August 2006 - mathfun@mailman.xmission.com

[math-fun] continued subscripts
by Marc LeBrun 10 Nov '06

10 Nov '06

Recent correspondence has inspired me that, analogous to stacking superscripts for iterated exponentiation, we may stack the subscripts that are vulgarly used as "base" notation. For example, we may apprehend, among 239's many sterling properties, the following gratuity: 10 11 12 13 14 15 16 = 239. 17 18 19 20 21 22 23 That is, working up from the bottom: ... ... ... 21 = 21 = 97 and so on = 239. 22 48 23 Question: If stacking superscripts produces a "tower", what are stacked subscripts? A "dungeon"? The sequence following this incrementing pattern of subscripts, namely, 10, 10, 10, 10, ... 11 11 11 12 12 13 has (if I've computed them correctly) the values 10 11 13 16 20 25 31 38 46 55 65 87 135 239 463 943 1967 4143 8751 18479 38959 103471 306223 942127 2932783 9153583 28562479 89028655 277145647 861652015 2675637295 10173443119 41132125231 168836688943 695134284847 2862242032687 11778342480943 48433422101551 199016451894319 817199416306735 3353334654921775 16351027752823855 82924577766468655 423717750455364655 2167310727003204655 11083411175259204655 56654591244123204655 289463880726363204655 1478277273827163204655 7546178967779163204655 38504861079779163204655 227971995605219163204655 1387065053878499163204655 8475364910242019163204655 51807613088766179163204655 ... (which I will submit to the OEIS as soon as I can craft a concise explanation!). Puzzle: the least-significant digits seem to be stabilizing ... why or why not? Of course we can write the numerals in the stack in other bases, such as binary: 10, 10, 10, 10, ... 11 11 11 100 100 101 and these have as their first few values 2 3 5 26 1370 9840770 851566070376026 ... but it grows rapidly--the next two terms have ~144 and ~450 digits respectively (and seem beginning to show a distinct tendency to end with the digits ...26 ?!) Question: What is a good approximate formula for a(n)? And of course other continued subscript patterns besides simply incrementing the terms are possible.

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[math-fun] Q-factorials
by Schroeppel, Richard 29 Sep '06

29 Sep '06

I've been playing with Q-factorials: QF(0) == 1 QF(N+1) == QF(N) * (1 + Q + Q^2 + ... + Q^N). When Q=1, this gives QF(N) = N!. The degree of QF(N) is Triangle(N-1) = N(N-1)/2. Q-binomials seem to be polynomials: QB(A,B) = QF(A+B) / (QF(A) QF(B)). The degree of QB(A,B) is AB. The degree of QB(A,B,C) = QF(A+B+C) / (QF(A)QF(B)QF(C)) is AB+AC+BC. Etc. Perhaps we can use some of our standard bag of tricks for defining QF(N) for non-integer N. Put QF(Z) ~= Q ^ [ Z(Z-1)/2 ] for RealPart(Z) large, and divide by (Q^Z-1)/(Q-1) to go from Q(Z) to Q(Z-1). There's a set of functional equations similar to X! (X+1/2)! = (2X+1)! * stuff. Presumably there's also a reflection equation with QF(Z) QF(-Z) = more-stuff that looks like pi Z / sin(pi Z). Rich

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P.S. Re: [math-fun] Chinese Mathematicians Prove Poincare Conjecture
by dasimov＠earthlink.net 08 Sep '06

08 Sep '06

P.S. I did not see this in the Ximhua press release, but the Asian Journal of Mathematics says (in the link provided by Ray) that the article in question finishes the proof of the Thurston Geometrization Conjecture (of which the Poincare conjecture is just one of many important consequences). Up to now, a few media (Scientific American in 1994, New York Times in the past year) have claimed the Poincare conjecture has been proved, although no respectable professional journal has reported any such thing (only comments like "time will tell.") And not even media outlets have reported that the Geometrization conjecture was proved. Hmm. I guess time will tell. --Dan

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Re: [math-fun] Poincare Conjecture & Perelman
by Daniel Asimov 31 Aug '06

31 Aug '06

Now that the Poincare conjecture appears to be solidly proven, I'd like to see the known corollaries. Over the years a number of theorems have proved "Either there's a counterexample to the Poincare conjecture, or else something very strange happens in 4 or 5 dimensions" (with the something very strange being specified). Very weird stuff is already known to happen in 4 dimensions. E.g., it's the unique dimension whose Euclidean space R^4 carries more than one differentiable structure, up to equivalence (diffeomorphism). There is a continuum of them! Also, given any finitely-presented group G, there is a 4-dimensional manifold fundamental group M whose fundamental group pi_1(M) is isomorphic to G (using topological operations that mimic the presentation using glueing of 1-handles & 2-handles, then capping it off with a conractible 4-manifold with the same boundary -- I think). Since a theorem of Markov says there is no algorithm which, given two arbitrary finite group presentations, can determine if they define isomorphic groups, it follows that there is no algorithm to classify all (even compact) 4-manifolds up to homeomorphism. (Okay, okay, this is also true in dimensions > 4, but 4 is the lowest case where it happens.) --Dan

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Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 31 Aug '06

31 Aug '06

Ed writes: << At a convention, 128 people all need to meet with each other, 1 on 1, for a paper exchange. Suppose a room is divided into a 8x8 square, and a clock is set up to ding 127 times, once every 30 seconds. Each person is given a map of their starting square, and a map of their route over the next hour and 4 minutes. Only two people can be in each cell. Ideally, most moves should be King moves. Not needing to move during a turn is acceptable. Simplicity is judged by the amount of text needed to explain the choreography, and the orderliness of the room. Alternately, the length of the maximum walk should be minimised. >> This is a lovely problem, even if we're still discussing its precise definition. I'm partial to the triangular lattice since it's the only symmetric 2D lattice such that any intersecting pair of Voronoi regions (in this case closed hexagons) intersect along an edge. (Though there is no necessary reason the network need be 2D; there is a natural generization in each dimension; in 3D it's the tiling by truncated octahedra). And I'm notorious in these quarters for preferring a torus over a bounded region, to avoid those pesky edge effects. Imo the most symmetric such tori of all are those arising from an iterative tiling by rosettes of 7 (of the preceding rosette, the first one just being 6 hexagons around 1 of them). Thus the number of nodes in these cases is a power of 7: 7, 49, 343, 2401, 16807, . . .. A torus of 49 seems a nice place to start. (Though 49 is clearly not as friendly a number as 120, which is my guess why Ed wanted to use that one -- its being both 5! and T_15.) Another nice feature of 7^n tori like this is that their network can be colored with 7 colors in a highly symmetric repeating pattern. A compromise would be considering the next most symmetrical such (triangular-lattice-based) tori, which imo are those which arise from an n x n rhombus; the nicest of reasonable size might be 12 x 12, since it's a both a square and a Fibonacci number, and also has more factors than any nearby number other than 120. (All these 60-deg. rhombic tori are in a strong sense "hexagonal" too, but the hexagon doesn't interact as nicely with the network when the # of nodes is not a power of 7.) But I'm interested in starting with the highly symmetric 7 x 7 hexagonal torus network of 49 nodes, each connected to 6 others. We can ask this question: Given 98 people, two on each node of this network at each stage, we want each one to have an itinerary (moving to an adjacent node, or staying put, on each step) such that each person shares a node with each of the other 49, such that the least total number of moves-to-an-adjacent-node are used. (Making no assumptions about the number of steps necessary.) This is clearly only one of many possible ways the details can be made precise. --Dan

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Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 31 Aug '06

31 Aug '06

<< I'm interested in starting with the highly symmetric 7 x 7 hexagonal torus network of 49 nodes, each connected to 6 others. We can ask this question: Given 98 people, two on each node of this network at each stage, we want each one to have an itinerary (moving to an adjacent node, or staying put, on each step) such that each person shares a node with each of the other ***49***, such that the least total number of moves-to-an-adjacent-node are used. (Making no assumptions about the number of steps necessary.) >> (asterisks added for emphasis) Of course, silly me, this should read 97, not 49. Gareth: Your solution looks promising, but I'm not sure exactly what you mean by n. --Dan

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OOPS Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 30 Aug '06

30 Aug '06

I see from the Subject line that Ed intended for one step to be to any other of the 64 squares, so I retract that question. --Dan

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Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 30 Aug '06

30 Aug '06

Ed asks: << At a convention, 120 people all need to meet with each other, 1 on 1. Suppose a room is divided into a 8x8 square, and a clock is set up to ring 60 times, once per minute. Each person is given a map of their starting square, and a map of their route over the next hour. What is the simplest set of maps? What is the length of the minimal walks people need to take? >> This is an intriguing question! One thing I'm not sure I understand is what kind of step one is allowed to take. Must it be N,S,E,W ? Or does it also include NE, SE, NW, SW ? Another question is how would one meet 119 other people in only 60 steps? I must be misunderstanding something. (Or is the real question how each person could meet 60 others in 60 steps?) A third question I have is: Is it obvious that (whatever the goal) there is necessarily some way to achieve it in the allotted time? ------------------------------------------------------------------------------------ I think that at any stage there are supposed to be two people in each of 60 squares, with 4 empty squares. (Is that right, Ed?) (Another possible avenue is to use a symmetrtic network with exactly 120 nodes, like T_15.) --Dan

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Re: [math-fun] Choreography of large complete graphs
by Jpr2718＠aol.com 30 Aug '06

30 Aug '06

Ed Pegg Jr. wrote: > At a convention, 120 people all need to meet with each other, 1 on 1. > Suppose a room is divided into a 8x8 square, and a clock is set > up to ring 60 times, once per minute. Each person is given a > map of their starting square, and a map of their route over the next > hour. Wouldn't each person need to have 119 meetings? Is it required that there are 60 meetings at a time (maybe this is optimal anyway)? John Robertson

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Re: [math-fun] Possibly naive question about algebraic numbers
by Daniel Asimov 29 Aug '06

29 Aug '06

Thanks a lot, Franklin and Gareth. I suspected that conditions (*) and (**) together would result in algebraic coordinates, mainlly because I'd expect to have heard of these numbers if they defined a larger class. The theorem Gareth proves seems very basic, so I wonder if it has a name and attribution. Also, given such an integer-polynomial function P: C^n -> C^n such that its Jacobian at each of its roots Z in C^n is non-singular: QUESTION: What can be said about the size of the (necessarily) 0-dimensional set of such roots? (Where of course a root of P means a Z in C^n where P(Z) is the origin in C^n.) --Dan

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