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[math-fun] primes in geometric progression and one conjecture
by Simon Plouffe 03 Jun '20

03 Jun '20

Hello, sometime ago, someone came up with 26 primes in arith. progression. The calculation was hard and took 6 months of intense CPU time. https://en.wikipedia.org/wiki/Primes_in_arithmetic_progression But what if the primes are in geometric progression instead like a(n) = { c^n } , where { } is the nearest integer. For example if c = 2.553854696... then the primes are : 3, 7, 17, 43, 109, 277, 709. Only 7 of them. Can we find better ? By using a Monte-Carlo + Simulated Annealing method, one can find much more. Here is a small table : Value of c : range : number of primes generated 2027.167168476491219434395... n = 1..97 : 97 primes 31622.7767185595693… n = 2..388 : 387 primes (see A332308) 55237.07504296764715433124 n = 2 ..633 : 632 primes (see A333127) A conjecture is then : if c is large enough, arbitrary long chain of primes can be generated. The data about this is here : http://plouffe.fr/NEW/ I could not find any mention of <primes in geometric progression> in literrature. Best regards, Simon Plouffe

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[math-fun] I have a function (nearly debugged) to convert an arbitrary composition of Möbius transformations
by Bill Gosper 03 Jun '20

03 Jun '20

into a non-regular continued fraction. E.g., threewise grouping the continued fraction for e, 1,0,1,1,2,1,1,4,1,1,6,1 ... gives the general matrix product Product[{{2 + 2 n, 1 + 2 n}, {1 + 2 n, 2 n}}], {n, ∞}] which converts to the non-regular continued fraction 2 - ContinuedFractionK[(2 n - 3) (2 n + 1), 8 n^2 - 4, {n, ∞}] == E which must converge exactly thrice as fast as the regular one: In[82]:= ContinuedFraction[%[[1]] /. ∞ -> 9] Out[82]= {2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, 14, 1, 1, 16, 1, 1, 19} In[83]:= Length@% Out[83]= 27 (Note that 19 = 18+1/1 .) This suggests that non-regular continued fractions could be abandoned in favor of matrix products. (Which will require the cooperation of computer algebraists. E.g., Mathematica has no non-commutative Product (E.g., Dot[...,{n,1,∞}]) expression. It doesn't even have Table (List) over a symbolic range.) —rwg

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Re: [math-fun] Anti-kissing number puzzle
by Dan Asimov 02 Jun '20

02 Jun '20

Thanks, Veit. I agree that "anti-kissing number" is the opposite of sexy. Did not know the word "saturated" in this context. Maybe n-dimensional (or nD) saturation number is good terminology. As for density, I am very impressed by how soon the density gets teensy. My calculation of the density for the 196560 23-spheres kissing the central 23-sphere in R^24 is roughly 0.0021977474957616015403222560573790967651611948928748253205373884442323921+, or in other words those 196560 tightly kissing sphere obscure barely over 1/500 of the surface of the 23-sphere. (Thanks, Wolfram Alpha online.) I'd appreciate if someone would care to check me on this calculation. —Dan Veit Elser wrote: ----- A configuration of unit spheres, either in Euclidean space or kissing a given unit sphere, is called “saturated” when it is not possible to add another unit sphere without intersecting one of the spheres. Torquato and co-workers have speculated that “random” saturated packings (sequentially select positions from the uniform distribution on the remaining volume) achieve the *densest* possible packing in the limit of high dimension. I do not know if anyone has looked into your variant of saturated packing, where one is interested in the *lowest* density (fewest number, in the kissing case). I doubt “anti-kissing number” will catch on. Maybe “blocking number”? The corresponding property of a graph is the “minimum size of a maximal independent set”. A(3) <= 6. When you place 6 spheres at the vertices of the regular octahedron no additional spheres can be placed since the positions closest to the central sphere are at the 8 octahedron faces. But these cannot be kissing since we know a 14 kissing configuration does not exist. I’ll guess A(3) = 6, because the most symmetrical 5-sphere configuration — the triangular bi-pyramid — does not work (3 spheres can be added at the equator). -Veit > On Jun 1, 2020, at 7:07 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > > The kissing number in n dimensions is the maximum number K(n) of non-overlapping unit spheres that can be placed tangent to the one centered at the origin. > > Known kissing numbers: > > dimension kissing number > —————————————————————————— > 1 2 > > 2 6 > > 3 12 > > 4 24 > > 8 240 > > 24 196560 > > No other kissing numbers are known. > > A related problem is to find the "anti-kissing number" in each dimension n: > the smallest number A(n) of non-overlapping unit spheres in n-space, all tangent to the unit sphere centered at the origin, so that there's no room for any additional non-overlapping unit spheres tangent to the central one. > > It's obvious that A(1) = 2 and easy to show that A(2) = 4. > > Puzzle: What is A(3)? -----

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[math-fun] Anti-kissing number puzzle
by Dan Asimov 02 Jun '20

02 Jun '20

The kissing number in n dimensions is the maximum number K(n) of non-overlapping unit spheres that can be placed tangent to the one centered at the origin. Known kissing numbers: dimension kissing number —————————————————————————— 1 2 2 6 3 12 4 24 8 240 24 196560 No other kissing numbers are known. A related problem is to find the "anti-kissing number" in each dimension n: the smallest number A(n) of non-overlapping unit spheres in n-space, all tangent to the unit sphere centered at the origin, so that there's no room for any additional non-overlapping unit spheres tangent to the central one. It's obvious that A(1) = 2 and easy to show that A(2) = 4. Puzzle: What is A(3)? —Dan

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Re: [math-fun] Anti-kissing number puzzle
by Dan Asimov 01 Jun '20

01 Jun '20

Aha. My copy (3rd edition) is not at hand. Can you please mention the page and/or chapter where that's found? —Dan ----- Conway's Sphere Packing, Lattices and Groups has an interesting (and surprising!) discussion of this particular case. -----

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[math-fun] The punctured torus
by Dan Asimov 31 May '20

31 May '20

It's well known that among the compact orientable surfaces (without boundary), the sphere S^2 can be given a metric of constant curvature K only if K > 0; the torus T^2 can be given one only if K = 0; and the surfaces of genera >= 2 can be given one only if K < 0. I had rarely thought about non-compact surfaces in such terms, but for these one would look for not just any metric of constant curvature but one that is *complete* — meaning that every geodesic curve can be extended indefinitely without "going off the edge" of the surface. So today I was mildly surprised to see that the torus with one puncture (or what's the same topologically, the torus with a disk removed) can be given a complete metric of constant negative curvature (say K = -1). What's more, this can be tiled with exactly two ideal triangles (<https://en.wikipedia.org/wiki/Ideal_triangle>). Which is pretty, since all ideal triangles are congruent. It cannot have a complete metric of constant 0 curvature, even though the unpunctured torus does, easily. Surprisingly, the cylinder (without boundary) can be given a complete metric of constant curvature either if K = 0 or else if K < 0 — in which case it, too, can be tiled with two ideal triangles. (Same goes for the Möbius band.) —Dan

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[math-fun] Chiral Genesis
by Brad Klee 31 May '20

31 May '20

See also: https://www.inaturalist.org/observations/47995301 . --Brad

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[math-fun] (2,3) and (3,4) Torus Knot Certificates
by Brad Klee 30 May '20

30 May '20

https://0x0.st/ifEP.txt https://0x0.st/ifEX.txt Meta-Data, by (line #): (1) (p,q) torus knot Parametric Eqs, (2) Z-linear surface, degree q, (3) Z-quadratic surface, degree 4, (4) dt/d\phi, (5) Annihilator of dt/du, an ODE in k, (6) Annihilator Certificate, A[dt/du]-d/dt(cert)=0. Imposing Symmetry leaves one degree of freedom. After trying many different constraints, I found that allowing, in fact requiring, an extra point (X,Y,Z)=(0,0,0) leads to smallest known proof data, see linked certificates above. For the first two cases (p,q) the Z-linear surface had degree q, and the ODE has q terms. This suggests apparent genus 1 and 2 for the first two torus knots, (2,3) and (3,4). When measuring genus by period ODE, it often happens that apparent genus is less than real genus due to hidden symmetry. I don't know about slice genus, but according to the proved Milnor conjecture, the numbers are 1 and 3, see corollary 1 of: https://arxiv.org/abs/math/0402131 I doubt that any representation of the stevedore knot would have a rational period function, but it is Lissajous, so it might possible to check. Unfortunately, I don't have anymore time this week. --Brad

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Re: [math-fun] Knot Problem
by Steve Witham 30 May '20

30 May '20

This has precious* little to do with Dan's method, but he spurred me locate the Grapher app on my computer (Utilities folder??), take a demo that comes with it (Torus-knot), turn it into a trefoil knot, and then fix the graphics that got broke. This page shows the settings as well as the picture: http://www.mac-guyver.com/switham/2020/05/Torus_knot <http://www.mac-guyver.com/switham/2020/05/Torus_knot/index.html> (As far as I know, the only way to get equations into Grapher is with its own editor. Their file format looks opaque. It can apparently import point sets, though...) --Steve *my precious. Fit for an elven king. Btw, summary of _The Hobbit_: Marty Feldman gets hold of an all-access key card to Cheyenne Mountain. > Dan Asimov <dasimov(a)earthlink.net> > Date: 5/25/20, 5:45 PM > > I tried to find equations in R^3 for the trefoil defined initially in > R^4 = {(x,y,z,w)}: > > (x+iy)^2 + (z+iw)^3 = 0 > > and > > x^2 + y^2 + z^2 + w^2 = 1 > > but which is then stereographically projected into > R^3 = {(X,Y,Z)} via > > X = x/(1-w), Y = y/(1-w), Z = z/(1-w). > > If my algebra is right, the final (two) equations are > > 4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 > & > 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0 > > where W is short for (X^2 + Y^2 + Z^2 - 1). > > I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher". > > Â—Dan > >

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Re: [math-fun] "Wrong Times Table"
by Dan Asimov 30 May '20

30 May '20

Who dat? —Dan ----- Joe Soap ... -----

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