January 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Squares and factorials
by Dan Asimov 09 Jan '11

09 Jan '11

I don't know -- very interesting question. But I wanted to chime in that 24 is my choice for most interesting integer, too. Partly because, among all the multiplicative groups of rings G_n := (Z/nZ)*, 24 is the largest n for which k | n implies that all elements of G_k are squares. Also because of its appearance in Dedekind eta. (Are the two things directly related?) --Dan << There are at least three solutions to the Diophantine equation: m² = n! + 1 namely (m,n) = (5,4), (11,5), and (71,7). As far as I know, no further pairs have been discovered. Is there a proof that these three solutions are the only ones in existence, or could there be others? 71 seems to exhibit other interesting properties, being the largest prime factor of the order of the Monster Group, and the largest supersingular prime. But 24 definitely wins as the most interesting integer... >> Those who sleep faster get more rest.

2 1

[math-fun] Squares and factorials
by Adam P. Goucher 09 Jan '11

09 Jan '11

There are at least three solutions to the Diophantine equation: m² = n! + 1 namely (m,n) = (5,4), (11,5), and (71,7). As far as I know, no further pairs have been discovered. Is there a proof that these three solutions are the only ones in existence, or could there be others? 71 seems to exhibit other interesting properties, being the largest prime factor of the order of the Monster Group, and the largest supersingular prime. But 24 definitely wins as the most interesting integer... Sincerely, Adam P. Goucher

2 1

Re: [math-fun] Exact rational 3D spacefill
by Bill Gosper 09 Jan '11

09 Jan '11

Jason> > I don't recall if someone already posted this sculpture of trilbert: > http://blog.makezine.com/archive/2010/11/math_monday_3d_hilbert_curve_in_st… > An 8^3 (not a 4^3, Steve) in plastic would be delicate, at best. The entire weight is everywhere supported by the thickness of a single pipe, often far offcenter. Neil Bickford built a 4^3 from construction cubes that you dasn't breathe on. And the illustrated 8^3 is a quivering mass of robot intestines. (Amazingly organic, for such a seemingly crunchy and artificial concept.) Neil was investigating the connection of Trilberts with Morse (= Stutter Free = "Square" Free) sequences, and found the spacefilling property enormously underconstraining. There seem to be 2^(n+1)-1 cubes of order n, corresponding to all the ways you can reflect a subcube in the symmetry plane that contains the edge it spans. In fact, Corey's first exact map had the inverse image of (1,1,1) at 85/126 instead of 19/28. I probably missed something, but am having difficulty finding a stutter-free (no straight angles) 8^4. It seems to require a lot of searching and manual reflecting instead of a nice, clean recursion. --rwg That's no hamster--it's an illegal gerbil! Get him!

2 2

[math-fun] Newton's cradle
by Bill Gosper 08 Jan '11

08 Jan '11

> I tried to do this by thinking of the balls as springs with a very big > constant instead of as rigid bodies. Then the question of space > between them doesn't seem so important any more, and I start to think > that the behavior of the system will depend more on the structure of > the big ball (how does it compress compared with the small balls). > > I still don't see how to predict exactly what will happen. In fact, > my cluelessness about the way the big spring acts makes me even less > clueful about what should happen. > > --Joshua Zucker Neil is cheating. He has this amazing physics simulator from http://www.phunland.com/wiki/Home that even models air resistance. The clone feature naturally placed the balls slightly apart, which we deemed a plus, but strangeness ensued! The middle balls soon started moving, then gradually calmed down, then gradually excited again, ... It turns out to depend on string length! If the strings are short and the gaps are large, the ball collisions are off center. So make the strings really long, and all should be clear. --rwg (Neil wishes to point out that his incorrect answer preceded the software experiment. He may be a poor theoretician but never a poor experimentalist.)

11 22

[math-fun] Fwd: Enumerating numbers of solid partitions -- new results
by rcs＠xmission.com 07 Jan '11

07 Jan '11

This seems like an interesting project. --Rich ----- Forwarded message from suresh.govindarajan(a)gmail.com ----- Date: Mon, 13 Dec 2010 20:35:08 -0600 From: Suresh Govindarajan <suresh.govindarajan(a)gmail.com> Reply-To: Number Theory List <NMBRTHRY(a)listserv.nodak.edu>, Suresh Govindarajan <suresh.govindarajan(a)gmail.com> Subject: Enumerating numbers of solid partitions -- new results To: NMBRTHRY(a)listserv.nodak.edu Dear number theorists, Recently, with some undergraduate students, I embarked on a project to enumerate the numbers of solid (three-dimensional) partitions of integers less than or equal to 100. The home page for the project is here: http://boltzmann.wikidot.com/solid-partitions For one and two-dimensional (plane) partitions, there exist generating functions due to Euler and MacMahon. However, MacMahon's guess for the generating function of solid partitions is known to be wrong and hence it is difficult to enumerate solid partitions. At the start of this project, the first 50 numbers were known (due to work by Knuth and subsequently by Mustonen and Rajesh -- see also http://oeis.org/A000293 ). Thanks to some nice innovations, so far the project has added 12 new numbers to this list thus extending the numbers to integers <=62. Roughly speaking, adding five numbers takes 10 times longer. The last run took about 30K CPU hours run on the IIT Madras super-clusters. Here are the twelve numbers (see http://boltzmann.wikidot.com/solidpartitions-62 for the complete results) 51 11622 14153 25837 57 2 41380 42828 01444 52 19379 44766 58112 58 3 97140 96826 33930 53 32238 23655 07746 59 6 52064 95439 12193 54 53505 67710 14674 60 10 68461 42257 15559 55 88603 33844 75166 61 17 47294 70062 57293 56 1 46400 93392 99229 62 28 51869 10933 88854 Mustonen and Rajesh also observed that the asymptotics of the wrong MacMahon generating function seemed to fit numerical data obtained through Monte Carlo simulations. In particular, they find that [n^{-3/4} log p(n )] --> 1.79 +- 0.01 , which is in agreement with the value one obtains from the asymptotic behaviour of the wrong MacMahon generating function. We observe that the asymptotic formula given by MacMahon's generating function seems to work even for small numbers like n=50/60. Using this formula (and its generalization), we were able to predict the numbers of solid partitions from 55-62 accurate to 0.02% -- around 3-5 digits. See http://boltzmann.wikidot.com/solidpartitions-62 for more details. I look forward to your comments and suggestions on these results and related issues. -- Suresh Govindarajan Professor Department of Physics Indian Institute of Technology Madras Chennai 600036 INDIA http://www.physics.iitm.ac.in/~suresh ----- End forwarded message -----

1 0

[math-fun] Exact rational 3D spacefill
by Bill Gosper 06 Jan '11

06 Jan '11

Last year I sent one named Treano and posted some graphs. It had three problems: 1) As Neil Bickford points out, it should have been named Trilbert. 2) It spanned a face diagonal instead of an edge, creating inconsistent textures when sampled at fixed frequencies. 3) It was so obscure that when I found the edge-spanning construction, I couldn't decipher and modify my own code. Fortunately, Corey Ziegler Hunts was up to both tasks: I3 = IdentityMatrix[3]; In[720]:= Vects = {{0, 0, 0}, {0, 0, 1}, {0, 1, 1}, {0, 2, 1}, {1, 2, 1}, {2, 2, 1}, {2, 1, 1}, {2, 0, 1}, {2, 0, 0}}/2; Mats = {{{0, 0, 1}, {0, 1, 0}, {-1, 0, 0}}.{{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}}, {{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}, {{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}, {{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}.{{1, 0, 0}, {0, -1, 0}, {0, 0, 1}}, {{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}.{{1, 0, 0}, {0, -1, 0}, {0, 0, 1}}, {{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}, {{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}, {{0, 0, -1}, {0, 1, 0}, {1, 0, 0}}.{{-1, 0, 0}, {0, 1, 0}, {0, 0, 1}}, {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}}/2; In[722]:= Trilbert[t_, mat_: I3, vect_: {0, 0, 0}] := (Trilbert[t, b1_: I3, b0_: {0, 0, 0}] = (Trilbert[t, matend_: I3, vectend_: {0, 0, 0}] := Inverse[matend - mat].(vect - vectend); Module[{t8 = 8*t, n}, n = Floor[t8] + 1; Vects[[n]] + Mats[[n]].Trilbert[FractionalPart[t8], mat.Mats[[n]], vect + mat.Vects[[n]]]])) (style-tweaked in ways Corey might disavow). The inverse images of the eight vertices are In[724]:= Trilbert /@ {0, 5/28, 9/28, 3/7, 4/7, 19/28, 23/28, 1} Out[724]= {{0, 0, 0}, {0, 0, 1}, {0, 1, 1}, {0, 1, 0}, {1, 1, 0}, {1, 1, 1}, {1, 0, 1}, {1, 0, 0}} (which you couldn't directly test without the exact rational trick). There are eight(!) inverse of the center point: In[729]:= {19/28, 1 + 23/28, 2 + 5/28, 3 + 23/28, 4 + 5/28, 5 + 23/28, 6 + 5/28, 7 + 9/28}/8 Out[729]= 19 51 61 107 117 163 173 205 {---, ---, ---, ---, ---, ---, ---, ---} 224 224 224 224 224 224 224 224 Trilbert /@ % Out[728]= 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 {{-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}} 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Every interior triple of dyadic rationals has eight inverses. Plots: http://gosper.org/Trilbertloops.png http://gosper.org/Trilbertcenters.png http://gosper.org/Trilbertcorners.png --rwg (While I was typing this, a popup said "Mathematica quit unexpectedly". When I clicked Restart, it said my license had expired, demanding a password, which I requested upon the first warning Christmas Eve. Why does WRI synchronize this with everyone's vacation?) It turns out the unexpected quit was due to a 7.0 bug eating the entire Apple VM while trying to convert to a .pdf the small .nb containing these three plots.

5 5

[math-fun] The Minsky-Gosper pump
by Bill Gosper 05 Jan '11

05 Jan '11

Still: http://gosper.org/pumpstill.gif I noted the 2D porism in high school after reading an article on Wankel engines. In 1965 I turned it into a "display hack" for the PDP-6 using uncorrected Minsky "circles" for both the shapes (curved dotted lines) and the motion. The eccentricities were not noticeable. As soon as Minsky saw it, he exclaimed "Helical twist!". Prior to that, I had tried to make it into a pump or an engine with various impractical kludges which I later learned were patented in the 1940s. I then devised the continuous deformation via "four point ellipses" to circles as a means of porting the intake and exhaust, and we always mentioned the twist and taper when we showed AILab visitors the crude 2D demo. I later learned that the twist idea without the taper to circularity was patented by someone in Florida a few months later. I've always suspected, by someone who stole the idea but didn't understand the taper description, which was difficult in those days to draw and very difficult to convey without a drawing, as I later learned by wasting several hundred dollars on inaccurate patent searches. When I first showed the 2D porism to Rich, he immediately remarked that three "Wankels" (Reuleaux triangles) do it too. Two other improvements which seemed unpatented at the time were to vary the pitch along the length to get an arbitrary compression profile (e.g., for Diesels), and to tessellate six or more rotors to make equally many pumping cells in opposite phases, which turns out deliver a constant, nonpulsing dVolume/dt (neat proof!) Alternatively, a single set of four blunt rotors could replace the "Ferris wheel" of rollers munching the compressible hose in a heart-lung machine. And I've always suspected that a sheet of small, low-friction rotors would form a "positive displacement wall" if driven just by the rotors along the edges, providing high volume and high pressure, but low velocity. With cones based on Lissajous arcs, it is even possible to mesh nonparallel rotors. I was never able to visualize the constant shape pumped by four fixed pitch rotors. Its ends are like twisted square pyramids with concave sides, but "bookending" them ought to approximate an octahedron, which is impossible, since there are only four smooth sides. Julian Ziegler Hunts to the rescue: http://gosper.org/pumpee.png . This is what provoked my msg about rwg> Good grief, tutor Julian has plotted two infinitely differentiable >surfaces intersecting in a simple arc that is not differentiable. Julian also determined the maximum removable rotor surface (after initially doubting that any reduction was possible): http://gosper.org/minpump.png His animation of four reduced rotors pumping a string of pumpees is weird. --rwg

1 0

Re: [math-fun] Adam's pi machine
by Bill Gosper 05 Jan '11

05 Jan '11

rcs> Gosper and I attended a workshop on Symbolic Math systems in NYC, c. 1986. The Chudnovskys presented some transcendence-related results, including approximation bounds on pi^2 and pi. IIRC, the pi result was that |pi - p/q| > 1/q^13 with p & q integers, q>0. This provides a weak upper limit on the length of a block of 0s or 9s: the block can't be longer than 12 times the number of digits preceding the block. A similar result would apply to any repeating pattern like 123123123... Rich What I remember most about that conference was the end of my talk, where, to show how weird things could get if you pushed my matrix methods far enough, I somewhat embarrassedly gave the 4F3[125/128] identity (d194) about 1/3 into http://www.tweedledum.com/rwg/idents.htm whose rhs is = csc(%pi/5)*(a/2-3/5)!*(a/2-1/5)!*2^(3*a-2/5)/((a/2-1/2)!*(a/2-3/10)!) This drew the expected freak-show grimaces from the audience, except for the Chudnovskys, who literally rushed me at the podium, offering congratulations and gratitude, assuring me it was just what they were looking for. --rwg

1 0

[math-fun] Adam's pi machine
by Bill Gosper 04 Jan '11

04 Jan '11

Some funsters might not recognize Adam as the the author of the spectacular Life configuration that computes pi in decimal (and even displays the digits). Maybe someday he can explain to us if and how he solved the ".999 problem", i.e., the absence of an a priori bound on the number of possible consecutive 9s or 0s, which, until they cease, leave the preceding digit unresolved (or equivalently, provoke negative or oversize "apology digits" following the premature estimate of an earlier one). The corresponding problem with continued fraction output is the unboundedness of a term. Both Neil's and my calculations simply delayed the output a few terms with a "smoother", but were capable of non-canonical outputs (0 or negative terms) following a term of truly newsworthy size. A continued fraction algorithm restricted to positive integer outputs would simply freeze on 1 2 2 2 ... times 1 2 2 2 ... . --rwg

4 4

Re: [math-fun] Cylinder puzzle
by Dan Asimov 03 Jan '11

03 Jan '11

P.S. This problem certainly didn't stump many people! In greater generality let a "unit slab" denote all points at a distance <= 1 from a hyperplane through the origin in R^N. Then if {S_k : k in Z+} is any countable family of unit slabs in R^N, their intersection has surface area = N * volume. (E.g., if in R^3 the countable family consists of only 3 distinct, mutually perpendicular unit slabs, one gets a cube of volume 8 and surface area 24.) --Dan Those who sleep faster get more rest.

1 0