[Anti-Crackpot]

Jeremy, just as a thought, here is an equation form that might be worth looking into just to see what comes of it. It's an "abuse" of the Ramanujan theta function, substituting primes p and q for a and b (where |ab| < 1) and setting the start of the range at 0, not negative infinity:

\sum^{∞}_{0}p^{T_n}q^{T_{-n}}

For example, set p to 2 and q to 3 and you get the below sequence as first differences:
2^(n(n+1)/2)*3^(n(n-1)/2)...
http://oeis.org/A081955
2, 24, 1728, 746496, 1934917632, 30091839012864, 2807929681968365568, 1572081206902992767287296...
# Divisors of first differences 2, 8, 28, 77... = (T_n + 1) * (T_-n + 1)

SUM = 2, 26, 1754, 748250, 1935665882, 30093774678746 ...
# divisors= 2, 4, 4, 32, 8, 32, 32, 32, 4, 128...

Curious if any interesting patterns would emerge. For instance, is that a coincidence that the # of divisors up above are all powers of 2? No idea actually, but it does seem a bit odd. Will have to see what happens with the 11th term at the very least :-)

If nothing else, it will get you thinking about the Ramanujan Theta Function in relation to your model. (Which I suspect is, in fact, related)

- AC

Note: Turns out the 11 th term has 16 divisors. Also a power of 2. The numbers are getting pretty big pretty fast, so not sure how many more terms I can check. Last term checked: 1.2872079903823517043082155685251539206949300954 × 10^46

 

[JeremyEbert]

This is a work of art Jeremy. Quite literally. And to think it's got the Dirichlet Divisor Sum embedded within. Very cool.

Its related to Julian Voss-Andreae's "Father and Mother"

"Father and Mother of the series Spin Family (2009) by physicist-turned-sculptor Julian Voss-Andreae"
http://en.wikipedia.org/wiki/File:"Father"_and_"Mother"_of_the_series_"Spin_Family".jpg

 

[Anti-Crackpot]

Its related to Julian Voss-Andreae's "Father and Mother"

"Father and Mother of the series Spin Family (2009) by physicist-turned-sculptor Julian Voss-Andreae"
http://en.wikipedia.org/wiki/File:"Father"_and_"Mother"_of_the_series_"Spin_Family".jpg

You read my mind. That's exactly the piece I was thinking of.

 

[JeremyEbert]

AC,

I'm working on some notes about this.

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]" https://dl.dropbox.com/u/13155084/Barycentric%20coordinates%20on%20triangles.txt

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5

relative intensities = k(n+1-k)

http://en.wikipedia.org/wiki/Quantum_number#Spatial_and_angular_momentum_numbers"the spin-raising operator is the square-root"

the square root produces the Klein four group from (n-k^2)/(2k)

AC has some findings for hbar (reduced Planck's constant) that fix right in as well I beleive.

interesting...

 

[JeremyEbert]

a binomial distribution link as well

http://epr.cm.utexas.edu/WhatIsEPR.html


"where N is the number of equivalent nuclei and I is the spin. It is important to note that this formula only determines the number of lines in the spectrum, not their

relative intensities. Coupling to a single nucleus with spin n/2 gives (n + 1) lines each of equal intensity.

2NI + 1 = 2(1)(n/2) + 1 = n + 1 lines

For example, coupling to a single vanadium nucleus (I = 7/2) will result in a spectrum of eight lines all of equal intensity.


Simulated EPR spectrum showing coupling to one nucleus (/I/ = 7/2)
Coupling to n equivalent nuclei, each with spin ½ again gives (n + 1) lines,

2NI + 1 = 2(n)(1/2) + 1 = n + 1 lines

but, since there are multiple nuclei interacting, the relative intensities of the lines follow the binomial distribution shown below.



# of Equivalent Nuclei Relative Intensities
1 1:1
2 1:2:1
3 1:3:3:1
4 1:4:6:4:1
5 1:5:10:10:5:1
6 1:6:15:20:15:6:1

An example of this is shown by the EPR spectrum of the radical anion of benzene, [C6H6•]-, in which the electron is delocalized over all six carbon atoms and therefore

exhibits coupling to six equivalent hydrogen atoms. As a result, the EPR spectrum shows seven lines with relative intensities of 1:6:15:20:15:6:1. Similar distributions

can be derived for n equivalent nuclei with spins greater than ½."

 

[JeremyEbert]

AC,

I'm working on some notes about this.

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]"


https://dl.dropbox.com/u/13155084/Barycentric%20coordinates%20on%20triangles.txt

Some great info on Angular Momentum States.


*"According to the laws of quantum mechanics spin angular momentum is
not only quantized in magnitude, but in orientation relative to the z axis in
physical space.Therefore, the vector representing spin can only assume certain
orientations relative to the z axis in spin vector space. The quantum number
Ms specifies the possible orientation of a given angular momentum in space."
...

*"Multiplicity, M, of state with quantum number S = 2S + 1"

*"Pythagorean Relationships and Spin Angular Momentum"

http://www.columbia.edu/itc/chemistry/photochem/spin/03.pdf

 

[JeremyEbert]

for a set of complex numbers Z

z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

make sense?

 

[Anti-Crackpot]

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5

relative intensities = k(n+1-k)

http://en.wikipedia.org/wiki/Quantum_number#Spatial_and_angular_momentum_numbers

Reverse the sign so that you're adding a square rather than subtracting it...

k(n+1+k)

Set a(0) = 0 and a(n + 1) = 2a(n) + n(mod 2)sgn(n-1)
a(n) = 0, 0, 0, 0, 1, 2, 5, 10, 21...
Without the 3 leading zeroes, that's this progression here: http://oeis.org/A000975

Then, for range 0 ≤ k ≤ 8 and n = a(n), you'll get the following set of values:

0, 2, 6, 12, 24, 40, 72, 126, 240

I haven't worked out the algebra, but you can be pretty sure the recursively defined progression up above is this one here (derived accidentally):
A029929 n(n+ceiling(2^n/12))
REFERENCES C. Muses, The dimensional family approach in (hyper)sphere packing..
http://oeis.org/A029929
Next values: 468, 960, 2002...

Via Wikipedia
http://en.wikipedia.org/wiki/ADE_classification
ADE Classification
Algebro-geometrically, McKay also associates E6, E7, E8 respectively with: the 27 lines on a cubic surface, the 28 bitangents of a plane quartic curve, and the 120 tritangent planes of a canonic sextic curve of genus 4.[14][15] The first of these is well-known, while the second is connected as follows: projecting the cubic from any point not on a line yields a double cover of the plane, branched along a quartic curve, with the 27 lines mapping to 27 of the 28 bitangents, and the 28th line is the image of the exceptional curve of the blowup. Note that the fundamental representations of E6, E7, E8 have dimensions 27, 56 (28·2), and 248 (120+128), while the number of roots is 27+45 = 72, 56+70 = 126, and 112+128 = 240.

Related progression:

A002336 Maximal kissing number of n-dimensional laminated lattice.
http://oeis.org/A002336

Imagine a crazy universe where lattices actually had something to do with physics and the optimal division of space had something to do with its converse, the optimal (recursively defined) packing of space. :-)

- AC

UPDATE:
k(n+1+k) = kn + k + k^2
Set n to 1/12*(2^k + 2(-1)^k - 6)
Thus, ceiling [k(1/12*(2^k + 2(-1)^k - 6) +1+ k)] = 0, 2, 6, 12, 24, 40, 72, 126, 240... etc.

NOTE that ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1+ n)] does not "equal" n(n+ceiling(2^n/12)), but you'd never know if you just looked at the integer solutions since the progressions with respect to that seem to be one and the same at least for n>-1.

Now reverse the sign back...
xxxxxx ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1- n] = 0, 0, -2, - 6, -8, -10, 0, 28, 112, 306...
... and take note of the 112 that appears in position 8. 112 + 128 = 240 (which follows since 128 = 2*8^2)

 

[Anti-Crackpot]

for a set of complex numbers Z

z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

make sense?

Can you walk me through this, Jeremy? Also, any graphics or specific numbers as examples would be great. I'm a bit of a dummy and need to see things visually at times.

 

[Anti-Crackpot]

UPDATE:
k(n+1+k) = kn + k + k^2
Set n to 1/12*(2^k + 2(-1)^k - 6)
Thus, ceiling [k(1/12*(2^k + 2(-1)^k - 6) +1+ k)] = 0, 2, 6, 12, 24, 40, 72, 126, 240... etc.

NOTE that ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1+ n)] does not "equal" n(n+ceiling(2^n/12)), but you'd never know if you just looked at the integer solutions since the progressions with respect to that seem to be one and the same at least for n>-1.

Now reverse the sign back...
xxxxxx ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1- n] = 0, 0, -2, - 6, -8, -10, 0, 28, 112, 306...
... and take note of the 112 that appears in position 8. 112 + 128 = 240 (which follows since 128 = 2*8^2)

Hopefully, it is apparent why I went into all that, Jeremy.
0, 2, 6, 12, 24, 40, 72, 126, 240
0, 0, -2, - 6, -8, -10, 0, 28, 112

What are the first differences between these two progressions?
0, 2, 8, 18, 32 for the first 5 values. aka 2*n^2 which, of course, gives the maximal number of electrons per shell in the context of relating that both to your formulas and sphere packings.

From the observational/heuristic standpoint, that's a heck of a coincidence don't you think? And the relationship followed organically from what you posted. You really might be on to something...

- AC

 

[JeremyEbert]

Can you walk me through this, Jeremy? Also, any graphics or specific numbers as examples would be great. I'm a bit of a dummy and need to see things visually at times.

Meee toooo...
here is the visual:
http://dl.dropbox.com/u/13155084/particle%20spin%205-2.png

 

[JeremyEbert]

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5

relative intensities = k(n+1-k)

http://en.wikipedia.org/wiki/Quantum_number#Spatial_and_angular_momentum_numbers


"the spin-raising operator is the square-root"

the square root produces the Klein four group from (n-k^2)/(2k)

AC has some findings for hbar (reduced Planck's constant) that fix right in as well I beleive.

interesting...

EDIT:Ms where spin S = ((n+1)/2)-0.5

relative intensities = k(n+1-k)

http://dl.dropbox.com/u/13155084/particle%20spin%205-2.png

 

[JeremyEbert]

Hopefully, it is apparent why I went into all that, Jeremy.
0, 2, 6, 12, 24, 40, 72, 126, 240
0, 0, -2, - 6, -8, -10, 0, 28, 112

What are the first differences between these two progressions?
0, 2, 8, 18, 32 for the first 5 values. aka 2*n^2 which, of course, gives the maximal number of electrons per shell in the context of relating that both to your formulas and sphere packings.

From the observational/heuristic standpoint, that's a heck of a coincidence don't you think? And the relationship followed organically from what you posted. You really might be on to something...

- AC

Very nice AC. This fits right in as well.

I noticed an interesting vector a while ago at 45 degrees or pi/4 radians.

parabolic coodinate values at pi/4 = q:

5.8284271247461900976033774484194
23.313708498984760390413509793678
52.455844122715710878430397035775
93.25483399593904156165403917471

q=((1+sqrt(2))*n)^2

u = ((n^2)*6) – q

n = (q*u)^(1/4)

m = ((n^2)*12) – 2q

(2q*m)^(1/2) = 2(n^2) = maximal number of electrons per shell.

visual:
http://dl.dropbox.com/u/13155084/45.png

 

[Anti-Crackpot]

I made a a typo in my recent post, Jeremy, but it's an interesting one...

WOLFRAM INPUT #1:
ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1- n)] - n(n+ceiling(2^n/12))

Where n is an integer, you get an integer (2(n*i)^2), else no. Descriptively, you get a series of arcs (think: "quantized" arcs) that trace the general form of a parabola. It's pretty cool. To get your "zeroes," then reverse sign:

WOLFRAM INPUT #2:
ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1+ n)] - n(n+ceiling(2^n/12))

Just thinking out loud here, but I can't help but wonder if there isn't possibly some significance to the 1st differences between these two progressions.

- AC

 

[Anti-Crackpot]

FOR REFERENCE
Feel free to add to this list Jeremy, ideally in your next post, so we can start to build a little resource to help you better communicate your ideas to professionals. No need for a thank you (You're welcome in advance.). - AC

Dirichlet Divisor Sum Algorithm
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Divisor Sum Formula + Dirichlet Divisor Sum as Quotient Table
http://math.stackexchange.com/quest...ntation-for-the-divisor-summatory-function-dx
(see bottom of page)
See: http://oeis.org/A006218

Prime UNSafe Periods (Related to Cicada Life Cycles)
http://dl.dropbox.com/u/13155084/prime.png
NOTE: This sequence + Dirichlet Divisor Sum totals a Triangular # (T_n)
See: http://oeis.org/A161664
This sequence + "safe" periods also equals a Triangular (T_(n + 1))
See: http://oeis.org/A002541
It follows that 2*unsafe + safe + D(n) is square.

The "Pythagorean Lattice" - Interactive Visual (2-D and 3-D simulation)
https://dl.dropbox.com/u/13155084/PL3D2/P_Lattice_3D_2.html
e.g. Click ok then key sequence = 4,d,space (3-D Pattern)
e.g. Click ok then key sequence = 3,v,space (2-D Recursion)

[m] = Control Menu
[ Space Bar ]=Start/Stop

[ 1 ]=NORMAL
[ 2 ]=REVERSE
[ 3 ]=RECURSION
[ 4 ]=PATTERN

[ d ]=3D ON/OFF
[ v ]=2D VIEWS

[ UP/DOWN ]=ZOOM
[ LT/RT ]=RECURSION +-

[ f ]=Faster
[ s ]=Slower
[ r ]=Reset

Circle Recusion
http://dl.dropbox.com/u/13155084/CircleRecusion.png First visual of hypothesized "Klein Four Group Geometry"
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

Divisor Symmetry n = 12
http://dl.dropbox.com/u/13155084/divisor%20semmetry.png Mapping to Particle Spin (m_{s})
http://dl.dropbox.com/u/13155084/particle%20spin%205-2.png RELATED LINKS

Divisor Summatory Function
http://en.wikipedia.org/wiki/Divisor_summatory_function
Coupon Collector's Problem
http://en.wikipedia.org/wiki/Coupon_collector's_problem

Planck's Constant
http://en.wikipedia.org/wiki/Planck_constant
Quantum Number
http://en.wikipedia.org/wiki/Quantum_number

EXCERPTS

I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png

Maybe its more than the Klein four group.

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."

I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = n

Z/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}

...

Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23}

...

In the Klein Four-Group the quadratic residues of (s+k)^2 and s^2 are permutations of the symmetric group of order 4 (S4).

https://www.physicsforums.com/showpost.php?p=3897960&postcount=4

Nice! The connection to the Fano plane opens up all kinds of wonderful things.

as far as the equivalence relation
s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s=cosine

this can be simplified to:
SIN(theta) = sqrt(n)/(s+k)
COS(theta) = s/(s+k)

which reduces to a very simple:

\frac{-k + i \sqrt{n}}{k + i \sqrt{n} }

for a set of complex numbers Z

z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

make sense?

here is the visual:
http://dl.dropbox.com/u/13155084/particle%20spin%205-2.png

ALSO...

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...

 

[Anti-Crackpot]

FOR REFERENCE
Feel free to add to this list Jeremy, ideally in your next post, so we can start to build a little resource to help you better communicate your ideas to professionals. No need for a thank you. (You're welcome in advance.) - AC

Dirichlet Divisor Sum Algorithm
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Divisor Sum Formula + Dirichlet Divisor Sum as Quotient Table
http://math.stackexchange.com/quest...ntation-for-the-divisor-summatory-function-dx
(see bottom of page)
See: http://oeis.org/A006218

Prime UNSafe Periods (Related to Cicada Life Cycles)
http://dl.dropbox.com/u/13155084/prime.png
NOTE: This sequence + Dirichlet Divisor Sum totals a Triangular # (T_n)
See: http://oeis.org/A161664
This sequence + "safe" periods also equals a Triangular (T_(n + 1))
See: http://oeis.org/A002541
It follows that 2*unsafe + safe + D(n) is square.

FYI, Jeremy, you may know this already, but others almost certainly don't...

Let S(n) = Safe Period
Let D(n) = Dirichlet Divisor Sum
Let U(n) = Unsafe Period

2S(n) + 1D(n) + 3U(n) is First Pentagonal
1S(n) + 2D(n) + 3U(n) is Second Pentagonal

Add and subtract at will in combination with n and 2n + 1 to generate equivalencies. It becomes clear, with a little experimentation, the recursive nature of the Dirichlet Divisor Sum as well as the likelihood of a relationship to Euler's summing technique for identifying primes based on Generalized Pentagonal Numbers (i.e. his hypothesis on that which I believe has never been proven is very likely to be true). What comes next depends on what came before and the "metronome" marking the "beat" is (at least partly) five-sided in nature...

- AC

These relationships follow since:
T_(n-1) = S(n) + U(n)
T_(n) = D(n) + U(n)
n^2 = S(n) + D(n) + 2U(n)

And, for good measure...
T_(n-1) + T_(n) + n^2 = 2n^2