Turtle 389,601 Posted May 9, 2012 Report Share Posted May 9, 2012 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, 136for n=1,2,...corresponding to your base=2,3,... There is also a table of values for n = 1..10000. See also: http://math.ca/crux/v23/n4/page224-241.pdf So no programming is needed in this case :) .Unless, of course, you want values beyond infinity... nice! :thumbs_up my base 10/mod9 mystery is solved at 60 and my work checked. no off-by-ones. :lol: so jess, can you squeeze something out of Pisano periods for your structures? can you work with the fuller graphic? will sally and mike travel in june? :) Quote Link to post Share on other sites

tetrahedron 16 Posted May 11, 2012 Report Share Posted May 11, 2012 Dunno anything about the stuff you've written as yet. Gotta empty my old head a little first. But, here's an interesting problem. You may remember my discovery (probably rediscovery) that when one stacks the equations describing the powers of the Metal Means one atop the other, the individual terms have powers and numerical coefficients that come directly out of the (2,1)-sided Pascal Triangle sister. I was discussing today with the fellow who just got his doctorate in Linguistics working on Phi-related behaviors in syntax, who is looking deeper into the issues involved. He now believes that a slew of Pisot-Vijayaraghavan numbers might describe some syntactic branching phenomena. My Metal Means are just those PV numbers where we have X^2-NX-1, where N is the N as in 'the Metal Mean of N' when one has the equation (N+sqrt(N^2+4))/2. What I want to know is this- if the (2,1)Pascal describes Metal Mean powers, might the OTHER Pascal sisters similarly describe other PV powers??? Jess Tauber Quote Link to post Share on other sites

Turtle 389,601 Posted May 11, 2012 Report Share Posted May 11, 2012 Dunno anything about the stuff you've written as yet. Gotta empty my old head a little first. But, here's an interesting problem. acknowledged. You may remember my discovery (probably rediscovery) that when one stacks the equations describing the powers of the Metal Means one atop the other, the individual terms have powers and numerical coefficients that come directly out of the (2,1)-sided Pascal Triangle sister. i haven't been able to "see", grok, get-a-grip-on or otherwise internalize exactly what a "Pascal Triangle sister" is or Metal Means are. here's where some drawings would find great utility. but otheriwse, are you just seeing/describing binomial coefficients? I was discussing today with the fellow who just got his doctorate in Linguistics working on Phi-related behaviors in syntax, who is looking deeper into the issues involved. He now believes that a slew of Pisot-Vijayaraghavan numbers might describe some syntactic branching phenomena. will have to find some info on those numbers as i'm unacquainted. :doh: do you have a list handy? but fear not, i just thought of a 3-dimension stacking phi-related behavior that i explored and craig added some computer calculations and observations on area & volume you will find interesting. it's short, but schweet. Fibonacci bricks My Metal Means are just those PV numbers where we have X^2-NX-1, where N is the N as in 'the Metal Mean of N' when one has the equation (N+sqrt(N^2+4))/2. What I want to know is this- if the (2,1)Pascal describes Metal Mean powers, might the OTHER Pascal sisters similarly describe other PV powers??? Jess Tauber PV? i've apparently missed the definition of that acronym. Quote Link to post Share on other sites

tetrahedron 16 Posted May 11, 2012 Report Share Posted May 11, 2012 Methinks that if you look these things up on Wiki (except for the sister Pascal stuff, which isn't there) most will be made clearer to you PV=Pisot-Vijayaraghavan numbers. Pretty famous. Don't fret- until a year or two ago I didn't know any math beyond advanced placement calculus from high school. Jess Tauber Quote Link to post Share on other sites

Turtle 389,601 Posted May 11, 2012 Report Share Posted May 11, 2012 Methinks that if you look these things up on Wiki (except for the sister Pascal stuff, which isn't there) most will be made clearer to you PV=Pisot-Vijayaraghavan numbers. Pretty famous. Don't fret- until a year or two ago I didn't know any math beyond advanced placement calculus from high school. Jess Tauber so too apparently is pisano "pretty famous", and previously unknown to either of us. :shrug: anyway, , seeing as all integers are "PV" it doesn't change anything regarding the modular arithmetic or the sets of polygonal and non-polygonal numbers as i have presented them. just as phi shows up in nature a great many ways, we peoples have devised a great many ways to represent it. some propositions, even if true, simply cannot be proved. don't fret on your short time in the field; there's plenty of time to learn to draw. ;) Quote Link to post Share on other sites

tetrahedron 16 Posted May 18, 2012 Report Share Posted May 18, 2012 Well, in the past days I've been exploring other connections between these numbers and Pascal Triangle sisters. So far I've found repeated motifs in the deep diagonals. For example, in analogues of the tetrahedral diagonal in higher Pascal systems, the values in the diagonals are themselves composed of original tetrahedrals in combination. One has summed contiguous pairs of classic tetrahedral numbers as its tetrahedral analogue values, making it act like Fibonacci numbers. The next higher analogue has a classical tetrahedral number added to TWICE the next contiguous tet number, making it analogous to Pell numbers. And so on. Thus while the actual Fiblike series generated by these Pascal sister triangles are all related only to the Golden Ratio, their deep diagonals relate more to the entire series of Metal Means. Even the classical Pascal Triangle, since its tetrahedral numbers sum with zero times the next contiguous one, this relates to the first Metal Mean, which is 1.00000.... All the Pascal sisters I've mentioned above have sides defined by numbers that are separated by one: (1,0),(2,1),(3,2) and so on. But if one were to use other differences between side values, would that mean that the internal diagonal values be related more to the PV numbers? The Golden Ratio is a Metal Mean, and the Metal Means are PV numbers, but in each case we see a broadening of classification, with reduction of shared prototypical characteristics with each larger class. I've found many relations in the periodic system to the Golden Ratio, a couple it seems to Metal Means. If there are higher level PV-related periodic behaviors, that would be very interesting indeed. Why should Nature limit herself to one-size-fits-all motivations? Jess Tauber[email protected] Quote Link to post Share on other sites

tetrahedron 16 Posted June 5, 2012 Report Share Posted June 5, 2012 (edited) Here's another weird coincidence- first the preamble... Years ago, while modeling subatomic particles, I hit on the idea of organizing charges of the fermions of the Standard Model as vertices of a cube. That is, we have matter and antimatter, so -3/3, +2/3, -1/3, +0/3 : +3/3, -2/3, +1/3, -0/3, with the sequences running from the analogues of the electron, upper quarks, lower quarks, and neutrinos respectively. You can then distribute these about the vertices of a cube. But how to get the coordinates to jibe with the charge values? I eventually came up with the idea of orienting the cube against a reference plane. First you draw a line through the center of a cube containing opposite vertices of the cube, still unmarked. Now you embed the line in a plane which can rotate about the line, or the cube against the plane. Two of the cube vertices are always in the plane, because the line embedded in the plane always runs through these vertices. I wondered whether one can drop projections, from the cube's vertices, onto the plane, and normal/perpendicular TO it, such that the relative distances to the plane along the projection segments fell in the ratios of 3,2,1,0, as seen as thirds of a charge above. And because of half the cube being above and below the plane, this would account for both the plus and minus values (mixed) in matter and antimatter. It turned out that a bunch of different angles would do the trick, symmetrically related to each other. The easiest to remember is gotten by taking ARCTAN(sqrt27)= 79.10660535...degrees. If one vertex is at this value rotated relative to the plane below, AND another vertex is embedded within the plane, then the relative lengths of the projection segments will be in the proper proportions. Well, I've been looking at the FSC again today, and just out of curiosity, trying this and that, I wondered what would happen if one took the ratio of this angle to the FSC: i.e. @137.0359991/79.10660535... The ratio is 1.73229528, which is strangely close to the square root of 3, 1.732050808. And the difference, from my admittedly inadequate handheld calculator, is .00024447244. I don't have a clue if this is significant- do the values in the difference string repeat (..2444....244)? Or is this just because of the limitations of the calculator? I'll have to see if other interesting things fall out of the other angles of the cube here. Jess Tauber Edited June 5, 2012 by pascal Quote Link to post Share on other sites

Don Blazys 26,415 Posted June 5, 2012 Author Report Share Posted June 5, 2012 (edited) That's very strange indeed. I too noticed that numbers very close to [math]\sqrt{3}[/math] do seem to occur more frequently than expected in equations involving the fine structure constant. Don Edited June 5, 2012 by Don Blazys Quote Link to post Share on other sites

tetrahedron 16 Posted June 5, 2012 Report Share Posted June 5, 2012 Also I didn't notice til now- remember that 27, whose square root relates to the angle in question, is 3^3. Is this significant? In my projection segment system, lengths from vertices having the values 3/2/1/0, 79... degrees is the highest angle, displaced the most from the plane- so its easy to figure out the others trigonometrically (less easy to work out the fractions behind them, though I did it, a while back, but one can always re-derive them). I've never actually worked out whether the in-plane (cosine) lengths have some relation to the properties of the particles in question. Maybe I can try that. What would be REALLY neat would be if these equivalent angular schemes (for ex. mirror 180-79..., etc.) had something to do with the different fermion generations- the sine values already relate to charges. Hmmmm... Jess Tauber Quote Link to post Share on other sites

Turtle 389,601 Posted June 6, 2012 Report Share Posted June 6, 2012 Here's another weird coincidence- first the preamble... snip... I wondered whether one can drop projections, from the cube's vertices, onto the plane, and normal/perpendicular TO it, such that the relative distances to the plane along the projection segments fell in the ratios of 3,2,1,0, as seen as thirds of a charge above. And because of half the cube being above and below the plane, this would account for both the plus and minus values (mixed) in matter and antimatter. It turned out that a bunch of different angles would do the trick, symmetrically related to each other. The easiest to remember is gotten by taking ARCTAN(sqrt27)= 79.10660535...degrees. If one vertex is at this value rotated relative to the plane below, AND another vertex is embedded within the plane, then the relative lengths of the projection segments will be in the proper proportions. Well, I've been looking at the FSC again today, and just out of curiosity, trying this and that, I wondered what would happen if one took the ratio of this angle to the FSC: i.e. @137.0359991/79.10660535... The ratio is 1.73229528, which is strangely close to the square root of 3, 1.732050808. And the difference, from my admittedly inadequate handheld calculator, is .00024447244. I don't have a clue if this is significant- do the values in the difference string repeat (..2444....244)? Or is this just because of the limitations of the calculator? I'll have to see if other interesting things fall out of the other angles of the cube here. Jess Tauber I may be beating a dead horse, but how about a drawing Jess? Did you make any drawings for yourself, or do you just see all this in your head? Quote Link to post Share on other sites

tetrahedron 16 Posted June 6, 2012 Report Share Posted June 6, 2012 (edited) Yeah, me and drawings, right. Yes, I did do some on paper. But really? Given that the first axis defined is through to vertices and the cube center, you can rationalize the six vertices as two sets of three each at the vertices of two equilateral triangles that also can rotate about the axis at their centers. That's what it boils down to. And the reference plane, in this reduction, is just a line going through each of the triangular centers. You rotate the triangles against this line, or vice versa, until the projections from the vertices normal to the line fall into relative proportions of 3:2:1. The fourth vertex, with projection of zero height, of each triangular set is on the plane itself, but not in the plane of the triangle. In fact, all four together form a tetrahedron- and as you know two co-centered tetrahedra may occupy the same cube's vertices. When I first looked into this I had no idea if it would work- why should the three vertices coordinate their projections at all to produce such a sequence as exactly 3:2:1? But it did. And you just add 120 and 240 degrees to ARCTAN(sqrt27) to get the other angles for this particular orientation. I'm sure you can figure out the others. Jess Tauber Edited June 6, 2012 by pascal Quote Link to post Share on other sites

Turtle 389,601 Posted June 6, 2012 Report Share Posted June 6, 2012 Yeah, me and drawings, right. Yes, I did do some on paper. But really? Given that the first axis defined is through to vertices and the cube center, you can rationalize the six vertices as two sets of three each at the vertices of two equilateral triangles that also can rotate about the axis at their centers. That's what it boils down to. And the reference plane, in this reduction, is just a line going through each of the triangular centers. You rotate the triangles against this line, or vice versa, until the projections from the vertices normal to the line fall into relative proportions of 3:2:1. The fourth vertex, with projection of zero height, of each triangular set is on the plane itself, but not in the plane of the triangle. In fact, all four together form a tetrahedron- and as you know two co-centered tetrahedra may occupy the same cube's vertices. When I first looked into this I had no idea if it would work- why should the three vertices coordinate their projections at all to produce such a sequence as exactly 3:2:1? But it did. And you just add 120 and 240 degrees to ARCTAN(sqrt27) to get the other angles for this particular orientation. I'm sure you can figure out the others. Jess Tauber yes really. what a cop out. good grief. :doh: Quote Link to post Share on other sites

tetrahedron 16 Posted June 6, 2012 Report Share Posted June 6, 2012 (edited) As Ripley said, acidly, to the Alien- there is no need for all this bad blood.... BTW, am writing off to specialist astrophysicists to see if any of them have any ideas about whether there are physical phenomena in the universe (like Hubble expansion/dark energy etc.) that might contribute to a speculative mismatch between the known low-energy scale value of the FSC and exactly 1/137, which as you may remember, seems to have very special qualities. Jess Tauber Edited June 6, 2012 by pascal Quote Link to post Share on other sites

CraigD 398,736 Posted June 6, 2012 Report Share Posted June 6, 2012 The easiest to remember is gotten by taking ARCTAN(sqrt27)= 79.10660535...degrees. If one ...Well, I've been looking at the FSC again today, and just out of curiosity, trying this and that, I wondered what would happen if one took the ratio of this angle to the FSC: i.e. @137.0359991/79.10660535...It’s important to be mindful that the degree unit – 1° = 1/360 full rotation – is an arbitrary convention owing its existence to a long string of human-factor decisions and happenchances. For example, had ca. 1800 European history played out different than it did, our favorite unit might be using gradians (AKA gons) – 1^{g} = 1/400 rotation, as well as having 10 hours days with each hour having 100 minutes and each minute 100 seconds. Personally, I’d prefer a nice, easy-to calculate 100,000 sec day vs. our present 86,400 one, but pity the denizen of an alternate history who must say “thirty-three-point-three continuing, sixty-six-point-six continuing, one” rather than our catchy “thirty, sixty, ninety” when referring to this common triangle. The point, is to not imagine special significance to the units we humans agreed, over the course of our long history and pre-history, to use. For angles, most math folk hold that the most natural unit is the radian –1/2[imath]\pi[/imath] in a full rotation, and usually don’t even consider at unit, but rather a sort of geometric convention for describing angles, so don’t denote it with any symbol or text [math]\tan^{-1}\left(\sqrt{27}\right) = 1.3806707234484298614903052164447686981316235229730262 \dots [/math] You can see lots more digits using a high-precision calculator, such as wolframalpha’s here. The ratio is 1.73229528, which is strangely close to the square root of 3, 1.732050808. And the difference, from my admittedly inadequate handheld calculator, is .00024447244. I don't have a clue if this is significant- do the values in the difference string repeat (..2444....244)? Or is this just because of the limitations of the calculator?It’s an artifact of your calculator, I believe. Using the higher-than-typical precision approximate value above and wolframalpha gives: [math]\frac{137.035999074}{1.3806707234484298614903052164447686981316235229730262 \frac{180}{\pi}} -\sqrt{3} = [/math] 0.0002444725305672958037591580738234641888349139491370… (for more digits) However, we don’t need nice calculators to know that a value involving division by an irrational number (ei: [imath]\tan^{-1}\left(\sqrt{27}\right)[/imath] or [imath]\sqrt{3}[/imath]) can’t be represented exactly as a repeating decimal, because any number exactly representable as a repeating decimal is rational, which contradicts that these value are irrational. Yeah, me and drawings, right. Yes, I did do some on paper. But really? Given that the first axis defined is through to vertices and the cube center, you can rationalize the six vertices as two sets of three each at the vertices of two equilateral triangles that also can rotate about the axis at their centers. That's what it boils down to.yes really. what a cop out. good grief. :doh:Really, Jess, pictures are nearly indispensable for communicating about geometry. As the old saying goes, each one really is worth a thousand words. Except for a people like some professional artists and designers, who have nice computer graphics tablets and the like, most people find pen/pencil and paper the best medium for drawing. Ten years ago, not many netzens could easily transfer an image from paper to a website. These days, thanks to ubiquitous digital cameras (most of them in mobile phones), most folk can, with a bit of self-training, pretty quickly snap a picture of a paper sketch (or almost anything else) and upload it to a forum like ours. I, and I’m sure Turtle (who’s a master of all things photographic, digital and analog) and others, would be glad to help you with any technical hurtles to seeing your graphical handiwork in these webpages. :) When it comes to simple geometry, even better than a picture can be a simple list of point (vertex) coordinates and lines. For example, I can quickly describe a unit cube like this:Points: 1:(0,0,0) 2:(1,0,0) 3:(0,1,0) 4:(0,1,1) 5:(1,0,0) 6:(1,0,0) 7:(1,1,0) 8:(1,1,1)Lines: 1:(1,2) 2:(1,3) 3:(1,5) 4:(2,4) 5:(2,6) 6:(3,4) 7:(3,7) 8:(4,8) 9:(5,6) 10:(5,7) 11:(6,8) 12:(7,8) Can you describe the figures you’ve been hinting at with words using this notation? Quote Link to post Share on other sites

tetrahedron 16 Posted June 6, 2012 Report Share Posted June 6, 2012 Points taken, so unless I want to join the rest of my Luddite dinosaur brethren in extinction, I guess I'd better adapt. By the way, if you folks remember (in the other discussion?) my stacking of generalized Fibonacci sequences over each other to generate the quantum number and periodic table constructional data, I'm starting to think the SAME stack of sequences may also define the relative amounts of Dark Energy, Dark matter, etc. in the universe. 73 and 23 are both terms in two contiguous sequence, separated by two spaces in line. Trying to make baryonic matter work, less luck- the next number (in the Lucas sequence) is 7, and we also get 1 from the classic Fib sequence above it. Jess Tauber Quote Link to post Share on other sites

Turtle 389,601 Posted June 6, 2012 Report Share Posted June 6, 2012 Points taken, so unless I want to join the rest of my Luddite dinosaur brethren in extinction, I guess I'd better adapt. By the way, if you folks remember (in the other discussion?) my stacking of generalized Fibonacci sequences over each other to generate the quantum number and periodic table constructional data, I'm starting to think the SAME stack of sequences may also define the relative amounts of Dark Energy, Dark matter, etc. in the universe. 73 and 23 are both terms in two contiguous sequence, separated by two spaces in line. Trying to make baryonic matter work, less luck- the next number (in the Lucas sequence) is 7, and we also get 1 from the classic Fib sequence above it. Jess Tauber you never miss a chance to denegrate orthodoxy, and yet without it what would you have to build on? seems fitting you've hitched your wagon to don's writings i suppose. i just don't get your reluctance to give any illustrations of your concepts. also, you keep saying the lucas "sequence" when from what i read there are many lucas sequences. perhaps you just mean lucas numbers or possibly the positive lucas numbers. maybe you could list the numbers you mean at least once? snarky little comments such as "i'm sure you can figure it out" just don't encourage or help anyone to understand what you are on about. Quote Link to post Share on other sites

tetrahedron 16 Posted June 7, 2012 Report Share Posted June 7, 2012 Here is the set of generalized Fib number sequences stacked one atop the other that I used previously with quantum number and periodic system constructional parameters: -1.....1.....0.....1.....1.....2.....3.....5.....8....13/0.....1.....1.....2.....3.....5.....8....13....21....34+1.....1.....2.....3.....5.....8....13....21....34....55+2.....1.....3.....4.....7....11....18....29....47....76+3.....1.....4.....5.....9....14....23....37....60....97+4.....1.....5.....6....11....17....28....45....73...118 Note differences between any two contiguous values within any column is fixed and Fib, and the differences from left to right grow as Fib.The fourth row is what I mean by Lucas sequence. Notice that in row 6 (bottom) we have value 73 in the 9th column, so marking the percentage of dark energy.In row 5, two moves back in column 7, we have 23, marking percentage of dark matter.For row 4, again two moves before that, in column 5, we have value 7, and in row 3, in column 3, we have value 2.Finally, in row 2, we get value 0 in column 1. Whether we need row 1, with column -1, value -3 (sequence (Fib with alternating +_/-)...3,-2,1,-1,0,-1,-1..(-Fib)) I don't know. If we can ignore row 1, then we end up with 2+7=9 covering percentage of 'everything else'. In the pie-chart of relative contributions to the universe baryonic matter is supposed to be around 4.6 percent (bit strange as the total becomes more than 100%). In any case 9% is too high. UNLESS one considers the missing antimatter. It's assumed that all of it is gone, and that we had a slight overabundance of normal matter as we see today. But what if that isn't true? The universe is a pretty big place. If there was merely a slight non-overlap between matter and antimatter after Inflation, etc., then far away out there beyond our horizon may be an antimatter half of the universe. One would presume they would have the same relative proportions of the differing constituents otherwise. 9/2=4.5, pretty close to the 4.6 percent of baryonic matter. How to distribute between 7 and 2 (or half of each) is another 'matter'. :-) Jess Tauber Turtle 1 Quote Link to post Share on other sites

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