February 2005 - mathfun@mailman.xmission.com

[math-fun] 100 digramms
by Eric Angelini 16 Feb '05

16 Feb '05

Record so far (Feb. 16th) is held by Jacques Tramu : 11 20 21 30 31 40 41 50 51 60 61 70 71 80 81 90 92 93 94 95 96 97 98 223 242 526 533 543 627 634 644 728 737 745 755 838 848 5667 8689 91001 sum : 115453 Best, É. __________________________________________________ http://angelink.be/?CeRaccourciFaitCinquanteSignes

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[math-fun] 100 digramms
by Eric Angelini 16 Feb '05

16 Feb '05

Record bettered a few minutes ago (by Jacques Tramu) : 90 91 92 93 94 95 96 97 98 100 112 130 203 223 314 340 415 424 450 516 525 535 606 617 626 707 718 727 808 828 3637 3846 4748 5758 6899 sum : 35658 « Show once and only once every couple of digits 00, 01, 02, 03, ..., 97, 98, 99 in the lightest finite strictly increasing sequence of integers. » ["Weight" of a sequence is given by the sum of all a(n)] Best, É. __________________________________________________ http://angelink.be/?CeRaccourciFaitCinquanteSignes

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[math-fun] 100 digrams
by Eric Angelini 15 Feb '05

15 Feb '05

Hello math-fun and seqfan; I'm trying to find a finite sequence like this one : 1 8 9 10 11 20 21 30 31 40 41 50 51 60 61 70 71 90 92 93 94 95 96 97 98 223 242 526 533 543 627 634 644 728 737 745 755 800 838 848 5667 8681 ... where one can read once and only once all the "di- grams" formed by the digits 0 to 9 [thus the digrams 00, 01, 02, 03, 04, ... 09 10, 11, 12, 13, 14, ... 19 ... 90, 91, 92, 93, 94, ... 99]. The "best" sequence would be the "lightest" one -- the "weight" of a finite sequence being the sum of all it's terms (the example here has a "weight" of 25111). The above sequence shows almost all the digrams -- only [99] is missing... Best, Éric Angelini -- Brussels, Belgium.

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[math-fun] 10,000 year gear train
by R. William Gosper 13 Feb '05

13 Feb '05

As Danny and I had already discussed offline, John McCarthy et al have described how to implement Steve Gray's solution via a giant game of billiards with Kuiper object cue balls. Gene's tidal deceleration might be handled by noncircular gears feeding a differential. The solution Danny posted (which is only good to 1.5 lsd, sorry) models n bodies with 4 n + 2 gears and no gear fanout, only shaft fanout. A partial solution using two gear fanouts, but lacking Earth and Mars, is 149 293 157 --- --- --- = 29.5350888532671 +.3 lsd 12 83 233 149 293 103 143 (--- ---) --- --- = 87.969257189543 +1 lsd 12 83 41 179 149 152 293 (---) --- --- = 224.700799133144 -.7 lsd 12 107 23 (Don't see how to reuse 293.) <Inter Venerem et Jovem planetas quaeremus.> 149 259 259 174 (--- --) --- --- --- = 4332.82012915184 +.1 lsd 12 83 13 31 149 259 259 321 183 (--- --- --- ---) --- --- = 10755.6986497112 +1 lsd 12 83 13 31 136 The idea here is to spend five gears followed by three. One avenue I didn't explore is to handle one or more intermediate planets with shaft fanout. --rwg ULTONIAN LUNATION, PRICELESS ECLIPSERS, TIME-SHARINGS MISSING-EARTH ASTROMETRIC MERITOCRATS

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[math-fun] Re: googol machine
by James Propp 12 Feb '05

12 Feb '05

>Maybe Google or somebody could commission a children's museum exhibit in >the form of Ganson's futernity machine extended to a length of fifty >100:1 gearboxes. The boxes would be labeled with the powers of 10 >and Conway names, and signs would label the first few shafts with, e.g., >"x turns per second," "x turns per minute," ..., "... years," "... millennia," >galactic rotations,..., solar nova consumes Earth, universe runs out of gas >(with a gratuitous universal joint as an inside joke), ..., googol. >--rwg Someone should create an applet that does this: "Give or take an unknown but fixed error, it is now 0000000000000000000000000000000000536112314159265359 seconds since the big bang." Clicking on each zero-digit would bring up a little text window telling you what the universe will look like when that particular 0 changes to a 1. Jim Propp

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[math-fun] Re: Rotor-router article
by James Propp 11 Feb '05

11 Feb '05

Michael Kleber wrote: >My publications list, with "Goldbug Variations" at the top, is at > > http://people.brandeis.edu/~kleber/pubs.html > >There's also a link to just download/view the blob picture alone. Does anyone have any ideas about how one could mechanically implement bugs and rotor-routers, in a fashion suitable for a classroom demo or a science museum display? (Hal Canary and Francis Wong created a software implementation of some of the variations; you can find an outdated version of the "1.0" release of their applet at http://www.math.wisc.edu/~propp/rotor-router-1.0/ . To play with the goldbug, click "The Applet", change "1-D Aggregation" to "1-D Walk", and click on "Paused (Press to Resume)".) Jim Propp

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[math-fun] ~s gears
by R. William Gosper 10 Feb '05

10 Feb '05

John McCarthy collects inventions that came too late to do much good, e.g., the non-pulsating candle wick. In this era of numerical machi- ning, we ought to be able to make interesting gears to solve mechanical problems of yesteryear. As a minor (and probably unoriginal) example, to minimize stripping, a smaller gear ought to have broader teeth and narrower notches than the larger gear it meshes with. This must be geometrically possible, at least sometimes, since you could delete alternate teeth from a(n even-toothed) larger gear and fill in the corresponding notches of the smaller one. However, if we view this as a deformation of gears with half as many teeth, we have lost something like half the tooth amplitude (proper term?), so there may be little advantage in desymmetrizing. We might get certain exotic ratios with differential planetaries where the ring and central gears differ by an odd number of teeth. This might be possible if those gears are in slightly different planes, slightly beveled, with slightly tilted planets spanning the gap. The planets or ring might have a hollow shaft coaxial with the central's, both driven (at different speeds) by one adjoining shaft bearing two gears. This would require "continuous" scalability of one gear and its mate on one of the coaxial shafts. I'd like to animate gears with improbably small numbers of teeth (3??, 4?). Can anyone save me the trouble of deriving the involute curves that avoid sliding friction? --rwg PS, apologies for my churlish Mahalingam ouburst. A few minutes of Google, or consulting with any Subcontinental would have taught me that it is a proud old name of long standing. E.g., Welcome to Shankar Mahalingam's homepage. Professor and Chair Department of Mechanical Engineering University of California, Riverside ... This gentleman has *absolutely no connection* with the villain in the up-and-coming Goldmember sequel. And all this is dinky by comparison with former African head-of-state Houphouet Boigny, which in his own language means "Pit of Excrement". "Oh Miss Rice, here's some xylo- caine in case you need to bite your cheeks during the introductions." SUBNARCOTIC OBSCURANTIC BUCCINATORS (You bet your mahalingam.)

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[math-fun] 10,000 year gear train
by Danny Hillis 07 Feb '05

07 Feb '05

I am building a mechanical clock that is intended to show the position of the visible planets and the phase of the moon over the next ten thousand years. Thus, I would like a gear train that computes all the following ratios using a small number of gears. 29.53058853, 87.96925718, 224.7007992, 365.2563554, 686.9795859, 4332.820129, 10755.69864 The gear train must use gears with tooth number not less than 12 and not greater that 400. I want to integrate daily rotations to show the orbital position within a few percent after 10,000 years, so I need accuracy to at least 6 places. After playing around with this problem using lattice methods, I realized that (a) there were like some very good solutions in the space and (b) that Bill Gosper would probably be better at finding them than I would. Indeed, RWG took up the challenge and kindly suggested the following solution (note the shared gears): RWG wrote: > 293 149 157 > (--- ---) --- = 29.5350888532671 +.3 lsd > 12 83 233 > > 293 152 358 101 61 > (--- --- ---) --- -- = 87.9692571945795 1.4 lsd > 12 107 295 67 44 > > 293 152 149 > (---) --- --- = 224.700799133144 -.7 lsd > 12 107 23 > > (Don't see how to reuse 149.) > > 293 152 358 269 > (--- ---) --- --- = 365.256355463807 +.6 lsd > 12 107 295 31 > > 293 152 358 101 263 267 > (--- --- --- ---) --- --- = 686.979585832448 -.7 lsd > 12 107 295 67 47 138 > > 293 152 358 101 263 322 281 272 > (--- --- --- --- --- ---) --- --- = 4332.8201300432 1 lsd > 12 107 295 67 47 103 107 183 > > (Don't see how to reuse 281.) > > 293 152 358 101 263 322 281 > (--- --- --- --- ---) --- --- = 10755.6986519239 1.2 lsd > 12 107 295 67 47 103 29 > > Beat that! > --Bill I post his solution here because it is lovely and also in the faint hope that someone on the list may be able to do better. I offer a $1 prize to the first person (including Bill) who can beat this solution, by reducing the number of gears size without decreasing the worst-case accuracy. -Danny

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[math-fun] A missing sequence from group theory?
by N. J. A. Sloane 04 Feb '05

04 Feb '05

To MathFun and SeqFans: A possibly new sequence? Could someone calculate it? Let t(G) = no. of unitary factors of the abelian group G, and let T(n) = Sum t(G) taken over all abelian groups of order <= n. There are several papers giving estimates for T(n), e.g. T(x) = c_1 x (log x + 2 gamma - 1) + c_2 x + ... My question is, how does the sequence T(n) begin? References and definitions follow, from MSN. NJAS MR1850168 Zhai, Wen Guang On the average number of unitary factors of finite abelian groups. II [MR1813450 (2001j:11092)]. (Chinese) Acta Math. Sinica 44 (2001), no. 4, 667--672. 11N45 MR1813450 (2001j:11092) Zhai, Wenguang On the average number of unitary factors of finite abelian groups. II. Acta Math. Sin. (Engl. Ser.) 16 (2000), no. 4, 549--554. (Reviewer: R. C. Baker) 11N45 (11L07 20K01) MR1641042 (99e:11124) Zhai, Wenguang; Cao, Xiaodong On the average number of unitary factors of finite abelian groups. Acta Arith. 85 (1998), no. 4, 293--300. (Reviewer: R. C. Baker) 11N45 MR1613290 (99f:11124) Wu, J. On the average number of unitary factors of finite abelian groups. Acta Arith. 84 (1998), no. 1, 17--29. (Reviewer: R. C. Baker) 11N45 (11L07 20K01) 1225427 (94k:11108) Schmidt, Peter Georg(D-MRBG) Zur Anzahl unitrer Faktoren abelscher Gruppen. (German) [On the number of unitary factors in abelian groups] Acta Arith. 64 (1993), no. 3, 237--248. Part of this review follows: Let $\scr A$ denote the set of all abelian groups. Under the operation of direct product, $\scr A$ is a semigroup with identity element $E$, the group with one element. $G_1$ and $G_2$ are relatively prime (German: teilerfremd) if the only common direct factor of $G_1$ and $G_2$ is $E$. We say that $G_1$ and $G_2$ are unitary factors of $G$ if $G=G_1\times G_2$ and $G_1,G_2$ are relatively prime. Let $t(G)$ denote the number of unitary factors of $G$, and let $T(x)=\sum_{G\in {\scr A}, |G|\le x} t(G).$ There are constants $A,B,$ and $C$ such that $\Delta(x)\colon =H(x)-Ax\log x - Bx -C\sqrt x=o(\sqrt x).$ In this paper, the author shows that $\Delta(x)\ll x^{3/8} \log^{7/2}x.$ The previous best estimate for $\Delta(x)$ had been $\Delta(x)\ll x^{31/82} \log^2 x$; this was given by H. Menzer \ref["Exponentialsummen und verallgemeinerte Teilerprobleme", Habilitationsschrift, Friedrich-Schiller-Univ., Jena, 1992; per bibl.]. Further references from the Wu (1998) paper: Many accented letters have been dropped! # R. C. Baker and G. Harman, Numbers with a large prime factor, Acta Arith. 73 (1995), 119--145. MR1358192 (97a:11138) # E. Cohen, On the average number of direct factors of a finite abelian group, ibid. 6 (1960), 159--173. MR0118764 (22 #9535) # E. Fouvry and H. Iwaniec, Exponential sums with monomials, J. Number Theory 33 (1989), 311--333. MR1027058 (91b:11097) # S. W. Graham and G. Kolesnik, Van der Corput's Method of Exponential Sums, Cambridge Univ. Press, Cambridge, 1991. MR1145488 (92k:11082) # C. H. Jia, The distribution of square-free numbers, Sci. China Ser. A 36 (1993), 154--169. MR1223084 (94f:11095) # G. Kolesnik, On the number of abelian groups of a given order, J. Reine Angew. Math. 329 (1981), 164--175. MR0636451 (83b:10055) # E. Krtzel, On the average number of direct factors of a finite abelian group, Acta Arith. 51 (1988), 369--379. MR0971087 (89m:11091) # H.-Q. Liu, The greatest prime factor of the integers in an interval, ibid. 65 (1993), 302--328. MR1259341 (95d:11117) # H.-Q. Liu, On some divisor problems, ibid. 68 (1994), 193--200. MR1305200 (96a:11089) # H.-Q. Liu, Divisor problems of 4 and 3 dimensions, ibid. 73 (1995), 249--269. MR1364462 (96k:11118) # P. G. Schmidt, Zur Anzahl unit\"arer Faktoren abelscher Gruppen, ibid. 64 (1993), 237--248. MR1225427 (94k:11108) # P. Sargos and J. Wu, Multiple exponential sums with monomials and their applications in number theory, Prpublications 97/ n37 de l'Institut lie Cartan, Universit Henri Poincar (Nancy 1). # J. Wu, On the distribution of square-full and cube-full integers, Monatsh. Math., to appear. cf. MR1657818 (2000a:11125)

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[math-fun] graph question
by Chris Landauer 04 Feb '05

04 Feb '05

hihi. all - let's start by stating the problem in a way that can be solved the graph certainly needs to be finite and connected in addition, you also have to limit the number of times each edge can be retraced, or else it is trivial to draw any connected finite graph with a continuous path, whether or not there are odd order vertices (pick any one vertex - connected means there is a path, so just draw all the paths from that vertex to the others and back) the usual restriction is to limit each edge to exactly one traversal (so the order of a vertex is the total number of arrivals and departures it encounters in the continuous path), which is called an eulerian path (the seven bridges of koenigsburg apparently brought the problem to his attention) there are no finite graphs with exactly one odd order vertex it is clear that any eulerian path in a graph with two odd order vertices will necessarily start at one of them and end at the other also, it is clear that any eulerian path for a graph with no odd order vertices must end at the same vertex at which it starts i expect there is a reasonably straightforward induction proof - it is certainly an old theorem more later, cal

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