December 2002 - mathfun@mailman.xmission.com

[math-fun] Lacings continued
by N. J. A. Sloane 13 Dec '02

13 Dec '02

I found the Nature article, which gives a formula but no proof (sequence A078601 below). I get a different answer (A078602)! NJAS %I A078601 %S A078601 1,3,42,1080,51840,3758400,382838400,52733721600,9400624128000,2105593491456000, %T A078601 579255485276160000,191957359005941760000,75420399121328701440000,34668462695110852608000000, %U A078601 18432051070888873171353600000,11223248177765618214764544000000,77593958120381337432427069440000 00 %N A078601 a(1)=1; for n > 1, a(n) = ((n!)^2/2)*Sum(binomial(n-k,k)^2/(n-k),k=0..floor(n/2)). %D A078601 B. Polster, What is the best way to lace your shoes?, Nature, 420 (Dec 05 2002), 476. %C A078601 The number of ways to lace a shoe that has n pairs of eyelets, assuming the lacing satisfies ce rtain conditions. %O A078601 1,2 %K A078601 nonn %A A078601 njas, Dec 11 2002 %Y A078601 Cf. A078602. %I A078602 %S A078602 1,2,21 %N A078602 Ways to lace a shoe that has n pairs of eyelets. %C A078602 The lace must pass through each eyelet exactly one, must begin and end at the extreme pair of e yelets, and cannot pass though three consecutive and adjacent eyelets that are in a line. %O A078602 1,2 %e A078602 a(3) = 21: label the eyelets 1,2,3 from front to back on the left side then 4,5,6 from back to front on the right side. The lacings are: 124356 154326 153426 142536 145236 132546 135246 and %e A078602 the following and their mirror images: 125346 124536 125436 152346 153246 152436 154236. %K A078602 nonn,bref,more %A A078602 njas, Dec 11 2002 %Y A078602 Cf. A078601.

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[math-fun] squares near cubes
by Richard Schroeppel 13 Dec '02

13 Dec '02

Here's an obscure sequence for Neil, from the Number Theory list. Rich rcs(a)cs.arizona.edu ---------------------- Date: Thu, 12 Dec 2002 23:16:36 -0500 Sender: Number Theory List <NMBRTHRY(a)LISTSERV.NODAK.EDU> From: Ismael Jimenez Calvo <ltqij04(a)pinar2.csic.es> Subject: Ten new "good examples of Hall's conjecture" M. Hall [Hall] conjectured that the nonzero difference k = x^3 - y^2 can not be less than Cx^{1/2}, for a constant C. His original conjecture, probably false, has been reformulated in the following way: "For any exponent e < 1/2, a constant K_e > 0 exists such that |x^3 - y^2| > K_ex^e" As much as , at present, this conjecture is neither proved nor disproved, it is interesting to enumerate the known cases when k < x^{1/2}, what I address as "good examples of Hall's conjecture" borrowing notation used for the related and more general ABC conjecture. A detailed theoretical and computational account on this subject can be found in [Elkies] where he published a table with the 25 items known. Simplifying, we can attribute 13 of them to Gebel Petho and Zimmer [GPZ], who developed an algorithm to search for all integer points in the elliptic curves y^2 = x^3 + k for |k| < 100 000. The other 11 items where found by N.D. Elkies using base reduction algorithms. German Saez, Javier Herranz and I have developed a new algorithm based in methods related to those used in [JS] that have found ten new "good examples of Hall's conjecture". Very briefly, the method starts from the fact that for x=t^2 and t integer, k is zero. If we consider rational values of t instead, and the integer values near t^2, we find that the points (x,k) correspond to the integer points of a set of cubic polynomials. Knowing the polynomials, we can select those which may contain a integer point with a small k . The algorithm is probabilistic in the sense that it seeks for "good examples of Hall's conjecture" where they can be found with higher probability. Nevertheless, the algorithm has found all the items known and 10 more which are displayed in the table bellow. One of them, the item #16, unnoticed for any reason in past computations, has been found using the new algorithm. The table contains the parameter r = sqrt(x)/k for each x value. The values of y and k are not listed because they can be easily computed from x taking y as the nearest integer to x^{3/2}. The algorithm was programmed as a PARI script that was translated to C and compiled with the utility GP2C. The executable was run in a Pentium III at 1000 Mhz during 30 days. For more information, you can see the web page: http://www.terra.es/personal9/ismaeljc/hall.html ---- Ismael Jimenez Calvo Good Examples of Hall's Conjecture ======================================== # x r ======================================== 1 2 1.41 2 5234 4.26 GPZ 3 8158 3.76 GPZ 4 93844 1.03 GPZ 5 367806 2.93 GPZ 6 421351 1.05 GPZ 7 720114 3.77 GPZ 8 939787 3.16 GPZ 9 28187351 4.38 GPZ 10 110781386 1.23 GPZ 11 154319269 1.08 GPZ 12 384242766 1.34 GPZ 13 390620082 1.33 GPZ 14 3790689201 2.20 GPZ 15 65589428378 2.19 E 16 952764389446 1.15 JHS 17 12438517260105 1.27 E 18 35495694227489 1.15 E 19 53197086958290 1.66 E 20 5853886516781223 46.60 E 21 12813608766102806 1.30 E 22 23415546067124892 1.46 E 23 38115991067861271 6.50 E 24 322001299796379844 1.04 E 25 471477085999389882 1.38 E 26 810574762403977064 4.66 E 27 9870884617163518770 1.90 JHS 28 42532374580189966073 3.47 JHS 29 51698891432429706382 1.75 JHS 30 44648329463517920535 1.79 JHS 31 231411667627225650649 3.71 JHS 32 601724682280310364065 1.88 JHS 33 4996798823245299750533 2.17 JHS 34 14038790674256691230847 1.27 JHS 35 372193377967238474960883 1.33 JHS GPZ - J. Gebel, A. Petho and H.G.Zimmer. E - Noam D. Elkies JHS - I. Jimenez, J. Herranz and G. Saez. References: [Elkies] Elkies, N.D.: Rational points near curves and small nonzero |x3 - y2| via lattice reduction. Pages 33--63 in Algorithmic Number Theory. Proceedings of ANTS-IV;W. Bosma, ed.; Berlin. Springer, 2000; LNCS 1838. [Hall] Hall, M.: The Diophantine equation x^3 - y^2 = k. Pages 173-198 in Computers in Number Theory (A. Atkin, B. Birch, eds.; Academic Press, 1971). [GPZ] Gebel, J., Petho, A., and Zimmer, H.G.: On Mordell's equation, Compositio Math. 110 (1998), 335-367. [JS] Jimenez Calvo, I. and Saez Moreno, G.: Approximate Power roots in Z_m. pages 310-323 in ISC 2001; G.I. Davida and Y. Frankel, eds.; Berlin: Springer, 2001; LNCS 2200.

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[math-fun] Re: five, six, and eight
by James Propp 12 Dec '02

12 Dec '02

I asked earlier about the Lyness sequence, given by the recurrence L(n+1) = (L(n)+1)/L(n-1). Regardless of the initial conditions, a sequence satisfying this recurrence will be periodic with period 5; this is easy to verify, but I wanted to know "why". Some illumination came from reading parts of the paper "Y-systems and generalized associahedra" by Sergey Fomin and Andrei Zelevinsky, available over the web at http://www.math.lsa.umich.edu/~fomin/Papers/ (to appear in Annals of Mathematics), which Michael Kleber and I looked at. First, we may switch over to thinking about composition of the two rational functions F and G where F(s,t) = (s,(s+1)/t) and G(s,t) = ((t+1)/s,t) . F and G are both involutions, but if we take all alternating compositions ...(F(G(F(...(s,t)...)))... we get a group of 10 operations. In particular, G-compose-F is of order 5. Second, noting that the expressions for F and G are subtraction-free, we may replace the maps F and G their "tropical analogues" f and g, replacing the constant 1 by the constant 0 and replacing the arithmetical operations +, *, and / by max, +, and -: f(s,t) = (s,max(s,0)-t) g(s,t) = (max(t,0)-s,t) Note that f and g are continuous piecewise linear operations on the plane. Each of f,g is of order 2, but their composition g-compose-f has order 5. Moreover, the orbit of (1,0) under the action of the composed map is (1,0), (0,1), (0,-1), (1,1), and (-1,0) which we can recognize as being related to the Laurent polynomials x, y, (y+1)/x, (x+y+1)/xy, (x+1)/y from the Lyness sequence when they are written as (1) / x^{-1} y^{0} (1) / x^{0} y^{-1} (1+y) / x^{1} y^{0} (1+x+y)/ x^{1} y^{1} (1+x) / x^{0} y^{1} To see what's going on with f and g geometrically, it's most helpful to think of (1,0) and (0,1) as making an angle of 120 degrees, and to divide the plane up into 60-degree sectors. However, the maps f and g do not permute these sectors (if they did, then their composition could not have order 5 without forcing f and/or g to be discontinuous). Instead, some sectors are mapped into sectors, some sectors are expanded into a union of two adjacent sectors, and some sectors are contracted into a part of a sector. A similar situation prevails for the 6-fold periodic recurrence and for the 8-fold periodic recurrence, and there is a Lie-theoretic story for what's going, involving the root-lattice. E.g., for the original 5-fold case, if you examine the list above you can check that if we throw out the first two denominators (the ones with negative exponents), the resulting vectors (1,0), (0,1), and (1,1) are the positive roots of the Lie algebra A_2 (relative to the basis of simple roots). This happens much more generally. By the way: Does anyone see a way to combine the 10-fold Lyness action with the 6-fold rotation action on the 6 sectors to get the alternating group A_5, or anything like it? Low-dimensional piecewise-linear maps might give faithful representations of interesting groups whose linear representations are all higher-dimensional. Jim Propp

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[math-fun] Ferguson faces deadline
by Richard Schroeppel 12 Dec '02

12 Dec '02

> From: "David H Bailey" <dhbailey(a)lbl.gov> > Date: Thu, 12 Dec 2002 07:55:07 -0800 http://www.washingtonpost.com/wp-dyn/articles/A43083-2002Dec11.html Running Out Of Time, Space Math-Inspired Artist Under Eviction Threat By Eugene L. Meyer Washington Post Staff Writer Thursday, December 12, 2002; Page B01 Helaman Ferguson is having a bad day in his unheated studio/barn. His 10-year-old compressor has died; the repairman says it's beyond fixing. The earliest December snowstorm in 20 years is about to delay the hauling of his nine-ton granite sculpture to a corporate client outside Philadelphia. To top it off, his landlord has declared him a squatter and is evicting him from the old dairy barn he converted -- and has toiled in for four years -- at the Maryland Science and Technology Center in Bowie. Others in Bowie regard Ferguson, 62, as a national treasure. He is the Tech Center's unofficial artist in residence, a world-renowned research mathematician who uses three-dimensional computer-imaging to turn algorithms into sculpture. His work has been exhibited in major cities, and he has received national awards. People from all over the world come to Bowie to see the 45-ton granite fountain he created from a formula devised 800 years ago by the Italian mathematician Leonardo Pisano, better known as Fibonacci. "Math is my design language," said Ferguson, a tall, lanky man with a trimmed white beard who also holds a world's record of sorts for juggling while jogging 50 miles. In the drafty barn, near two large granite pieces, equations are scrawled on sheets of insulation and a chalkboard. The classical music he listens to while he works can be heard hundreds of yards away, blaring from inside the barn across the snowscape. Ferguson's predicament is the latest wrinkle in the saga of the 466-acre office park at Routes 50 and 301. Formerly affiliated with the University of Maryland, it remains largely undeveloped after two decades, having suffered through economic downturns, political problems and foreclosures that landed two former developers in jail. The land for the office park is under covenants that limit development to research and technology tenants. But Ferguson, who was working in the basement and garage of his Howard County home, negotiated the use of the barn a few years ago. Ferguson had been hired by one of the park's handful of tenants, the Institute for Defense Analyses, a private think tank affiliated with the National Security Agency. In return for rights to the barn, which he spent $80,000 converting into a studio, he agreed to design and build the fountain that sits in the middle of a reservoir at the center, dubbed Lake Fibonacci. He has worked since without a written agreement but with an unwritten pledge from one of the property owners that he could stay while completing his other large granite pieces. "He's the most determined guy I know. He's done some wonderful things," said Dean Morehouse, the Tech Center partner who suggested to Ferguson that he use the barn. Morehouse spent "a little over $500,000" on the fountain, which paid for materials, some equipment and 28 concrete pilings driven into the reservoir floor. Two years ago, Morehouse sold his managing stake in the Tech Center and about half the property -- including the barn -- to the Baltimore-based MIE Co., a builder of office parks and shopping centers. MIE specializes in warehouse-style "flex space," which is what it wants to build where the barn is. Ferguson was given until September to complete his work and leave. When the deadline came, he said he needed more time and MIE let him stay until the end of the year. Now, he is between an 11-ton rock -- his latest commission, from Macalester College in Minnesota -- and a hard place. The new property owners say they will turn off his electric service at the end of this month and then tear down the barn. "He's not staying there," said Ramon Benitez, MIE's development manager. "He's, frankly, a squatter at this point." Ferguson is no typical squatter. The piece he has just finished but not yet delivered to pharmaceutical giant Merck, an abstract sculpture of hands almost shaking, drew a commission "in the low six figures," Ferguson said. Without being told, a viewer would not know its mathematical origins. Ferguson starts with a mathematical equation or theorem, usually of his own design. He uses computers and stereoscopic perspective to translate it into a three-dimensional image, which he then projects onto a block of stone. With a variety of tools -- huge milling machines, water-spouting diamond saws, hammer and chisel -- Ferguson turns the block into its final form. The fountain on Lake Fibonacci is based on a 13th-century mathematical problem: Starting with one newborn pair of rabbits, how many pairs of rabbits will you have after a year if each pair begets another pair every month from the pair's second month on? (Each pair has to wait a month to begin reproducing.) Fibonacci's solution involves a numerical sequence (1,1,2,3,5,8,13,21, etc.) in which each number is the sum of the preceding two. Place those numbers on a chart, and you have the design of the fountain. The 18-foot-tall sculpture (official name: "Fibonacci Fountain: Essential Singularity II") resembles a graph curve, rising from an elongated foundation with 14 jets of water shooting up from the base. It is lit up at night. "For me," Ferguson said, "it's taking these math forms no one's ever seen, felt or touched and bringing them into where they can be seen and felt and touched." Audrey E. Scott, the former Prince George's County Council member and three-term Bowie mayor, is among those who have tried to help him stay. She recalled a meeting with Benitez: "I said, 'You have here on your premises at no cost to you a national treasure, and it seems to me you would be going out of your way to capitalize on this gift you inherited.' But they don't see it that way." Ferguson keeps chipping away at the granite and the problem. MIE had said he could rent some of its flex-space, but then decided that his work would be too noisy and messy for adjoining tenants. Morehouse and Robert G. Depew, both limited partners with MIE, still own other land at the Tech Center and have told Ferguson that he could relocate to their property. Ferguson considered, then rejected moving the barn as too costly. He is now thinking of building a new barn on a site Depew owns at the Tech Center, he said, but he has no idea how to pay for it or what it would cost. "If somebody comes along with a big fat commission, that's where the funding comes from," Ferguson said. "Needless, to say, I've been beating the bushes." And then there is the problem of what to do with all his material and equipment on Jan. 1. So far, he said, he has no answer but to hope for "another reprieve." (c) 2002 The Washington Post Company

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[math-fun] Re: unimodality question
by James Propp 12 Dec '02

12 Dec '02

Hunter Snevily (snevily(a)uidaho.edu) wrote back to me: >Hi- Jerry Griggs found a counterexample 2,2,2,-1,-1,-1,-1,-1,-1 >I have not checked this out but Jerry is reliable. sorry hs Also, David Callan (callan(a)stat.wisc.edu) wrote: >The set {1,1,1,-1,-1,-1} has {r1,r2,...,rn} = {3, 12, 10, 12, 3, 1}, >not unimodal. >DC Jim Propp

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[math-fun] Re: math on tv
by Dean Hickerson 12 Dec '02

12 Dec '02

Jud McCranie wrote: > A google search finds this (it is about half way down) ... > and this, the script: > > http://communicationsoffice.tripod.com/3-14.txt Based on the script, I don't think we're supposed to believe that there actually is such a thing as a Fibonacci opening. It appears to me that Bartlet made it up to psych out Sam, and Sam realized it: BARTLET Ah, the Fibonacci opening. Very interesting. SAM You're just gonna mess with me this whole time, right? BARTLET Yeah. Dean Hickerson dean(a)math.ucdavis.edu

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[math-fun] math on tv
by Richard Schroeppel 11 Dec '02

11 Dec '02

Last week's episode of the West Wing TV show included the "President" playing two games of chess. In one, he recognized his opponent's opening was the Evans Gambit based on the opponent's first move (!); in the other, he immediately recognized the Fibonacci Opening. I haven't played serious chess for years -- some of my friends might say Never -- but I don't recall anything named after Fibonacci. His era antedates Ruy Lopez by about 3 centuries. Does anyone know of a Fibonacci opening? Rich rcs(a)cs.arizona.edu

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Re: [math-fun] Thought up while drawing polygons ...
by asimovd＠aol.com 11 Dec '02

11 Dec '02

Steve Gray writes: << I don't see that the events [A lying in BCD, B lying in ACD, C lying in ABD and D lying in ABD, where A,B,C,D are (randomly chosen) points in a square] are disjoint. >> Sure they are, except for a set of measure 0 which we'll ignore. If e.g. A is in the interior of triangle BCD, then e.g. D cannot be in the interior of the triangle ABC. Etc. --Dan

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Re: [math-fun] Thought up while drawing polygons ...
by asimovd＠aol.com 11 Dec '02

11 Dec '02

I found a math discussion site on the web where someone calling themself Poliwrath posed and solved the question of what is the probability that if A,B,C,D are random points in a square, then AB and CD intersect? Poliwrath's clever reasoning is as follows: If X,Y,Z are any three random points in the unit square, let T be the expected area of the triangle XYZ. (This just happens to be 11/144.) Hence the probability of (A lying in BCD) = T, and same for B lying in ACD, C lying in ABD and D lying in ABD. Since these 4 events are disjoint (!), the probability that one of them occurs is 4*T, and so the probability that none of them occurs, i.e. the probability that the convex hull of A,B,C,D is a quadrilateral, is 1-4T. Among the 24 equally likely ways that A,B,C,D can be assigned to the corners of any given quadrilateral, just 1/3 of them involve AB interesecting CD (since wherever A is, B can be adjacent with prob. = 2/3, or opposite, with prob. = 1/3). This shows that P = [the probability AB and CD intersect] is (1-4T)/3. Assuming T = 11/144, this gives P = 25/108. (I don't know the calculation giving T = 11/144, but both this figure and the 25/108 are well approximated by my computer simulations, so I believe them.) --Dan

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[math-fun] Taken, not stirred
by Marc LeBrun 11 Dec '02

11 Dec '02

You might be amused that the Fibonacci sequence made a cameo appearance last night on the SciFi Channel's "Steven Spielberg Presents Taken" mini-series (a well-produced junk-science UFO drama a la "X Files", "Close Encounters" etc). I wasn't quite following the random dialog, but it got my attention when one of the characters scribbled 1 1 2 5 8 13 on a pad and held it up, mentioning that each is the sum of the two previous. Then there was something about 55 (number of landing sites?) and finally they claim that there were exactly 46368 alien abductees, exulting "24th Fibonacci!"

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