# An Exact Value For The Fine Structure Constant.

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### #103 Turtle

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Posted 01 May 2012 - 11:45 PM

...

And this might go to polygonal numbers. What would a polygon with 2 sides look like, or 1 side, or zero? A 2 sided polygon would have no area. Would this be analogous to the flight of an s electron? Something to think about....

Jess Tauber

if using the polygonal number representation of gnomons, then the geometric represntation of 2-sided numbers is a line of dots. a line has length,but no area; correct. of 1-sided number(s), there would be a single dot. no area, no length. no 0-sided numbers exist. i have attached the gnomon for 3-sided numbers for reference.

algebraically, P = 1/2((n2*s) - 2*n2 - n*s + 4*n) is the generalized expression/equation for polygonal numbers. n & s must be retricted to >=3, else all natural numbers would be polygonal. for all n=1, s=1. for all n=2, P=s. for all s=1, P=1. for all s=2, P=n. for there to exist any differentiation between polygonal numbers and natural numbers, then we exclude the "trivials" derived when s & n <=2. this restriction also gives rise to the set of non-polygonal numbers, i.e. natural numbers with no integer solutions n & s for the generalized expression.

i will leave it to you to decide how or if this fits your electron.

Edited by Turtle, 01 May 2012 - 11:48 PM.

### #104 Turtle

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Posted 02 May 2012 - 02:26 AM

...
algebraically, P = 1/2((n2*s) - 2*n2 - n*s + 4*n) is the generalized expression/equation for polygonal numbers. n & s must be retricted to >=3, else all natural numbers would be polygonal. for all n=1, s=1. for all n=2, P=s. for all s=1, P=1. for all s=2, P=n. for there to exist any differentiation between polygonal numbers and natural numbers, then we exclude the "trivials" derived when s & n <=2. this restriction also gives rise to the set of non-polygonal numbers, i.e. natural numbers with no integer solutions n & s for the generalized expression.
...

following up, it is this distinction, this bifurcation, of natural numbers into 2 disjoint sets that first led me to take the count of how many of each cast were in a given range of natural numbers from modest's list of non-polygonal numbers <1000. for all practical purposes it is sufficient to say that ~64% of all natural numbers (>=6) belong to the polygonal set and ~36% belong to the non-polygonal set.

Edited by Turtle, 02 May 2012 - 02:41 AM.

### #105 Don Blazys

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Posted 02 May 2012 - 08:43 AM

To:Lars(LBg),

Running the counting function beyond $x=10^{15}$
we find that its upper and lower bounds are as follows:

For $x<10^{19.0324}$,

the upper bound is $\varpi(10^{n})=.640362739400577657*10^{n}$

which is in exellent agreement with your findings.

(Actually, the more fitting word here would be remarkable,
because that upper bound occurs slightly after $x=10^{19}$
which is ten thousand times beyond where we are now!)

For $x>10^{19.0324}$,

the lower bound is $\varpi(10^{n})=.64036273685156*10^{n}$

Don.

Edited by Don Blazys, 02 May 2012 - 06:53 PM.

### #106 pascal

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Posted 02 May 2012 - 09:09 PM

I've been thinking that relativistic effects in the atomic electronic system might shift packing/growth norms through the various Metal Means (though it might be some other function along these lines)- this does appear to work in the nucleus with regard to nucleons N vs P in the curve of these.

But of course C is involved as well in the FSC. What if C is not an arbitrarily value? What if it is exactly what it is due to a need to successfully shift through Metal Means for both electronic and nuclear systems?

What I've just found regards the Phi-based Fiblike sequences and construction of the periodic tabularization, with each sequence associated with different structural motifs in the periodic system's orbitals, where one simply reads off information through the sequences, the same columns throught them marking the same kinds of information. All quantum based- I've now located both n and l in columns 4 and 3 respectively, seemingly hinting that l outprioritizes n. It also means that (n+l), which marks periods, is column 5. Some columns perform more than one function!

All of the above says nothing about relativistic effects which change the masses and so orbits/energies of electrons, the sizes of atoms/ions (which alters surface behaviors), etc. I'm hoping perhaps to find that higher level Pascal Triangle analogues are involved here, whose diagonal relationships can culminate in other periodicities that have been missed, as well as the other Metal Means. One might stack such triangles (either in-line, or as a torus, etc.) and graph trajectories through the volume that actually do follow real elemental behaviors, where these deviate from purely quantum level predictions.

Jess Tauber

Edited by pascal, 02 May 2012 - 09:12 PM.

### #107 pascal

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Posted 03 May 2012 - 08:04 PM

Further results: I tried back-engineering the Pascal Triangle sisters based on the 'seed' pairs giving different Fibonacci sister sequences. And I also looked at sequences earlier than the ones I laid out here the other day, which start below (-1,1). All have the second digit as 1.

What I found is that there is pairing between sequences on the positive and negative sides of the Fib matrix, and on opposite sides of the same sister Pascal Tiangle. This is fantastically unexpected. So for example the sister Pascal with (2,1) sides gives Fibonacci on one side by summing along shallow diagonals, and the Lucas sequence on the other (seeds 1,2 versus 2,1).

My research partner, Valery Tsimmerman (visit his site at www.perfectperiodictable.com), who had independently rediscovered my first tetrahedral rendering of the periodic system, has recently come up with an equation describing the layout of the periodic table. In exploring it into negative number territory we've wondered whether it can describe antimatter elements, in a sort of anti-periodic table. He's already shown that the layout of this table is not a recapitulation of the one we all know and love. It prioritizes things differently. I'm now thinking perhaps the two periodic systems get their motivations from opposite sides of the same Pascal Triangle sisters. If this works then it has interesting implications for how elements and antielements might interact. Perhaps some combinations might be stable against annihilation under some circumstances? Hello warp drive!

Jess Tauber

### #108 Don Blazys

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Posted 06 May 2012 - 03:50 AM

To: Jess (Pascal),

I checked out Valery Tsimmerman's site. Very impressive! Please give him my regards.

You see, I finally got a chance to do a little research on periodic tables,
magic numbers, shell theories, the theoretical "island of stability", etc.
and what I found is that both the fine structure constant and
the counting function for polygonal numbers of order > 2
do indeed seem to tie into those things.

For instance,
Quoting Valery Tsimmerman:

Those who are familiar with the energy levels obtained in nuclear shell theory
know that there are 22 energy levels, 7 of which yield the magic numbers!
Moreover, there are 7 more numbers generated by the tetrahedral sphere packing that
coincide with the nuclear shell energy levels 6, 14, 38, 40, 58, 64, 100.
That is 14 out of 22 numbers (64%) of the energy level match with the sphere packing results.
Most of the remaining 8 numbers deviate slightly.

Well, our polygonal number counting function shows that 64.036273685156% of all positive integers
are polygonal numbers of order greater than 2, so I wonder if there might be some kind correlation here.

Edited by Don Blazys, 06 May 2012 - 07:45 AM.

### #109 Turtle

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Posted 06 May 2012 - 10:59 AM

To: Jess (Pascal),

I checked out Valery Tsimmerman's site. Very impressive! Please give him my regards.

You see, I finally got a chance to do a little research on periodic tables,
magic numbers, shell theories, the theoretical "island of stability", etc.
and what I found is that both the fine structure constant and
the counting function for polygonal numbers of order > 2
do indeed seem to tie into those things.

For instance,
Quoting Valery Tsimmerman:

Those who are familiar with the energy levels obtained in nuclear shell theory
know that there are 22 energy levels, 7 of which yield the magic numbers!
Moreover, there are 7 more numbers generated by the tetrahedral sphere packing that
coincide with the nuclear shell energy levels 6, 14, 38, 40, 58, 64, 100.
That is 14 out of 22 numbers (64%) of the energy level match with the sphere packing results.
Most of the remaining 8 numbers deviate slightly.

Well, our polygonal number counting function shows that 64.036273685156% of all positive integers
are polygonal numbers of order greater than 2, so I wonder if there might be some kind correlation here.

if by "our polygonal number counting function" you mean your equation(s), then again, without someone else actually counting how many polygonals in a range, your equations are meaningless. if by if by "our polygonal number counting function" you mean to imply you had a hand in conceiving of taking the natural density of polygonal/nonpolygonal numbers, then again that was my conception.

the first case is just a silly claim; the second i consider theft. stop taking credit for my work Don.

following up, it is this distinction, this bifurcation, of natural numbers into 2 disjoint sets that first led me to take the count of how many of each cast were in a given range of natural numbers from modest's list of non-polygonal numbers <1000. for all practical purposes it is sufficient to say that ~64% of all natural numbers (>=6) belong to the polygonal set and ~36% belong to the non-polygonal set.

### #110 pascal

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Posted 06 May 2012 - 06:42 PM

Turtle ti:kama:nude: 'for all practical purposes it is sufficient to say that ~64% of all natural numbers (>=6) belong to the polygonal set and ~36% belong to the non-polygonal set.'

An interesting split, in and of itself. If one creates tetrahedra composed of close packed spheres by jacketing a tetrahedral core of 20 spheres on all faces to make the next larger, one finds that the inner core consists of 4+16=20 spheres, and the outer jacket consists of 36+64=100 spheres. Further full jackets always will contain numbers of spheres that are the sums of the squares of the next two even integers. Thus core= 2sq+4sq, first jacket=6sq+8sq, next jacket=10sq+12sq or 244, etc. These pairs, by the way, in tetrahedral periodic table models, are equivalent to four periods worth of elements.

Jess Tauber

Edited by pascal, 06 May 2012 - 06:43 PM.

### #111 pascal

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Posted 06 May 2012 - 06:55 PM

By the way, Don, the quote you posted was originally from research done by someone else trying to rationalize the nuclear shell filling in a quite bizarre 'double tetrahedral' model, which doesn't correspond to the way real nuclei are organized. In actual fact the nuclear shell filling pattern is more like the one I posted about some months back, at least ideally, without regard for energy level inversions between magic numbers.

My current understanding is that such inversions are actually quite common in natural sequences- in genomes, in languages, etc. Inversion has to be imposed upon a simpler underlying order pattern, though it can be reified, and the original part of the pattern lost (just as scaffolds and forms are taken away after building structures). There may be a maximal or 'climax' inversion patterning, when everything that can be turned on its head has been. Perhaps this has its own idealized mathematical description. But less fully developed systems are going to be mathematically dualic, at least (since there can be inversions atop inversions, and so on).

Jess Tauber

### #112 Turtle

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Posted 06 May 2012 - 08:41 PM

Turtle ti:kama:nude: 'for all practical purposes it is sufficient to say that ~64% of all natural numbers (>=6) belong to the polygonal set and ~36% belong to the non-polygonal set.'

An interesting split, in and of itself. If one creates tetrahedra composed of close packed spheres by jacketing a tetrahedral core of 20 spheres on all faces to make the next larger, one finds that the inner core consists of 4+16=20 spheres, and the outer jacket consists of 36+64=100 spheres. Further full jackets always will contain numbers of spheres that are the sums of the squares of the next two even integers. Thus core= 2sq+4sq, first jacket=6sq+8sq, next jacket=10sq+12sq or 244, etc. These pairs, by the way, in tetrahedral periodic table models, are equivalent to four periods worth of elements.

Jess Tauber

ti:kama:nude: ? you lost me on that one. i googled the phrase and found a reference to an earlier post of yours. >>

Just toying around with my original FSC proportion here:

All the numbers together sum to exactly 600. The numerators sum to 310, the denominators sum to 290. That is, 300 + or - 10.
Then the left numbers sum to 265 and the right to 335. That is, 300 + or - 35. Finally, the right numerator minus the left denominator sums to 301 and the left numerator plus the right denominator sums to 299, or, 300 + or - 1.

This might SEEM to be a jumble of random differences, but look again: 1, 10, 35 are every other Pascal Triangle TETRAHEDRAL number, the fuller set being 1, 4, 10, 20, 35, 56, 84, 120. More specifically, 1, 10, 35 are the running sums of squares of odd integers 1, 1+9, 1+9+25... (the complement of what is found as atomic numbers in the Janet Left-Step periodic table, where we find the running sums of the squares of even integers being the atomic numbers of the right edge, s2 configuration elements).

Jess Tauber

i'm afraid i am not familiar enough with the periodic table to visualize your descriptions, and i'm trying to visualize your packing description but it's not clear to me either. can you post some drawings?

anyway, to the stuff i do understand. first, when i started the non-figurate thread i was unclear on the proper use of the term "figurate". i have thought about having it changed to "non-polygonal numbers" as it technically should be, but i rather think that might confuse the issue even more. in truth, there are no "non-figurate" numbers.

so, the [standard] polygonal numbers are a subset of figurate numbers. (not that you didn't know, only that i want to clarify. ) this being the case, polygonal numbers are 2-dimensional in their geometric representations, known as gnomons. as i pointed out in the post previous to what you just quoted, for there to be a 64/36 split, one can only consider the 2-dimensional geometric standard non-polygonals numbers -and so non-polygonal numbers- to find that proportion. (i say "standard" to differentiate from "centered polygonal numbers"; another menagerie of beasties altogether.)

one can get your tetrahedral numbers (a special case of figurate numbers) without pascal's triangle by summing the triangular numbers from the 2-dimensional polygonal schema.

triangular numbers:1 3 6 10 15 21 28 36 45 55 66... :2-dimensional geometry
tetrahedral numbers:4 10 20 35 56 84 120 165 220 286... :3-dimensional geometry?

one can continue this to infinity & beyond, and all such sets are figurate numbers. [sum the pentagonal numbers, sum the hexagonal numbers, etcetera] i guess we might call the sum of tetrahedral numbers hyper-tetrahedral, but beyond that i think we would want to say n-tuple triangular numbers or some such a matter.

then too, we can sum the square numbers to get "cubic" numbers, but of course these aren't the perfect-cubes so that is confusing too. also, i don't think, as far as i recall, that these cubic numbers correspond to any 3-dimensional geometry. perhaps you know better?

square numbers:1 4 9 16 25 36...
"cubic" numbers: 5 14 30 55 91...

notice that 55 is both "cubic" and triangular, as is 91.

anyway, i hope i clarified something there and gave you some ammunition as it were. let me know.

~

Edited by Turtle, 06 May 2012 - 08:42 PM.

### #113 pascal

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Posted 07 May 2012 - 01:01 AM

Triangular numbers from the classical Pascal Triangle give numbers of spheres in a triangular close-packed matrix, like billiard balls in the rack. Summing them to give tetrahedral numbers is like stacking the triangular matrices of spheres layer by layer. The next diagonal represents the pentatope numbers, which unfortunately for visualization give a four dimensional figure (time might work as the fourth dimension)- see the Wiki page on the topic. Anyway, for the (2,1) sided sister Pascal Triangle, we get going down one side first 2's replacing 1's, then odd integers replacing natural numbers, square numbers replacing triangulars. So by analogy in the next diagonal we stack the square matrices of spheres (no longer close packed, but all at right angles) giving not a cubic matrix, but a square pyramidal one, just like the pyramids of the Giza Plateau, except without mummies.

However, I have noticed that the sister Pascal diagonal seem to do double duty- the square mapping isn't the only one. Rhombi of close-packed spheres contain exactly the same numbers as squares do. One can stack these as well in the next diagonal. My tetrahedral model of the periodic table is actually a compromise here- it is composed of rhombi, bent up at the minor axis to a tetrahedral dihedral angle, then these are stacked. BUT only the rhombi based on squares of even integers stack, it doesn't include those based on odds. So you end up with a tetrahedron (related to the classical Pascal Triangle) but its made of rhombi (related to the (2,1) Pascal). I haven't looked at the 1's side of the (2,1) system, perhaps I should. It still gives the classical Fibonacci sequence on its side (versus the Lucas on the 2's side), but the numbers in the diagonals are offset, as is the Fibonacci sequence. Maybe the compromise is from the SAME Pascal system here? The (2,1) system IS special, as it relates directly to the equations that give the powers of all the Metal Means- I don't know if other Pascal sisters have other related Metal Means functions but I wouldn't doubt it. Too much to learn in one lifetime, and I was hoping to do more linguistics tonight.

Which brings me to ti:kama:nude: This is a Yahgan language word, composed of t- a prefix making reference to something related to the verb following (instrument, goal, location, etc. i.e. various case relations, but on the VERB, so not case precisely. This is called by the field a circumstantial voice marker. Relatively rare in the world's languages). Then comes the verb i:kama:na which means to write or decorate with linear features, itself composed of (y)i:ku: meaning 'to scrape, scratch', and (a)ma:na meaning 'not to strike true, to graze, etc.', and then -(u)de: which is simple past tense. A vowel with : is 'long' as in English, without it 'short'. I've studied the language for around a decade now. There is only one fluent speaker left, very elderly, living in Tierra del Fuego. The Yahgans were people that Darwin encountered on the voyage of the Beagle, where the ship was returning a couple of the people after they were trained in English culture back in England itself. The Brits wanted to win souls for the Anglican Church. Of course it all turned out nicely. After many people were 'civilized', clothed in western clothing and learning western job skills, western diseases came in and killed most of them- the survivors didn't fare too well either, suffering lifelong debilitating effects from the illnesses. All those who were literate in their language died at this time- and that was quite a few. Today the region is a mecca for tourism of the environmentally correct kind, and there is plenty of money for conservation and sightseeing, not very much to help save the last embers of Yahgan culture and language. Hey, we killed them fair and square, right?

Jess Tauber

Edited by pascal, 07 May 2012 - 01:03 AM.

### #114 Turtle

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Posted 07 May 2012 - 01:26 AM

Triangular numbers from the classical Pascal Triangle give numbers of spheres in a triangular close-packed matrix, like billiard balls in the rack.
snip...
Hey, we killed them fair and square, right?

Jess Tauber

so that's a no on any drawings then? but hey, i guess that's fair & square right?

### #115 Don Blazys

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Posted 08 May 2012 - 09:03 AM

Quoting "turtle":

If by "our polygonal number counting function" you mean your equation(s)...

Here is the polygonal number counting function:

$\varpi(x)\approx\left(\left(\sqrt{\left(\left(1-\frac{1}{\left(\alpha*\pi*e+e\right)}\right)*x\right)}-\frac{1}{4}\right)^{2}-\frac{1}{16}\right)*\left(1-\frac{\alpha}{\left(6*\pi^{5}-\pi^{2}\right)}\right)$

where:

$\alpha=\left(\left(A^{-1}*\pi*e+e\right)*\left(\pi^{e}+e^{\left(\frac{-\pi}{2}\right)}\right)-\frac{\left(\left(\pi^e+e^{\frac{-\pi}{2}}+4+\frac{5}{16}\right)*\left(\ln\left(x\right)\right)^{-1}+1\right)}{\left(6*\pi^{5}*e^{2}-2*e^{2}\right)}\right)^{-1}$

and here is how it compares to the empirical data (observed results) which we have designated $\varpi(x)$:

$x$_______________________$\varpi(x)$_________________ $B(x_{F\alpha})$_________Difference
10_______________________3______________________5___________________2
100______________________57_____________________60__________________3
1,000____________________622____________________628_________________6
10,000___________________6,357__________________6,364________________7
100,000__________________63,889_________________63,910_______________21
1,000,000________________639,946________________639,963______________17
10,000,000_______________6,402,325______________6,402,362_____________37
100,000,000______________64,032,121_____________64,032,273____________152
1,000,000,000____________640,349,979____________640,350,090____________111
10,000,000,000___________6,403,587,409__________6,403,587,408__________-1
100,000,000,000__________64,036,148,166_________64,036,147,620_________-546
1,000,000,000,000________640,362,343,980________640,362,340,975________-3005
10,000,000,000,000_______6,403,626,146,905______6,403,626,142,352_______-4554
100,000,000,000,000______64,036,270,046,655_____64,036,270,047,131_______476
200,000,000,000,000______128,072,542,422,652____128,072,542,422,781______129
300,000,000,000,000______192,108,815,175,881____192,108,815,178,717______2836
400,000,000,000,000______256,145,088,132,145____256,145,088,130,891_____-1254
500,000,000,000,000______320,181,361,209,667____320,181,361,208,163_____-1504
600,000,000,000,000______384,217,634,373,721____384,217,634,374,108______387
700,000,000,000,000______448,253,907,613,837____448,253,907,607,119_____-6718
800,000,000,000,000______512,290,180,895,369____512,290,180,893,137_____-2232
900,000,000,000,000______576,326,454,221,727____576,326,454,222,404______677
1,000,000,000,000,000____640,362,727,589,917____640,362,727,587,828_____-2089

In all of mathematics, this is the only counting function for polygonal numbers of order greater than 2
or regular figurative numbers as they are otherwise known, that actually converges with the empirical data.

Note that I didn't even bother to calculate the relative error because it is now virtually non-existant and
quickly approaching zero. Also note that this incredible convergence between the counting function and
the empirical data is not possible without the fine structure constant which, in turn,
involves the prime number generating Blazys constant $A=2.5665438321713888444675291...$.

Quoting "turtle":

Without someone else actually counting how many polygonals in a range, your equations are meaningless.

Well, for one thing, people are counting polygonals. As a matter of fact, just last August,
the great mathematician Lars Blomberg counted them all the way to $10^{15}$.

That's the world record!!!

Moreover, I am getting e-mails all the time, asking me for "code", to which I invariably respond that
I am not a programmer or "coder", and that I don't even own a computer. A lot of them "crashed" their
computers, and a lot of them gave up because their computers are too slow, but to my knowledge,
there are still several ongoing attempts to break Lar's record. From this I can only assume that
counting order >2 polygonals is an interesting programming challenge in and of itself, and that
efforts to penetrate the secrets of this incredibly fascinating erratic sequence will continue.

It doesn't matter much to me because the counting function is now so accurate that any further adjustments
would be minor indeed. The form is definitely correct, and that, to me, is by far the most important thing.

The rest are just details.

I did my job, and delivered what I promised.

A super accurate counting function for polygonal numbers of order greater than 2.

Quoting "turtle",

If by if by "our polygonal number counting function" you mean to imply you had a hand in conceiving of taking
the natural density of polygonal/nonpolygonal numbers, then again that was my conception.
The first case is just a silly claim; the second i consider theft. stop taking credit for my work Don.

"Taking the natural density of polygonal numbers" occurs the moment someone puts them in one to one correspondence
with the natural numbers. In other words, when somebody "numbers them" as follows:

1____6
2____9
3____10
4____12
5____15
6____16
7____18
8____21
9____22
10___24

then it is easy to see at a glance that there are 3 order >2 polygonals less than or equal to 10.

I certainly don't want to "steal" any "credit" for that "achievment" which rightfully, does indeed belong to
both you and the ancient Greeks.

Don

### #116 Turtle

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Posted 08 May 2012 - 11:45 AM

[quote name='Don Blazys' timestamp='1336489436' post='317849']
Quoting "turtle":[quote name='Turtle' timestamp='1336323569' post='317797']
if by "our polygonal number counting function" you mean your equation(s), ...[/quote]

if your verbiage was clear, i wouldn't have challenged it.

[quotename='Don Blazys']...
In all of mathematics, this is the only counting function for polygonal numbers of order greater than 2
or regular figurative numbers as they are otherwise known, that actually converges with the empirical data.

Note that I didn't even bother to calculate the relative error because it is now virtually non-existant and
quickly approaching zero. Also note that this incredible convergence between the counting function and
the empirical data is not possible without the fine structure constant which, in turn,
involves the prime number generating Blazys constant $A=2.5665438321713888444675291...$. [/quote]

not quite Don.

[quote name='LBg' timestamp='1316082987' post='310481'] Investigation.pdf: ...
Conclusion
The values of  (x) calculated so far can fairly well approximated by:
f 0.640362740245
2
( ) 1 3 0 0 0  x  xf  xf 
Reasoning purely from the data available, f0 cannot be associated with "Fine
structure constant".
Of course it is possible to create expressions involving f0 together with e, pi,
proton to electron mass ratio, or any number of other dimensionless
constants to get a result slightly above 137.
But that would be just a play with numbers which does not prove anything. [/quote]

[quotename='don']Quoting "turtle":[quote name='Turtle' timestamp='1336323569' post='317797']
...
then again, without someone else actually counting how many polygonals in a range, your equations are meaningless.
[/quote]
Well, for one thing, people are counting polygonals. As a matter of fact, just last August,
the great mathematician Lars Blomberg counted them all the way to $10^{15}$. [/quote]

yes i know about Lars' calculation, as well as several more here and at least one elsewhere. again, they are doing it because I suggested it, nothwithstanding that you took my suggestion to others.

[quotename='don blazys']That's the world record!!!

Moreover, I am getting e-mails all the time, asking me for "code", to which I invariably respond that
I am not a programmer or "coder", and that I don't even own a computer. A lot of them "crashed" their
computers, and a lot of them gave up because their computers are too slow, but to my knowledge,
there are still several ongoing attempts to break Lar's record. From this I can only assume that
counting order >2 polygonals is an interesting programming challenge in and of itself, and that
efforts to penetrate the secrets of this incredibly fascinating erratic sequence will continue. [/quote]

yes it's an interesting programming challenge. so too is finding a new prime. for the programmers, that's all it is.

[quotename='don blazys']It doesn't matter much to me because the counting function is now so accurate that any further adjustments
would be minor indeed. The form is definitely correct, and that, to me, is by far the most important thing.

The rest are just details.

I did my job, and delivered what I promised.

A super accurate counting function for polygonal numbers of order greater than 2.

Quoting "turtle", [quote name='Turtle' timestamp='1336323569' post='317797']
... if by if by "our polygonal number counting function" you mean to imply you had a hand in conceiving of taking the natural density of polygonal/nonpolygonal numbers, then again that was my conception.

the first case is just a silly claim; the second i consider theft. stop taking credit for my work Don.
[/quote]

"Taking the natural density of polygonal numbers" occurs the moment someone puts them in one to one correspondence
with the natural numbers. In other words, when somebody "numbers them" as follows:

1____6
2____9
3____10
4____12
5____15
6____16
7____18
8____21
9____22
10___24

then it is easy to see at a glance that there are 3 order >2 polygonals less than or equal to 10.

I certainly don't want to "steal" any "credit" for that "achievment" which rightfully, does indeed belong to
both you and the ancient Greeks.

Don
[/quote]

find one reference that actually points out the natural density of polygonals/non-polygonals and that predates mine don. i dare say i have done more research on this than you and i have seen no such reference, whether ancient greek or contemporary geek. your cavalier discounting of such details is sloppy at least and deceptive and dishonest at worst. if you were to profit by your use of my work, i daresay it would then be illegal and a matter for our attorneys.

Edited by Turtle, 08 May 2012 - 01:31 PM.

### #117 LBg

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Posted 08 May 2012 - 12:45 PM

... the great mathematician Lars Blomberg ...

Don, you flatter me.
I am not a mathematician, neither great nor small,
but I enjoy creating efficient programming solutions to mathematical problems.

/LBg

### #118 pascal

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Posted 08 May 2012 - 01:57 PM

Gosh darn it I leave the room for two seconds!

I've been thinking about the interesting fact that 1/89 generates the sequence of Fib numbers distributed over negative powers of ten, at least when expressed in BASE ten. I'm wondering whether other sister series, such as the Lucas, can be similarly distributed, say, over negative powers of some other number in some other base (I've tried base 10 without success), and also still be a simple fraction in that system. Would this have any relation to the fact that the classical Pascal Triangle's rows give powers of 11 (though expressed in base 10)- the sister Pascal Triangles also give powers of 11 plus increments defined by the 'seeds' of their sides that begin sister Fib series.

Jess Tauber

Moderator note: 11 replies to this post have been moved to their own thread, "the 1/89 lemma", because they're they're about a different topic than Don's FSC formula.

Edited by CraigD, 20 May 2012 - 01:05 PM.

### #119 LBg

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Posted 09 May 2012 - 02:31 AM

Lars and Phillip, or any of you lurking math programming junkies, i invite, if not challenge, you all to unleash your enthusiasm & skills 2.0 by checking and then expanding my table to infinity and beyond.

See http://oeis.org/A001175:

A001175 Pisano periods (or Pisano numbers): period of Fibonacci numbers mod n.
1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, 16, 30, 48, 24, 100, 84, 72, 48, 14, 120, 30, 48, 40, 36, 80, 24, 76, 18, 56, 60, 40, 48, 88, 30, 120, 48, 32, 24, 112, 300, 72, 84, 108, 72, 20, 48, 72, 42, 58, 120, 60, 30, 48, 96, 140, 120, 136
for n=1,2,...