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[math-fun] The minimal polynomial for e^(i π/170) has a "balanced ternary" coefficient list.
by Bill Gosper 10 May '19

10 May '19

In[87]:= CoefficientList[MinimalPolynomial[E^(I π/170), x], x] Out[87]= {1, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, -1, 0, -1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, -1, 0, -1, 0, -1, 0, -1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, -1, 0, -1, 0, -1, 0, -1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, -1, 0, -1, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1} (Presumably unexceptional.) In[88]:= FromDigits[%, x] (1 - x^2 + x^4 - x^6 + x^8 + x^34 - x^36 + x^38 - x^40 + x^42) // Expand Out[88]= 1 + x^170 —rwg

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[math-fun] Fwd: Re: 170
by Bill Gosper 09 May '19

09 May '19

Interestingly, In[78]:= Arg[(((Sqrt[5] + 1)*(-Sqrt[Sqrt[17] - 3]* Sqrt[4*Sqrt[17] - 2*Sqrt[2*Sqrt[17] + 34]] + Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 1)/ 64) - ((Sqrt[5 - Sqrt[5]]* Sqrt[Sqrt[Sqrt[17] - 3]* Sqrt[2*Sqrt[2*Sqrt[17] + 34] + 4*Sqrt[17]] - Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 17])/16))* I + (((Sqrt[5] + 1)* Sqrt[Sqrt[Sqrt[17] - 3]* Sqrt[2*Sqrt[2*Sqrt[17] + 34] + 4*Sqrt[17]] - Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 17])/(16* Sqrt[2])) + ((Sqrt[ 5 - Sqrt[5]]*(-Sqrt[Sqrt[17] - 3]* Sqrt[4*Sqrt[17] - 2*Sqrt[2*Sqrt[17] + 34]] + Sqrt[2*Sqrt[17] + 34] + Sqrt[17] + 1))/(32*Sqrt[2]))] // FullSimplify // tim During evaluation of In[78]:= 9.430501 [seconds], 2 [multiplicands] Out[78]=π/170 is a constructible angle, whereas In[79]:= UnitConvert[Quantity[π/180, "Radians"], "Degrees"] /.n_Integer :> IntegerName@n Out[79]= Quantity["one", º] is not. —rwg -------- Original Message -------- Subject: Re: [math-fun] 170 Date: 2019-05-08 08:50 From: James Propp <jamespropp(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> That's perfect! I see that in particular the Gray-code-based puzzle "The Brain Puzzler" takes 170 moves to solve: https://everything2.com/title/The+Brain+Puzzler Thanks, Michael! Jim On Wed, May 8, 2019 at 11:06 AM Michael Kleber <michael.kleber(a)gmail.com> wrote: > Membership in http://oeis.org/A000975 is definitely the way to go. The > description of the sequence makes it clear that it's just saying 170 > is 10101010 in binary, but the comments section offers plenty of > examples > of cool things about those alternating-0s-and-1s numbers. > > --Michael > > > On Wed, May 8, 2019 at 10:41 AM Mike Stay <metaweta(a)gmail.com> wrote: > > > http://oeis.org/search?q=170 > > > > On Wed, May 8, 2019 at 8:29 AM James Propp <jamespropp(a)gmail.com> wrote: > > > > > > I’ve agreed to host Carnival of Mathematics #170 next month, and I’m > > > wondering whether the number 170 has any especially nice properties > that > > > might deserve mention. > > > > > > The number 170 has its own Wikipedia page, but none of the mathematical > > > properties listed there appealed to me. > > > > > > I might say a little about writing numbers as sums of two squares, but > > I’d > > > prefer something a little more exciting. > > > > > > Jim Propp > > > ________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > -- > Forewarned is worth an octopus in the bush.

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[math-fun] Egan's _Dichronauts_
by Keith F. Lynch 09 May '19

09 May '19

I've mentioned Greg Egan's "Orthogonal" trilogy (_The Clockwork Rocket_, _The Eternal Flame_, and _The Arrows of Time_) in which the interval between two events isn't x^2 + y^2 + z^2 - t^2 as in our universe, but x^2 + y^2 + z^2 + t^2. In other words, there is no distinction between the time and space dimensions. The perceived time dimension is simply the direction of your worldline, which can point in literally any direction. As in our universe, people traveling at close to the same speed will almost perfectly agree on the direction of the time axis. He has since written _Dichronauts_, in which the interval between two events is x^2 + y^2 - z^2 - t^2. Roughly speaking, there are two space axes and two time axes. But people experience linear time, just as in our universe and in the "Orthogonal" universe. The main difference from our universe is that the three axes that aren't your personal time axis don't all act alike. For instance objects can't rotate around a vertical axis by 45 degrees or more. Smaller rotations result in the object elongating. Some people are born facing east, others facing west. To look in the opposite direction you need to bend way forwards or way backwards. There would be no point in facing north or south even if you could, since no light travels in those directions. Also, energy works rather differently than in our universe. See http://www.gregegan.net/DICHRONAUTS/00/DPDM.html for more information on the physics. There's a similar page for the physics of the Orthogonal trilogy, http://www.gregegan.net/ORTHOGONAL/00/PM.html I could be wrong, but I think that swapping all the signs makes no difference, so there are only the three possibilities, all the same signs (Orthogonal), three and one (our universe), and two and two (Dichronauts). Those novels are perfect for people who enjoyed learning physics so much that they wish they could do it all again.

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Re: [math-fun] Impossible triangle raytracer
by Dan Asimov 09 May '19

09 May '19

The impossible triangle most definitely exists, in that there is a solid object (many!) in space that looks exactly like the drawings of it (in one view). —Dan ----- how can you find the intersection of a ray and an object that can't exist? -----

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[math-fun] ROITMAN: Zen math will never add up.
by bradklee＠gmail.com 09 May '19

09 May '19

https://www.lionsroar.com/it-will-never-add-up/ Re: Kong-an about dog—does Buddha nature have a dog? And if yes, does Buddha nature have a Viola Cryana? I am worried, very worried, terribly worried, about what the latest ipbes Report on biodiversity implies to our understanding of this “Shunyata”. d/dx F(x)=-k*F(x) — BrAd.XR

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[math-fun] DGF --> OGF?
by Mike Stay 09 May '19

09 May '19

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? -- Mike Stay - metaweta(a)gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com

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Re: [math-fun] Integers of the form K*K +K*L + L*L
by Keith F. Lynch 08 May '19

08 May '19

Dan Asimov <dasimov(a)earthlink.net> wrote: > 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, .... I wouldn't think so. There are lots of examples of two-dimensional integer sequences, aka tables, where there's a simple pattern to every row and to every column, but no pattern to all the numbers sorted in order. Examples include: * The multiplication table (excluding the 1s row and column). The entries are the composite numbers (A002808), which are of course exactly as patternless as the primes. * Power numbers (A001597), i.e. the union of the squares, cubes, etc. Only in 2002 was it even proven that 8 and 9 are the only adjacent members of the set. * Numbers in Pascal's triangle (other than the first two rows and columns), aka the binomial coefficients (A006987). * Nontrivial polygonal or figurate numbers, i.e. the union of the triangular numbers (other than 1 and 3), the squares (other than 1 and 4), the pentagonal numbers (other than 1 and 5), etc. (A090466). * Numbers that are a power of 2 plus a power of 3 (A004050). Similarly with any other two primes. * Numbers that are a power of 2 times a power of 3 (A003586). Similarly with any other two primes. Can anyone think of a nontrivial example of such a table sequence that *does* have a simple formula for finding the next one? I can't. (By "trivial," I think I just mean tables that contain all numbers or that contain only a small number of distinct numbers.)

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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 07 May '19

07 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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