December 2010 - mathfun@mailman.xmission.com

Re: [math-fun] [seqfan] The Fibonomial Triangle
by Richard Guy 04 Dec '10

04 Dec '10

Maximilian, That's more like it! Even better, I now see, is A055870. I wonder if all the cross-references between all the sequences mentioned are mentioned? R. On Sat, 4 Dec 2010, Maximilian Hasler wrote: > some more (or less) information might be in: > http://oeis.org/A010048 : Triangle of Fibonomial coefficients. > > Maximilian > > On Sat, Dec 4, 2010 at 5:20 PM, Richard Guy <rkg(a)cpsc.ucalgary.ca> wrote: >> Dear all, >> How much of the following is known to those >> who well know it? I haven't yet been able to consult >> Knuth, vol.1, p.85, so it may be there. Or in Duke >> Math J 29(1962) page numbers need correcting in some? >> of [A000012, A000045, A007598], A056570--4, A056585--7. >> >> The characteristic polynomials for these sequences are >> >> x - 1 >> x^2 - x - 1 >> x^3 - 2x^2 - 2x + 1 >> x^4 - 3x^3 - 6x^2 + 3x + 1 >> x^5 - 5x^4 - 15x^3 + 15x^2 + 5x - 1 >> x^6 - 8x^5 - 40x^4 + 60x^3 + 40x^2 - 8x - 1 >> x^7 - 13x^6 - 104x^5 + 260x^4 + 260x^3 - 104x^2 - 13x + 1 >> x^8 -21 -273 +1092 +1820 -1092 -273 +21 +1 >> x^9 -55 -1870 +19635 +85085 -136136 -85086 +19635 +1870 -55 -1 >> x^10 -89 -4895 +83215 +582505 -1514513 -1514513 +582505 + - - + >> >> which (it is known by some) factor into >> x-1 >> x^2-x-1 >> (x+1)(x^2-3x+1) >> (x^2+x-1)(x^2-4x-1) >> (x-1)(x^2+3x+1)(x^2-7x+1) >> (x^2-x-1)(x^2+4x-1)(x^2-11x-1) >> (x+1)(x^2-3x+1)(x^2+7x+1)(x^2-18x+1) >> (x^2+x-1)(x^2-4x-1)(x^2+11x-1)(x^2-29x-1) >> (x-1)(x^2+3x+1)(x^2-7x+1)(x^2+18x+1)(x^2-47x+1) >> (x^2-x-1)(x^2+4x-1)(x^2-11x-1)(x^2+29x-1)(x^2-76x-1) >> (x+1)(x^2-3x+1)(x^2+7x+1)(x^2-18x+1)(x^2+47x+1)(x^2-123x+1) >> >> where the middle coefficients are, of course, Lucas numbers. >> >> If we make a triangle of the coeffs of the unfactored >> polynomials, we find that, apart from signs, they are >> >> e (F0/F1) >> >> e F1/F1 >> >> e F2/F1 F2F1/F1F2 >> >> e F3/F1 F3F2/F1F2 F3F2F1/F1F2F3 >> >> e F4/F1 F4F3/F1F2 F4F3F2/F1F2F3 F4F3F2F1/F1F2F3F4 >> >> e F5/F1 F5F4/F1F2 F5F4F3/F1F2F3 .... >> >> e F6/F1 F6F5/F1F2 F6F5F4/F1F2F3 F6F5F4F3/F1F2F3F4 '''' >> >> e F7/F1 F7F6/F1F2 F7F6F5/F1F2F3 F7F6F5F4/F1F2F3F4 .... >> >> where e = 1 is the empty product, and F7F6F5/F1F2F3, >> for example, is 13*8*5/1*1*2 = 260, a sort of `binomial coefficient' >> of Fibonacci numbers F1=F2=1, F3=2, F4=3, ... >> >> Of course, these sequences are all divisibility sequences. R. >> >> >> _______________________________________________ >> >> Seqfan Mailing list - http://list.seqfan.eu/ >>

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Re: [math-fun] superbinary numbers
by Robert Munafo 04 Dec '10

04 Dec '10

That's neat. It's very much like the Knuth -yllion naming system that uses a basis of 10, 10^2, 10^4, 10-^8, 10^16, etc. http://mrob.com/pub/math/largenum-2.html#yllion I wonder if you can devise a system that makes use of representations like "132" to shorten the typical length of representations. I naively suggest that 1 be left out whenever it leaves a descending pair of digits. You can also leave off the initial 1 if it is followed by another digit (which is always a 2 or more). Then you would count: 1, 2, 21, 3, 31, 32, 321, 23, 231, 232, 2321, 213, 2131, 2132, 21321, 4, 41, 42, ... On Sat, Dec 4, 2010 at 12:14, Steve Witham <sw(a)tiac.net> wrote: > Sometimes people describe 10^9 as a "thousand million" and use "billion" to > refer to 10^12. [...] > > Here is a minimal system starting with base two. Counting: > 1, 12, 121, 13, 131, 1312, 13121, 123, 1231, 12312, 123121, 1213, 12131, > 121312, 1213121, 14... > The number 123 (eight) is analogous to "one thousand million"-- the first > with two power numbers in a row. [...] -- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com

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[math-fun] superbinary numbers
by Steve Witham 04 Dec '10

04 Dec '10

Sometimes people describe 10^9 as a "thousand million" and use "billion" to refer to 10^12. In that way of speaking, you can have a number like "one thousand two billion three thousand four million five thousand six." I'm leaving out the interesting but inconsistent hundreds to make the job easier for the Martian in the next paragraph. A Martian might infer that our number system was based on 10^3, 10^6, 10^12... and possibly extrapolate 10^24, 10^48, etc. Strange system, but maybe we have a population of a million families, each with a million members, each with a thousand tentacles, each with a thousand suction cups, or something gloriously rational and expansive like that. Here is a minimal system starting with base two. Counting: 1, 12, 121, 13, 131, 1312, 13121, 123, The number 123 (eight) is analogous to "one thousand million"--the first with two power numbers in a row. 1231, 12312, 123121, 1213, 12131, 121312, 1213121, 14... So, 1=1, 12=2, 13=4, 14=16, 15=256, 16=65536, 17=2^32, and so forth. Notice that '0' hasn't shown up. Let's say that 0 means zero, just like in our own number word system, where "zero" only shows up by itself. 18 = the size of the address space of a 64-bit processor, still less than Avogadro's number. 19 = 2^128, about 3x10^38 123456789 = 2^255, about 6x10^76, more than the diameter or age of our universe in Planck units, but less than the volume (about 10^184 Planck volumes(?)), or a googol, of course. The holographic principle puts the number of bits in the universe at something like 4 x the area of an enclosing sphere in Planck areas, about 4x10^58(??), 2^195, or 181219. Who needs a number greater than 123456789 then, since, as every Martian schoolchild knows, it could only be used in a question if asked the answer to which could not be contained in the Whole. Martians might speculate that, due to some evolutionary accident, humans started with base five, then progressed towards enlightenment by multiplying by two, then again by squaring to get 100. It's hard to guess why we took the detour to 10^3, perhaps it had something to do with the nearness of 2^10. But then we got back on track with a squaring trend. Lately we've been mentioning "memories" in sizes 16, 17 and 18, which must be gratifying for Martians to hear. (Martian adolescents have a rude rhyme: 1213121412131215121312141213121612131214121312151213121412131217 1213121412131215121312141213121612131214121312151213121412131218 1213121412131215121312141213121612131214121312151213121412131217 1213121412131215121312141213121612131214121312151213121412131219 1213121412131215121312141213121612131214121312151213121412131217 1213121412131215121312141213121612131214121312151213121412131218 1213121412131215121312141213121612131214121312151213121412131217 121312141213121512131214121312161213121412131215121312141213121 --!!) --Steve And remember, "Zero means zero."

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Re: [math-fun] Guess the mystery plot
by mwolf＠ift.uni.wroc.pl 04 Dec '10

04 Dec '10

It can be a Fourier transform with peaks corresponding to prominent harmonics. Am I right? Marek

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[math-fun] A step closer to neutrino communications
by Henry Baker 04 Dec '10

04 Dec '10

I was listening to a podcast from the Economist, and I learned that physicists are now using the ocean to look for neutrinos. Previously, they built very expensive large volumes of very clear water & looked for flashes. Now, they are looking only for certain types of sounds that the flashes would have caused. All that is necessary is to filter out every other sound, including whale music. Since sound propagates extremely well in water, the scientists can now utilize entire oceans to look for their neutrinos, dramatically enhancing their ability to detect them. Soooo, set up a neutrino source on one side of the Earth that can be modulated, & utilize the ocean to listen in from the other side of Earth. Since the neutrinos travel at ~c, while the speed of light in fiber optics is currently ~2/3c, and the fiber optics must also follow the Earth's curvature, one could conceivably beat the speed of fiber optics (not on speed, but latency). Of course, if the signal has to travel very far in the water before being detected, this would definitely slow things down, so it would be better to have lots of microphones connected via the best fiber optic cable. The number of bits per second doesn't have to be very high to be valuable -- e.g., talking to submerged submarines or breaking news for stock markets. Of course, speeds in fiber optics can also be improved by using vacuum (or even air) cores instead of glass to get to ~c, but they still can't avoid following the Earth's surface.

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[math-fun] Guess the mystery plot
by Jason 03 Dec '10

03 Dec '10

Since Mike blew us all away last time, I'm leaving this plot unlabeled to start with, and I'll reveal clues over time until sometbody gets it: http://lh3.ggpht.com/_oXIW_jM0QDA/TPgYrMHiH3I/AAAAAAAAW2k/1u8HphQg8wM/myste…

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[math-fun] pendulum harmonics
by Bill Gosper 02 Dec '10

02 Dec '10

(Retreating to the relative safety of mathematics.) In the special case of "half overswing", i.e., horizontal extrema, the Fourier series comes out particularly simple (though not particularly simply): theta[t]== 2*JacobiAmplitude[(Sqrt[g/L]*t)/Sqrt[2], 2] == 4*Sum[(Sech[(1/2)*(-1 + 2*k)*Pi]* Sin[((-1 + 2*k)*Sqrt[g/L]*t*Gamma[-(1/4)]^2)/(16*Sqrt[Pi])])/(-1 + 2*k), {k, Infinity}] g Sqrt[-] t L 2 JacobiAmplitude[---------, 2] == Sqrt[2] g 1 2 (-1 + 2 k) Sqrt[-] t Gamma[-(-)] 1 L 4 Sech[- (-1 + 2 k) Pi] Sin[---------------------------------] 2 16 Sqrt[Pi] 4 Sum[------------------------------------------------------------, -1 + 2 k {k, Infinity}] Does anyone recognize this identity? The Fourier series for general swing amplitude looks a bit tougher. --rwg Incredibly, Plouffe's ISC misses 4*sech(pi/2), even with smart lookup. (The actual Fourier integral scaled this by some Gammas. Rotsa ruck.)

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Re: [math-fun] correct name for 4-D donut?
by Dan Asimov 02 Dec '10

02 Dec '10

The simplest example of this induction step is showing that T^2 embeds in R^3, assuming that T^1 embeds in R^2. Let T := T^1 be embedded in R^2 as the circle of radius 2 about (0,0). At time s, for -1 <= s <= 1, let P_s := { p in R^2 | dist(p, T) = sqrt(1-s^2) } P_s is one circle for s = +-1. Otherwise P_s is the union of two circles, of radii 2 +- sqrt(1-s^2). Now consider the subset X of R^2 x R^1 = R^3 defined by X := {(p,s) in R^2 x R^1 | -1 <= s <= 1 and p in P_s}. It's easy to visualize exactly what X looks like since it's in R^3. Clearly, X is a 2-torus embedded in R^3. That's because what we have here is a T^2 x C, where C is the circle obtained by identifying two intervals by glueing their left endpoints together, and also glueing their right endpoints together. (As a movie, C is just a single point that immediately separates into two points that reach their maximum separation, then retrace their steps by approaching each other until they coincide.) Noting that a T^1 is just (S^1)^1 = S^1: But one may also imagine the new dimension as time, in which case X is a movie of a varying point-set in R^2. This set at time s is just X_s := P_s x {s}. For s = -1 this is one T^1, which immediately separates into two T^1's that move away from each other, reach a maximum separation at s = 0, and then the two T^1's approach each other until coinciding at s = 1. The last paragraph generalizes to showing that if T^n embeds in R^(n+1), then T^(n+1) embeds in T^(n+2). --Dan Robert Munafo wrote: << I'm not sure I get it. I tried to use that description to show that the familiar torus T^2 can exist in R^3, but the resulting surface intersects itself. Placing the initial T^1 (which is a circle) on the X-Y plane, and treating the Z axis as the "time" dimension, and using the X axis as the direction in which the two copies get pushed while time is moving forward, I end up with the surface intersecting itself at various points in the Y-Z plane. I could use a 4th dimension to prevent this self-intersection, but then it's a bit hard to explain how you can twist half of the result around to embed the surface into R^3 in a non-self-intersecting way, or even to explain whether or not it is a Klein bottle. On Wed, Dec 1, 2010 at 18:40, Dan Asimov <dasimov(a)earthlink.net> wrote: << This is a less elegant but maybe more transparent way to see that T^n embeds in R^(n+1): ---------- The induction step is typified by showing T^3 embeds in R^4, assuming T^2 embeds in R^3. Start from a T^2 = T_0 embedded in R^3. Letting the next dimension be "time", we make a short movie that starts with the T^2 embedded in R^3, which immediately separates into two T^2's that move apart to a maximum separation. (Just push T_0 off itself in R^3 in both directions.) This is the first half of the film; the rest of it is just the first half in reverse time order. ---------- The result can be thought of as two copies of T^2 x [0,1], with both copies of T^2 x {0} identified with each other, and also both copies of T^2 x {1} identified with each other. This is easily seen to be T^2 x S^1 topologically. >> >> Those who sleep faster get more rest.

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Re: [math-fun] correct name for 4-D donut?
by Dan Asimov 02 Dec '10

02 Dec '10

This is a less elegant but maybe more transparent way to see that T^n embeds in R^(n+1): ---------- The induction step is typified by showing T^3 embeds in R^4, assuming T^2 embeds in R^3. Start from a T^2 = T_0 embedded in R^3. Letting the next dimension be "time", we make a short movie that starts with the T^2 embedded in R^3, which immediately separates into two T^2's that move apart to a maximum separation. (Just push T_0 off itself in R^3 in both directions.) This is the first half of the film; the rest of it is just the first half in reverse time order. The result can be thought of as two copies of T^2 x [0,1], with both copies of T^2 x {0} identified with each other, and also both copies of T^2 x {1} identified with each other. This is easily seen to be T^2 x S^1 topologically. ---------- As has been mentioned, (S^1)^3 with its natural embedding in (R^2)^3 = R^6 has the natural metric of R^3/Z^3 (at least up to uniform scaling). So in general each "cubical n-torus" Q^n := R^n/Z^n has a natural flat metric with a lot of symmetry. QUESTION: What is the lowest dimension F(n) of a Euclidean space in which this Q^n = R^n/Z^n can be embedded isometrically? Clearly n+1 <= F(n) <= 2n, but what more can be said about F(n) ??? (Obviously F(1) = 2 and F(2) = 4, but what about for n >= 3 ? In particular, is F(n) == 2n for all n ?) --Dan Robert Munafo wrote: << Hey Bill, that makes a lot of sense. If I read you right, you're saying that just as we can expand a circle into a set of epsilon-sized circles which together make up a 2-torus (the surface of a donut whose "donut hole" is almost as big as the original circle) we can do the same thing in 4 dimensions to get a 3-torus surface, and so on. I still like to embed the 2-torus in 4 dimensions, because then nothing gets stretched. Probably just the way my mind works. Bill Thurston wrote: << The standard name for an n-cube with opposite sides identified via translation = (S^1)^n is the n-torus, T^n. It's also equivalent to the quotient of the additive group of vectors in R^n by the integer lattice subgroup, T^n = R^n/Z^n. In general, the n-torus can be embedded in R^{n+1}. For the circle, this is immediate. Everything else can be done by induction: if the n-torus is smoothly embedded in R^{n+1}, then the boundary of an epsilon neighborhood in R^{n+2} gives a smooth embedding of T^{n+1} in R^{n+2}. >> >> Those who sleep faster get more rest.

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[math-fun] Integers shaped like right triangles
by Eric Angelini 01 Dec '10

01 Dec '10

Hello Math-Fun, Write an integer somewhere on a grid, one digit per square: +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | 1 | 3 | 4 | 0 | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ The digits will now move vertically: - odd digit j will climb j squares - even digit k will sink k squares - zero stays were it is In our example, the digits of 1340 move like this: +---+---+---+---+ | | 3 | | | +---+---+---+---+ | | | | | +---+---+---+---+ | 1 | | | | +---+---+---+---+ | . | . | . | 0 | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | | | 4 | | +---+---+---+---+ Encircle now those new digits with an elastic band. The resulting shape is not a right triangle - which is a pity because we are precisely looking for integers moving into right triangles! The additional constraint being that ALL the new digits must be ON the elastic band. [you can see this more clearly there: http://www.cetteadressecomportecinquantesignes.com/RightTriangleNumbers.htm ] Here are a few good examples: Integer 130 (the right angle is on 1): +---+---+---+---+ | | 3 | | | +---+---+---+---+ | | | | | +---+---+---+---+ | 1 | | | | +---+---+---+---+ | . | . | 0 | | +---+---+---+---+ Integer 313333 (right angle on 1): +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | 3 | | 3 | 3 | 3 | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | 1 | | | | | +---+---+---+---+---+---+ | . | . | . | . | . | | +---+---+---+---+---+---+ Integer 2201022 (right angle on 1): +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | | | 1 | | | | +---+---+---+---+---+---+---+ | . | . | 0 | . | 0 | . | . | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | 2 | 2 | | | | 2 | 2 | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ Here is a bad example: Integer 21000 (yes we have a triangle, yes there is a right angle --on 1--, but the band doesn't pass through ALL digits): +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | | 1 | | | | | +---+---+---+---+---+---+---+ | | . | . | 0 | 0 | 0 | | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ | | 2 | | | | | | +---+---+---+---+---+---+---+ | | | | | | | | +---+---+---+---+---+---+---+ Questions: a) Find the 40 first integers producing a right triangle b) Find the biggest integer producing a right triangle c) Find the smallest integer producing a rectangle. [Answers here in December 2010] Best, É.

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