October 2003 - mathfun@mailman.xmission.com

[math-fun] Mrs Perkins' Quilt; new Fermat Composite
by Richard Schroeppel 12 Oct '03

12 Oct '03

I'd like to encourage posters to present a little bit more background when they mention a problem. In the case of Mrs. Perkins' Quilt, the puzzle is to dissect an integer sized square into smaller integer squares, with non-integer-square pieces forbidden. (And the gcd of the piece sizes must be 1, to exclude dissecting a 6^2 into four 3^2.) In the case of "Lens Spaces", there's probably no short definition, so we're stuck. But the recreational problems are usually easy to state. It's easy for those of us with high-speed net connections to regard giving a URL as equivalent, even better, than summarizing background material. But to some folks, those URLs represent minutes of download time each, often loading useless decorations of no information value. (I've been on both sides of this fence, so I have some sympathy for the bandwidth challenged.) Rich rcs(a)cs.arizona.edu --- In other news, John Cosgrave &al have a new Fermat Composite. -- Rich Date: Sun, 12 Oct 2003 11:59:12 -0400 Sender: Number Theory List <NMBRTHRY(a)LISTSERV.NODAK.EDU> From: John Cosgrave <john.cosgrave(a)spd.dcu.ie> Subject: F[2478782], a new largest known composite Fermat number To: NMBRTHRY(a)LISTSERV.NODAK.EDU Dear colleagues, In Ferbuary this year I announced the discovery of a new largest known composite Fermat number: the 645817-digit prime (3*2^2145353 + 1) is a factor of F[2145352] = 2^(2^2145352) + 1, only [then] the second prime factor - with more than 100,000 digits - of a Fermat number. See www.spd.dcu.ie/johnbcos for complete details and further links. Now - in a slight state of shock, and delight - I wish to announce a further computational advance: the 746190-digit prime (3*2^2478785 + 1) is a factor of F[2478782] = 2^(2^2478782) + 1, and is now the fourth more-than-100,000 digit factor of a Fermat number (a 'smaller' one - now the third in ranking - was found by Takahiro Nohara on August 4th this year; for details see Wilfrid Keller's Fermat numbers site at http://www.prothsearch.net/fermat.html) There will most likely be some GFN results following, which I will put up at my web site over the next several days. (3*2^2478785 + 1) was discovered on M?ire N? Bhaoill's computer (a 2.6GHz Pentium IV) - pre-sieved with Paul Jobling's newpgen and PRP-ed with George Woltman's program - and then subjected to Yves Gallot's Proth (primality testing) and ProthF (generalised Fermat number testing). She (M?ire) is one of my supportive St Patricks College Proth-Gallot group members (http://www.spd.dcu.ie/johnbcos/proth-gallot_group_(spd).htm) and the new number actually appeared as a PRP-probable prime on her machine on June 22nd of this year. I didn't notice it until Tues 7th October... The Fermat divisibility result - "3*2^2478785+1 Divides GF(2478782, 2)" - appeared on my office screen at 11:02 today (the independent primality test was done at home), minutes before my double 2nd year number theory lecture was due to begin. Of course I had to tell my students, and one of them suggested we should all go across the road for coffee and cake, which is exactly what we did. We were there for almost two and a half hours, during which time they came to know of Euclid (and perfect numbers), ... , Mersenne, Fermat, Euler, Landry, ... , Brillhart-Morrison, CRAY, Wilfrid Keller, Jeffrey Young, Richard Crandall, Richard Brent, Proth, Yves Gallot, George Woltman, Paul Jobling, ... I will put up digital photos of computer screens sometime next week on my home page. John Cosgrave

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[math-fun] Re: Can you help to me
by Richard Guy 12 Oct '03

12 Oct '03

In this connexion, has anyone discovered any more spoof odd perfect numbers, such as Descartes's 2 2 2 2 3 7 11 13 22021 which is perfect if you pretend that 22021 is prime ? R. On Thu, 9 Oct 2003, Jud McCranie wrote: > At 07:20 AM 10/9/2003, f.firoozbakht(a)sci.ui.ac.ir wrote: > > >a(n) is the n-th odd number that sigma(n)=2n+6. > > > >a(1)=8925, a(2)=32445, a(3)=442365, a(4)=? > > > >I know a(4) is greater than 2*10^9 . > > There are no terms between 2*10^9 and 6.5*10^9. > >

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[math-fun] Mrs. Perkins Quilt -- Orders 89, 90 improved over UPIG
by Ed Pegg Jr 12 Oct '03

12 Oct '03

Lance Gay has cracked a problem in Unsolved Problems in Geometry. Quoting RK Guy -- known records with "some slightly increasing lack of confidence" are as follows: {1 >> 1}, {4 >> 2}, {6 >> 3}, {7 >> 4}, {8 >> 5}, {9 >> 6,7}, {10 >> 8,9}, {11 >> 10-13}, {12 >> 14-17}, {13 >> 18-23}, {14 >> 24-29}, {15 >> 30-39,41}, {16 >> 40,42-50}, {17 >> 51-66,68,70}, {18 >> 67,69,71-87}, {19 >> 88-100} RK can pencil in a correction for that last bit. Lance Gay (Lance.Gay(a)ngc.com) found order-18 solutions for side 89 and 90 Mrs. Perkins Quilts. These appear to be better than the current known values. Side=89 Order=18 - [48 41][5 5 11 20][2 8][41 9][2 9][3 7][12][8 28][20] Side=90 Order=18 - [49 41][8 12 21][41 9 7][3 9][2 8][11][2 28][5 5][21] Lance also asks for a source for the best know solutions for side 100-250. The only source I knew of was Duijvestijn, A. J. W. ftp://ftp.cs.utwente.nl/pub/doc/dvs/TableI. -- but this seems to be down. Other pertinent refs: http://mathworld.wolfram.com/MrsPerkinsQuilt.html http://mathworld.wolfram.com/PerfectSquareDissection.html I must humbly take blame for botching a diagram on Eric's site -- the order 7 square there is wrong (but the sequence right under is correct). Nick Gardner pointed out the mistake to me. For an exercise, try to dissect the order 7 square into 9 smaller squares. An Ascii solution is below, along with the Duijvestijn code. --Ed Pegg Jr, www.mathpuzzle.com (Fixed width Ascii art) ___ ___ ___ ___ ___ ___ ___ | | | | | | | | | | | | | |___ ___|___ ___| | | | | | |___ ___ ___|___|___| | | | | | | |___|___ ___| | | | | | | | | | | | | | | | |___ ___ ___ ___|___ ___ ___| [3 2 2][1 1 2][4 1][3]

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Re: [math-fun] The spatial universe is finite !!!???
by asimovd＠aol.com 11 Oct '03

11 Oct '03

John asks: << Is there a shape for space-time that makes space compact but that doesn't have an invariant factorization as space x time? >> The lens spaces L(7,1) and L(7,2) (compact 3-manifolds) are known to be topologically distinct but homotopy equivalent. I wonder if L(7,1) x R is homeomorphic to L(7,2) x R. IF this is the case, then L(7,1) x R = L(7,2) x R is a topological 4-manifold with a non-unique factorization as (compact 3-manifold) x R. (Is this what you meant, John?) --Dan

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[math-fun] Hodge conjecture
by Jon Perry 10 Oct '03

10 Oct '03

The Hodge conjecture, right, is nothingmore than visualising infinite space as a finite product of our imagination, given the neccessary and sufficient conditions that all is well and good with the owrld. In other words, every that will be will be, and nothing that won't be can't be. Or, we exit to think and thinking is existence. Personally, I see the Hodge conjecture as the rationalization of the human problem into reality, with the human influence removed. 7/10 this is correct, but it is the 3/10 times that this is incorrect that really buggers things up. Hopefully the second world in better than the first world. Peace to all hippies. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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Re: [math-fun] Rational Group
by asimovd＠aol.com 10 Oct '03

10 Oct '03

Jon Perry writes: << ... if we consider n-manifolds as spaces constructed from n-1-manifolds, then the n-manifold is formed by a continous expansion of varying n-1-mainfolds. ... >> There are any number of things that this could mean. Could you spell out what you mean by each of these two clauses? Thanks, Dan

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RE: [math-fun] The spatial universe is finite !!!???
by Thomas Colthurst 10 Oct '03

10 Oct '03

Perhaps the first thing to say is that this is experimental physics, so everything is model dependent and everything has error bars. For example, one of the central pieces of Weeks et al.'s evidence for the dodecahedral model are the low quadrupole and octopole WMAP amplitudes. But a recent paper by G. Efstathiou (http://xxx.lanl.gov/abs/astro-ph/0310207) shows that if you do a better statistical estimation of the poles that is more robust to foreground galaxy (aka the Milky Way) contamination, then the pole amplitudes are higher than originally estimated: ~200 uK^2 for the quadropole (vs. the original 123 uK^2) and ~1000 uK^s for the octopole (vs. the original 611 uK^2). [Naive explanation: if you do the simple thing and just "cut out" the Milky Way portion of the WMAP from the sky, you are going to heavily bias the low spherical harmonics toward low energies. The cleverer you are, the more you are going to remove that bias.] These numbers make both the dodecahedral model and the standard flat universe model fit the quadrople better (though the dodacehdral is still much closer), and makes the flat universe model fit the octopole better than the dodecahedral model. The probability of the data given the flat universe model used to be around 0.2%; now it's 3-4%. Depending on your priors for the two models, that difference can be a big deal. Another example: the WMAP data will never prove (in the Popperian sense) that the universe is R^3; in fact, that's the one and only geometry that it can't conclusively rule out! An R^3 universe will never show circles on the sky or other topological effects, and it's flatness will never be measured to be "1 exactly", merely as "1 + 1.x * 10^(-n) +/- y * 10^(-m)" for large n and m. Now, if your prior on the universe said that there was a 33% chance of flatness, then n need only be 2 or 3 before you are pretty sure that the universe is flat. On the other hand, if your prior said that curvature = 1 exactly was just as probable as curvature = 1.0000000007 exactly, then you still wouldn't know. The second thing to say is that a rebuttal to the docahedral model by Tegmark et al. is up on the arXiv: http://xxx.lanl.gov/abs/astro-ph/0307282 It's mostly about testing (and ruling out) a toroidal universe model with one dimension (in the direction of Virgo) smaller than the horizon scale. It's also the first circles in the sky analysis of the WMAP data that I've seen. The fact that it's been over half a year and the first analysis appeared just yesterday should give some indication of the computational and modelling dificulties involved: there is foreground noise again (that darn Milky Way!), the late integrated Sachs-Wolfe effect, the Doppler effect, detector noise, pixelization effects, six dimensional search space, ... Anyway, Tegmark et al. only look for "back-to-back" matched circles, and they don't find any. The Weeks et al. dodecahedral model predicts matched circles, so that should be enough to rule it out. Still, the difficulties of actually finding even matched circles is big enough that I wouldn't take this as the last word just yet. David Spergel of the WMAP team is also going to post a rebuttal, including the "definitive" circles in the sky analysis, on the arXiv soon, but it's not there yet. Finally, I should point out that the presence of a distinguished spatial section (aka, the big bang) makes it hard to physically motivate a factorization of large scale space-time into anything other than space times (some interval of R). I'm not saying it can't be done; I'm just saying it's hard to do in a way that a cosmologist would take seriously. My own question for math fun about manifold factorizations is this: let's say we take the string theorists seriously for a second and pretend that space is locally ten dimensional. Are there any "nice" ten dimensional manifolds that factorize "naturally" as a very very large dodecahedral (say) four manifold times a very very small Calabi-Yau six manifold? -Thomas C

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Re: [math-fun] The spatial universe is finite !!!???
by asimovd＠aol.com 10 Oct '03

10 Oct '03

Megachiropteran cinematographer: marchantiaceous centauromachias. Bedroom boredom? Brighten berthing! << Oh! -- but which direction you get rotated in invariant! That is, the Poincare Dodecahedron is not its own mirror image. So such a universe would have a large-scale handedness. Wouldn't that be neat... >> I have the impression that the Poincaré dodecahedral manifold D created by a 1/10 rotation identifying opposite faces, and the one using a -1/10 rotation, are isometric to each other -- much as a right- and a left-handed H opf fibration of the 3-sphere are isometric. What I'm wondering now is whether D double-covers a non-orientable manifold. * * * By the way, a cool thing about D is that it was the first counterexample to something Poincaré thought he had proved -- that any 3-manifold M with trivial (i.e., zero) first homology group H_1(M) must be homeomorphic to the 3-sphere. He discovered that the fundamental group pi_1(D), when abelianized (to obtain H_1(D)) becomes the zero group. So Poincaré, chastened, then suggested that *maybe* if the fundamental group of a 3manifold is trivial, *then* it must be homeomorphic to the 3-sphere. (He didn't actually couch this as a conjecture, but only as a possibility.) --Dan

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Antwort: Re: [math-fun] The spatial universe is finite !!!???
by dirk.lattermann＠bgs-ag.de 10 Oct '03

10 Oct '03

Hello, forgive my naive thoughts, but wouldn't the argument used in the article of wavelengths needing to fit in the space (don't even know if this is a commonly agreed one, even if it sounds plausible) prevent non-orientable finite spaces at all? At least, it would only allow an uneven number of zero-crossings of a wave, whereas an orientable finite space allows only an even number? Could this be astronomically observed somehow? (Just tried to draw some waves on a moebius band) Greetings, Dirk Lattermann [...] Boy, I'd be delighted if the universe were a Poicare Dodecahedron! (Though I'd been hoping for a nonorientable space: if Weeks is right, then you could go in a straight line for 74 billion light-years and get back to where you started, just rotated by 2pi/5. I'd been hoping you'd get back to where you started mirror imaged.) --Michael Kleber kleber(a)brandeis.edu _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] The spatial universe is finite !!!???
by Shel Kaphan 09 Oct '03

09 Oct '03

At 08:48 PM 10/9/2003 -0400, Michael Kleber wrote: >if Weeks is right, then >you could go in a straight line for 74 billion light-years and get back to >where you started, just rotated by 2pi/5. ... but after 74 billion years (minimum) you'd hardly recognize the place! In fact, doesn't it become a bit difficult even to define what is meant by "where you started"?

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