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Re: [math-fun] Suppose you have N congruent non-overlappin ellipses in a circle ...
by Dan Asimov 06 May '19

06 May '19

I'm still not sure, but it looks as though for N = 3 ellipses in a unit disk if you chop the inscribed hexagon into 3 60º - 120º rhombi of side = 1, then possibly the largest ellipses in such a rhombus add up to more area than the largest congruent circular disks that can all 3 exist disjointly in D. But that *might* be the only case... —Dan James Buddenhagen wrote: ----- I don't seem to have good intuition on this, but do you have an example where N congruent ellipses covers more than N congruent circles? What about very eccentric ellipses whose major axes are like spokes of a wheel? On Mon, May 6, 2019 at 12:44 PM Dan Asimov <dasimov(a)earthlink.net> wrote: > I probably should have asked if E(N) is monotonic *decreasing*, > since E(1) = 1 and for N > 1, E(N) < 1. But after N = 2 it's > hard to see what happens. > > Also, for N large almost all ellipses will not be near the boundary > circle, so it seems likely that as N —> oo the configurations will > approach close-packed disks in the plane, so the density will approach > π/sqrt(12). > > —Dan > > ----- > Let E(N) be the maximum possible fraction of a 2-disk D's area > that is occupied by N non-overlapping, congruent ellipses in D. > > OK, the supremum. > > E(N) seems pretty hard to determine explicitly for each N. But > one thing that seems not immediately obvious to me is: > > Question: > --------- > Is E(N) an increasing function of N ??? > ----- -----

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Re: [math-fun] Suppose you have N congruent non-overlappin ellipses in a circle ...
by Dan Asimov 06 May '19

06 May '19

I probably should have asked if E(N) is monotonic *decreasing*, since E(1) = 1 and for N > 1, E(N) < 1. But after N = 2 it's hard to see what happens. Also, for N large almost all ellipses will not be near the boundary circle, so it seems likely that as N —> oo the configurations will approach close-packed disks in the plane, so the density will approach π/sqrt(12). —Dan ----- Let E(N) be the maximum possible fraction of a 2-disk D's area that is occupied by N non-overlapping, congruent ellipses in D. OK, the supremum. E(N) seems pretty hard to determine explicitly for each N. But one thing that seems not immediately obvious to me is: Question: --------- Is E(N) an increasing function of N ??? -----

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[math-fun] Suppose you have N congruent non-overlappin ellipses in a circle ...
by Dan Asimov 05 May '19

05 May '19

Let E(N) be the maximum possible fraction of a 2-disk D's area that is occupied by N non-overlapping, congruent ellipses in D. OK, the supremum. E(N) seems pretty hard to determine explicitly for each N. But one thing that seems not immediately obvious to me is: Question: --------- Is E(N) an increasing function of N ??? * * * And in more generality, ask the same question for the supremum over all N non-overlapping, congruent, ellipsoids inside a d-dimensional disk D (bounded by a (d-1)-dimensional sphere): ----- If E_d(N) is the supremum, over all such arrangements, of *the fraction of D's area lying inside any of the N ellipsoids* then: Question_d: ----------- Is E_d(N) a strictly increasing function of N ??? ----- If this is obvious, I'm not sure why yet. —Dan

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[math-fun] Integers of the form K*K +K*L + L*L
by Dan Asimov 05 May '19

05 May '19

Integers of form K*K + K*L + L*L arise a lot in connection with the triangular lattice, and so with the ring Z[tau], tau = exp(2πi/6) whose elements form a triangular lattice. A particularly nice way they arise is for hexagonal tessellations of the hexagonal torus T_H. T_H is defined by identifying opposite sides of a regular hexagon H. A hexagonal tessellation of T_H is just a tiling of T_H by congruent regular hexagons, as usual required to meet vertex to vertex and edge to edge. Then: ----- The number N is the number of tiles in some hexagonal tessellation of T_H if and only if For some K, L in Z: N = K*K + K*L + L*L. ----- Basic question: --------------- I made a table of the Loeschian numbers Lo(K,L) for K >= L. and then tried to *connect each one with the next largest one* with a line segment on paper. But it was hard to see any pattern in how they connect (below). So the question boils down to: Is there a concise formula for the "next largest number" in a set like Lo = {Lo(K,L) | K, L in Z} ??? I.e., f(0) = 1, f(1) = 3, f(3) = 4, ... in 0—1—3—4—7—9—12—13—16—19—21—25—27—28—31-36-39-43-48—49—.... ??? K| 0 1 2 3 4 5 6 7 8 9 10 L | ——————————————————————————————————————————————— 0 | 0 1 4 9 16 25 36 49 64 81 100 | 1 | 3 7 13 21 31 43 57 73 91 111 | 2 | 12 19 28 39 52 67 84 103 124 | 3 | 27 37 49 63 79 97 107 129 | 4 | 48 61 76 93 112 133 156 | 5 | 75 91 109 129 151 175 | 6 | 108 127 148 171 196 | 7 | 147 169 193 219 | 8 | 192 207 224 | 9 | 243 261 | 10 | 300 —Dan ————— PS The set Lo+ of numbers that are represented more than once by K*K + K*L + L*L (e.g. 49) has a particularly peculiar definition: OEIS A118886 says Lo+ consists of integers whose factorization into prime factors contains * at least two factors equal to 1 modulo 6, and * an even number of factors equal to -1 modulo 3.

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Re: [math-fun] Smullyan paradox about belief
by Dan Asimov 02 May '19

02 May '19

What's the mathematical version of this question? Because anyone can be mistaken about anything, including their own beliefs, it would seem. —Dan Jim Propp wrote: ----- I recall encountering in one of Raymond Smullyan’s books a thought experiment that convinced me that it is possible to be mistaken about what one believes. That is, one can say “I believe X” and be speaking falsely even though one is not intentionally lying. Does anyone know what thought experiment I’m dimly recalling, and which of his books it appears in? -----

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[math-fun] I'm a grampa
by David Wilson 02 May '19

02 May '19

Forgot to tell all my math friends, I am now a grandfather. August Scott Moreno, born April 27, 2019.

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[math-fun] Factoring near-factorials
by Tomas Rokicki 02 May '19

02 May '19

So, I recently came across this: 4!+1 = 5^2 5!+1 = 11^2 7!+1 = 71^2 Is there any other n such that n!+1 is a perfect square? A perfect square of a prime? Also, n!+1 seems to generate a fair number of primes. For instance, 73!+1 appears to be prime. But this is apparently well known. -- -- http://cube20.org/ -- http://golly.sf.net/ --

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[math-fun] Smullyan paradox about belief
by James Propp 02 May '19

02 May '19

I recall encountering in one of Raymond Smullyan’s books a thought experiment that convinced me that it is possible to be mistaken about what one believes. That is, one can say “I believe X” and be speaking falsely even though one is not intentionally lying. Does anyone know what thought experiment I’m dimly recalling, and which of his books it appears in? Thanks, Jim Propp

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Re: [math-fun] Cubulated three-sphere puzzle
by Dan Asimov 01 May '19

01 May '19

Now that I've thought about it for a moment, it's clear that even the condition (*) (that each k-face is determined uniquely by its set of vertices) is too weak. Two squares could each share 3 of their vertices with the other. For a change I will attempt to hold my peace while thinking. —Dan

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Re: [math-fun] Cubulated three-sphere puzzle
by Dan Asimov 01 May '19

01 May '19

Good points, Tom and James! Let X (like the S^3 in question) be the abstract union of unit n-dimensional cubes Q = [0, 1]^n, where any two intersect only along entire common k-faces, for some k. Let Vert(X) denote its set of vertices (i.e., coming from the n-cubes Q). By definition, a "subcube" of X is any k-dimensional cubical face of one of its component n-cubes Q, for any k. I've been trying to think of what additional constraint(s) will fix this problem of too many examples. A natural condition is to require that: (*) Any subcube is uniquely determined by its vertices. (See below for another related condition.) Assuming this condition: ------------------------ This rules out the examples of Tom and James, and leads to the 6 square faces of a cube as the unique smallest possible cubulated S^2. (Sorry for having overlooked the need for this, in my headlong, willy-nilly rush to post my half-baked idea. I hope this time it's at least 3/4-baked.) ----- Question: --------- Under the additional constraint of I) above, what is a (the?) cubulated 3-sphere S^3 with the fewest cubes Q = [0, 1]^3 ??? ----- —Dan ————— *PS Also consider this condition: (**) X embeds in some Euclidean space R^n with Vert(X) = X ∩ Z^n. I.e., there is a mapping f : X —> R^n that is a topological equivalence with its image f(X), such that the vertices of X are precisely X ∩ Z^n. It seems clear to me that (**) implies (*). --------- QUESTION: Does (*) imply (**) ??? --------- —D. --------------------------------------------------------------------------- James Buddenhagen wrote: ----- More generally for 2D squares, let n be any positive integer greater than 2. Form a (say regular) n-gon as an equator of S^2 and join all vertices to both the north and south pole. Now erase the equatorial polygon and you are left with a topological S^2 composed of n topological squares. On Tue, Apr 30, 2019 at 10:38 PM Tom Karzes <karzes(a)sonic.net> wrote: > For the case with 2D squares, can't you join three squares at a common > vertex, as in the corner of a cube, and then do the same thing with the > opposite corners to obtain a sphere from 3 squares? > > Also, can't you join two squares at an edge to obtain a rectangle, then > join two rectangles (as in the two square case) to obtain a topological > sphere from 4 squares? Or is there a constraint that I missed that > these examples don't satisfy? > > Tom > > Dan Asimov writes: > > Suppose the sphere S^2 is made of (2D) square pieces of rubber sewn > together > > abstractly along common edges. (I.e., the result is topologically > equivalent > > to a sphere.) > > > > According to this rule, there is a trivial case where only 2 squares are > > sewn together, by identifying corresponding edges — this is indeed a > > sphere topologically. The fact that [in 3-space any two planar squares > > with the same boundary are the same] is not our concern. > > > > It's seems clear that — except for this trivial example — the smallest > number > > of squares in such a thing is 6. > > > > Now suppose the 3-sphere S^3 = {x in R^4 | ||x|| = 1} is, similarly, > > built abstractly out of (solid) cubes, copies of Q = [0,1]^3, that are > allowed to intersect > > only along entire common square faces, edges, or vertices. > > > > Puzzle: > > ------- > > What is the smallest number of cubes with which it's possible to build > something > > topologically equivalent to S^3 this way? > > ------ > > > > I think I know, but I haven't tried to prove it yet. -----

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