May 2019 - mathfun@mailman.xmission.com

Re: [math-fun] DGF --> OGF?
by Dan Asimov 08 May '19

08 May '19

Does an "arithmetic function" take its values in Z ? If so, there will be lots of Dirichlet series that have non-integral coefficients, the simplest of which might be A(s) = (1/2) 1/2^s = 1/2^(s+1). Note that if a Dirichlet series with real coefficients equals the 0 function, then all its coefficients are 0. So A(s) above won't have any *other* such series than 1/2 * 1/2^s. It seems like a good question to ask which functions A(s) *do* have a Dirichlet series with integral coefficients, but hard. It's interesting that the set D_I of all Dirichlet series with integer coefficients are closed under addition and multiplication, so form a subring of the ring D formed by all such series. Then we can say the "obstruction" to a function A(s)'s Dirichlet expansion having integer coefficients is the image phi(A(s)) in the quotient ring D / D_I by the quotient map phi : D —> D / D_I. —Dan Mike Stay wrote: ----- Can every suitably nice function A(s) be "inverted" to give an arithmetic function a(n) such that A(s) = sum_{n >= 1} a(n)/n^s? (E.g. by Perron's formula?) In other words, given ordinary generating functions A_i(x) = sum_{n >= 0} a^i_n x^n, I can combine them into a Dirichlet generating function A(s) = prod_{i >= 1} A_i(p_i^{-s}). My question is whether given a suitably nice but otherwise arbitrary function of s and think of it as a Dirichlet generating function, can I decompose it into a collection of ordinary generating functions? If so, what constitutes "suitably nice"? Merely convergence for Re(s) > sigma? -----

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[math-fun] Chess questions
by Dan Asimov 08 May '19

08 May '19

Someone told me recently that the traditional estimate of the size of [any advantage that White may have over Black] has decreased of late. He even suggested that some think *Black* has an advantage. Googling suggested the answers might be findable only with a lot more research, so I'm just asking y'all. 1. Does anyone know about this? What are the White/Black win/loss percentages when a top chess engine like AlphaZero plays itself? His point was that recent computer games have altered prior perceptions. 2. I also wonder what the latest belief is on whether either player has a winning advantage with optimal play, or whether optimal playon both sides can lead only to draws. Likewise, statistics from top chess engine self-play might be revealing. —Dan

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Re: [math-fun] Suppose you have N congruent non-overlappin ellipses in a circle ...
by Dan Asimov 07 May '19

07 May '19

It turns out that the largest ellipse in any rhombus always occupies π/4 of the area of the rhombus. (For, mapping the rhombus linearly onto a square multiplies all areas by the same factor, so the inverse mapping carries the maximum ellipse in the square — namely, a circle — onto the maximum ellipse in the rhombus. The fraction of area covered is the same in both cases.) So the N=3 case, the three 60º-120º rhombi of side = 1 in the unit disk D can cover the fraction π/4 of the area of the inscribed hexagon, itself having area = 6 sqrt(3)/4 = sqrt(27/4), for a total area of π*sqrt(27/64) and a total ratio of (area of 3 ellipses / area of circle) = sqrt(27/64) = 0.6495190+. This *might* be the maximum for N = 3. Meanwhile as n —> oo, the fraction —> π/sqrt(12) = 0.9068996+ (approaching e.g. a triangular arrangement of circular disks). So the fraction must increase eventually. —Dan I wrote: ----- 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... -----

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Re: [math-fun] Suppose you have N congruent non-overlappin ellipses in a circle ...
by Dan Asimov 07 May '19

07 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 06 May '19

06 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 03 May '19

03 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 03 May '19

03 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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