August 2016 - mathfun@mailman.xmission.com

[math-fun] Correction: Man, some of the subtriangles in Tom's permuted pentatriacontagons are mighty tiny.
by Bill Gosper 20 Aug '16

20 Aug '16

On Fri, Aug 19, 2016 at 2:42 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > (E.g., area .00013.) gosper.org/00013.pdf > (Bug in Mathematica's pdf renderer: Windings are iored vs xored. But > you can see the little sucker near the Middle of [the] B&W Graphics.) > --rwg > gosper.org/00013.png > Also, Julian bummed my winding # function: > > turns[L_List] := Total@Mod[RotateLeft[#] - # &@Arg@L/π, 2, -1]/2 > (Mathematica has run all afternoon trying to FullSimplify the exact expression for the area (=.00013...) to Root[5316527372659441 - ... - 182556260829602250752 #1^22 + 675821419082285056 #1^24 &, 13]

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[math-fun] New sequence
by Henry Baker 19 Aug '16

19 Aug '16

ace02468bdf13579... Could be hex; not sure what comes next -- might require solving an equation.

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[math-fun] Man, some of Tom's permuted pentatriacontagons are mighty tiny.
by Bill Gosper 19 Aug '16

19 Aug '16

(E.g., area .00013.) gosper.org/00013.pdf (Bug in Mathematica's pdf renderer: Windings are iored vs xored. But you can see the little sucker near the Middle of B&W Graphics.) --rwg Also, Julian bummed my winding # function: turns[L_List] := Total@Mod[RotateLeft[#] - # &@Arg@L/π, 2, -1]/2

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[math-fun] Periodic chart of the superabundant numbers
by Allan Wechsler 19 Aug '16

19 Aug '16

I have not tried hard to prove any of the assertions I make here. I figured, before I tried, to find out how much of this is well-known. The "abundancy index" of a natural number is the sum of its divisors, divided by the number itself. For classical perfect numbers, the abundancy index is 2; for multiperfect numbers in general, the abundancy index is an integer. The "superabundant numbers" (https://oeis.org/A004394) are record-setters for abundancy index; a number is in A004394 if its abundancy index exceeds that of all smaller numbers. It is easy to show that the superabundant numbers are a subsequence of the "prime signature leaders" (https://oeis.org/A025487) which I call "generalized primorials" for reasons that I think I have sketched here in the past. (A quick search fails to find the occasion, though.) The foregoing is uncontroversial. My speculations start here. Every superabundant number can be factored uniquely into primorials; this property is inherited from the generalized primorials. Apparently, the largest primorial factor of the superabundant numbers ascends smoothly, without ever retreating. It is 1 for n=1, 2 up through n=3, 6 up through n=8, 30 up through n=14, and 210 up through n=22 (I think). A less mysterious way to say this is that if a superabundant number is divisible by a prime p, all larger superabundant numbers are also divisible by p. Let p# be the largest primorial dividing a superabundant number n. Then n/p# is a "cofactor"; the cofactors of the superabundant numbers in order are: 1,1,2,1,2,4,6,8,2,4,6,8,12,24,4,6... This is not in OEIS. It looks like these cofactors all belong to another sequence, 1,2,4,6,8,12,24,48,72,120; within each primorial "period", the cofactors march up this sequence in order, resetting to a lower value when a new period starts. This sequence of cofactors is also not in OEIS. If these speculations are true, then the sequence of superabundant numbers can be compressed by giving the starting index of each period, and the starting index of the cofactor sequence. But this leaves the cofactor sequence inadequately characterized.

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[math-fun] Medium res zero area icosapentagon
by Bill Gosper 17 Aug '16

17 Aug '16

gosper.org/icosapentagon.pdf I'll still bet there's a sexy way in Mathematica to get the exact polygons and their colors. There's also a positive area icosapentagram. Scroll the black one up and down vigorously to see an excited sea urchin. It is happy because the winding number plotter mysteriously provided it with a nice sandy bottom. --rwg

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Re: [math-fun] Reentrant polygons with area zero
by Bill Gosper 17 Aug '16

17 Aug '16

On 2016-08-17 08:30, Tom Rokicki wrote: > Wow, Mathematica can do that? Yeah but it took 9 minutes. > Thanks, Bill, very nice! > It's 20 times faster if you do it floatingly, but sometimes you get dandruff: gosper.org/icosaisagon.png And sometimes you don't. > On Wed, Aug 17, 2016 at 7:18 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > >> Pending the discovery of some high-powered Mathematica solution, you >> could always display the full-res B&W polygon next to an el-crudo winding >> number color legend: gosper.org/pentadecagon.png (Where >> turns[L_List] := Total[Mod[RotateLeft[#] - # &@Arg@L + π, 2π] - π]/2/π .) >> --rwg >> >> On 2016-08-16 06:27, James Propp wrote: >> > I agree with Allan: pictures that show the different winding numbers >> around >> > different regions should be in the final write-up, and using color might >> > give some very nice pictures. But there's a non-trivial programming >> problem >> > here. Are there any off-the-shelf graphics packages that play nicely >> > with reentrant polygons? >> > >> > Tom's pictures remind me of the Klein disk model, and so I am led to >> > wonder: do all these closed paths that encircle signed area zero retain >> the >> > zero-signed-area property if we replace the line segments by circular >> arcs >> > perpendicular to the boundary? (I guess there are two flavors of this >> > question, depending on whether you measure Euclidean area or hyperbolic >> > area.) >> > >> > Jim Propp >> > >> > On Wednesday, August 10, 2016, Allan Wechsler <acwacw(a)gmail.com> wrote: >> > >> >> It would be glorious if you could color-code the different >> winding-numbers, >> >> so we could see the +1's and -1's balancing (as well as the occasional >> 2's >> >> and 3's). >> >> >> >> On Wed, Aug 10, 2016 at 1:19 PM, Tom Rokicki <rokicki(a)gmail.com >> >> <javascript:;>> wrote: >> >> >> >> > The images show in my email, not sure why you aren't seeing them. >> >> > >> >> > The following link should work: >> >> > >> >> > https://docs.google.com/document/d/1G3FL1Pcc8RgZ0DZOxibSonMGSJtrYOc6OUwXjTh… [Bogus linebreak removed. And probably reinstalled by xmission.com.] --rwg >> >> > >> >> > -tom

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[math-fun] buying gas last night
by Bill Gosper 17 Aug '16

17 Aug '16

The pump stopped > $3 short of the $20 I gave the Desi clerk, so I squeezed it up to $17 and, remembering that a new employee at Empress of India had once accepted one, demanded a $3 bill with Bill Clinton's picture on it. The clerk said $2.71. In disbelief, I walked back to the pump. It said $17.29. The clerk denied that 1729 meant anything to him, but became very enthusiastic when I mentioned Ramanujan. I still don't understand where that 29¢ came from. There now was actually a little gas spilled on the ground. --Bill

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Re: [math-fun] Reentrant polygons with area zero
by Bill Gosper 17 Aug '16

17 Aug '16

Pending the discovery of some high-powered Mathematica solution, you could always display the full-res B&W polygon next to an el-crudo winding number color legend: gosper.org/pentadecagon.png (Where turns[L_List] := Total[Mod[RotateLeft[#] - # &@Arg@L + π, 2π] - π]/2/π .) --rwg On 2016-08-16 06:27, James Propp wrote: > I agree with Allan: pictures that show the different winding numbers around > different regions should be in the final write-up, and using color might > give some very nice pictures. But there's a non-trivial programming problem > here. Are there any off-the-shelf graphics packages that play nicely > with reentrant polygons? > > Tom's pictures remind me of the Klein disk model, and so I am led to > wonder: do all these closed paths that encircle signed area zero retain the > zero-signed-area property if we replace the line segments by circular arcs > perpendicular to the boundary? (I guess there are two flavors of this > question, depending on whether you measure Euclidean area or hyperbolic > area.) > > Jim Propp > > On Wednesday, August 10, 2016, Allan Wechsler <acwacw(a)gmail.com> wrote: > >> It would be glorious if you could color-code the different winding-numbers, >> so we could see the +1's and -1's balancing (as well as the occasional 2's >> and 3's). >> >> On Wed, Aug 10, 2016 at 1:19 PM, Tom Rokicki <rokicki(a)gmail.com >> <javascript:;>> wrote: >> >> > The images show in my email, not sure why you aren't seeing them. >> > >> > The following link should work: >> > >> > https://docs.google.com/document/d/1G3FL1Pcc8RgZ0DZOxibSonMGSJtrY >> > Oc6OUwXjThEj2c/edit?usp=sharing >> > >> > -tom >> > >> > On Wed, Aug 10, 2016 at 10:12 AM, Mike Stay <metaweta(a)gmail.com >> <javascript:;>> wrote: >> > >> > > Where can we see the images? >> > > >> > > On Wed, Aug 10, 2016 at 11:00 AM, Tom Rokicki <rokicki(a)gmail.com >> <javascript:;>> wrote: >> > > > Yes, oops. Apologies. >> > > > >> > > > On Wed, Aug 10, 2016 at 9:57 AM, Allan Wechsler <acwacw(a)gmail.com >> <javascript:;>> >> > > wrote: >> > > > >> > > >> Tom, where you say "polynomial" above, do you always mean "polygon"? >> > > >> >> > > >> >> > > >> On Wed, Aug 10, 2016 at 12:24 PM, Tom Rokicki <rokicki(a)gmail.com >> <javascript:;>> >> > > wrote: >> > > >> >> > > >> > Consider a possibly re-entrant polynomial on n equally-spaced >> points >> > > >> around >> > > >> > a circle. James Propp asks, with a signed winding-number >> definition >> > > of >> > > >> > signed area, for what n can we create a polynomial of zero area? >> > > >> > >> > > >> > For even n > 2, an easily solution is to connect the points in the >> > > order >> > > >> > (1, 2, 3, ..., n/2, n, n-1, n-2, ..., n/2+1). >> > > >> > >> > > >> > Adam P. Goucher showed that for prime n, no solution is possible. >> > > >> > >> > > >> > This note completes the solution for all n by showing that, for >> all >> > > >> > composite odd n, a solution exists. We give an explicit >> > construction. >> > > >> > >> > > >> > To explain our construction, there are two key insights, and then >> a >> > > bit >> > > >> of >> > > >> > engineering. The first insight is a remarkably simple equation to >> > > >> compute >> > > >> > the area. The second is identification of a specific sum of >> > triangles >> > > >> that >> > > >> > always sum to zero. >> > > >> > >> > > >> > Normally when we compute the area of a polygon using the normal >> > > >> > winding-number definition of area, we add a series of trapezoids >> > from >> > > >> the x >> > > >> > axis, as follows: >> > > >> > >> > > >> > 1/2 sum_i (x_{i+1}-x_i) (y_{i+1}+y_i) >> > > >> > >> > > >> > where the +1 is interpreted modulo n and the i ranges from 0 to >> n-1. >> > > But >> > > >> > there's nothing special about using trapezoids from the x axis; we >> > > can as >> > > >> > easily just sum triangles from the center of the circle around >> which >> > > the >> > > >> > points are splayed. The area of a triangle with an angle of theta >> > > >> between >> > > >> > two legs of length 1 is simply 1/2 sin(theta), so the area can be >> > > >> specified >> > > >> > as just >> > > >> > >> > > >> > 1/2 sum_i sin(theta_{i+1}-theta_{i}) >> > > >> > >> > > >> > Let us define the original set of points p_0..p_{n-1} as >> > > >> theta=p_i*tau/n, >> > > >> > with x=cos(theta), y=sin(theta). The polygon can be defined by a >> > > >> > permutation of these points (s_0, s_1, ..., s_{n-1}), so for >> > instance >> > > for >> > > >> > n=4 the polygon (0, 1, 3, 2) connects in sequence p_0, p_1, p_3, >> > p_2 >> > > >> and >> > > >> > then goes back to p_0. The relevant angles can be computed from >> the >> > > gap >> > > >> > sequence, which is just the sequence of differences between the >> > point >> > > >> > indices, going around the polygon, so in this case the gap >> sequence >> > is >> > > >> (1, >> > > >> > 2, -1, -2). The area is just half the sum of the sines of these >> > > values. >> > > >> > Since the sine function is odd, this sum of the sines of this gap >> > > >> sequence >> > > >> > is clearly zero. >> > > >> > >> > > >> > This brings us to our second insight. The sum of the sines of a >> set >> > > of >> > > >> > angles equally spaced around the circle is zero. That is, for any >> > n>1 >> > > >> and >> > > >> > any a, >> > > >> > >> > > >> > sum_i sin(a+i*tau/n)=0. >> > > >> > >> > > >> > As an example, sin(5)+sin(5+tau/3)+sin(5+2*tau/3) is zero. If >> this >> > > is >> > > >> not >> > > >> > obvious to you, visualize the center of mass of a series of point >> > unit >> > > >> > masses distributed evenly around a circle of radius 1; the sine >> > > function >> > > >> > gives the y component of the moment imparted by a particular mass. >> > > >> > >> > > >> > So if we can arrange that the gaps can be partitioned into subsets >> > so >> > > >> that >> > > >> > the angles derived from those gaps in each subset are equally >> > > distributed >> > > >> > around zero, the total area will be zero. >> > > >> > >> > > >> > [image: t-9.png] >> > > >> > >> > > >> > Let us consider the solution that Propp provided, for n=9. This >> > > solution >> > > >> > (after suitable reflection/rotation) visited the points in the >> order >> > > (0, >> > > >> 1, >> > > >> > 5, 3, 4, 8, 6, 7, 2), which generates the gap sequence (1, 4, 7, >> 1, >> > > 4, 7, >> > > >> > 1, 4, 7). The area is then >> > > >> > >> > > >> > 3/2 (sin(tau/9)+sin(4 tau/9)+sin(7 tau/9)) >> > > >> > >> > > >> > Since the three points at tau/9, 4 tau/9, and 7 tau/9 are evenly >> > > >> > distributed around the circle, the sum of the sines is zero and >> thus >> > > so >> > > >> is >> > > >> > the total area. >> > > >> > >> > > >> > So all we need to do is find a gap sequence that can be >> partitioned >> > > into >> > > >> > subsets each consisting of angles that are evenly distributed >> around >> > > the >> > > >> > circle. That gap sequence must be such that every point in the >> set >> > is >> > > >> hit >> > > >> > exactly once and joins back to the first point. >> > > >> > >> > > >> > As a warmup we will cover all odd squares greater than 1. Let n >> be >> > > such >> > > >> a >> > > >> > square, and call the square root a. The gap sequence (1, 1+a, >> 1+2a, >> > > ..., >> > > >> > 1+(a-1)a) repeated a times generates a polygon that hits each >> point >> > > once >> > > >> > and joins back to the first point. We'll prove the second part >> > first. >> > > >> > >> > > >> > The sum of the points 1, 1+a, 1+2a, ..., 1+(a-1)a is simply >> > > >> > a+a(a)(a-1)/2=1/2a(a^2-a+2). >> > > >> > >> > > >> > The sum of the full gap sequence is a times this. Since a^2-a+2 >> is >> > > >> always >> > > >> > even, this final sum is always divisible by a^2 and thus always >> > > brings us >> > > >> > back to the first point. >> > > >> > >> > > >> > To prove the first part, we divide the points visited into a >> subsets >> > > each >> > > >> > with a points. The first subset is {p_0, p_a, p_{2a}, ...}. The >> > > second >> > > >> > subset is {p_1, p_{a+1}, p_{2a+1}, ...}. The polygon always >> visits >> > a >> > > >> point >> > > >> > in the first subset, then a point in the second subset, and so on >> > and >> > > so >> > > >> > forth, because each gap is equal to 1 modulo a, and thus s_i = i >> > (mod >> > > a). >> > > >> > So we visit a points in each subset. We need to show those points >> > are >> > > >> > distinct. >> > > >> > >> > > >> > The points we visit in the first subset are always separated >> around >> > > the >> > > >> > polygon by the sum of the gaps modulo n, and this sum is >> > > 1/2a(a^2-a+2). >> > > >> If >> > > >> > this visits all the points in the first subset, then it will visit >> > all >> > > >> the >> > > >> > points in all subsets, because the visit order in the other >> subsets >> > > are >> > > >> > just a rotation of the visit order in the first subset. >> > > >> > >> > > >> > At time ai we will visit point 1/2ai(a^2-a+2) which is in the >> first >> > > >> > subset. If 1/2(a^2-a+2) is relatively prime to a, we will visit >> all >> > > the >> > > >> > points. This formula is equal to a(a-1)/2+1. Since a is odd, a-1 >> > is >> > > >> > divisible by 2, and thus 1/2(a^2-a+2)=1 modulo a. >> > > >> > >> > > >> > For example, for n=25, the sequence is (1, 6, 11, 16, 21) >> repeated 5 >> > > >> times, >> > > >> > and the resulting figure is >> > > >> > >> > > >> > [image: t-25.png] >> > > >> > >> > > >> > This is a construction for odd squares that always generates a >> > > polynomial >> > > >> > that is rotationally symmetric of order a. It is straightforward >> to >> > > >> extend >> > > >> > this to any odd composite that has a repeated factor; we omit the >> > > details >> > > >> > here. We can extend this construction to work for all odd >> composite >> > > >> > numbers. We do this by adding an even number of additional >> subsets >> > > and >> > > >> > engineering a workable order of those additional points. >> > > >> > >> > > >> > For the odd composite numbers, we will let a be the smallest prime >> > > factor >> > > >> > of n; since a is odd, that factor will also be odd. We define b >> to >> > be >> > > >> > n/a. We create b subsets of the n points; the first subset is >> {p_0, >> > > p_b, >> > > >> > p_{2b}, ...}, and so on. Let u be the sequence (1, b+1, 2b+1, >> ..., >> > > >> > (a-1)b+1). Let s*i for a sequence s and an integer i be the >> > sequence >> > > s >> > > >> > repeated i times. Let d=(b-a)/2 (note that it is always an >> > integer). >> > > >> The >> > > >> > sequence we use is >> > > >> > >> > > >> > (u, (1, 1)*d, u, (b+1, (a-1)b+1)*d, u, (2b+1, (a-2)b+1)*d, ..., >> u, >> > > >> > ((a-1)b+1, b+1)*d) >> > > >> > >> > > >> > This generates b gaps each of 1, b+1, 2b+1, ..., (a-1)b+1. We add >> > > pairs >> > > >> of >> > > >> > gaps such that each pair is exactly 2 mod n, so our earlier >> argument >> > > that >> > > >> > we cover all of the points in the first subset still works. We >> can >> > > split >> > > >> > the additional subsets of points into pairs. For the first subset >> > of >> > > >> each >> > > >> > pair we always visit the points in exactly the same order as we do >> > of >> > > the >> > > >> > first subset overall. In the second subset of each pair, we visit >> > the >> > > >> > points in a reversed-and-shifted order, starting with the first >> > point, >> > > >> then >> > > >> > the last, then the one before the last, and on down to the second >> > > point. >> > > >> > >> > > >> > For example, the total sequence for n=15, with a=3 and b=5 is >> > > >> > >> > > >> > (1, 6, 11, 1, 1, 1, 6, 11, 6, 11, 1, 6, 11, 11, 6) >> > > >> > >> > > >> > In this case, u is (1, 6, 11), and the pairs we add are (1, 1), >> (6, >> > > 11), >> > > >> > and (11, 6). The resulting image is >> > > >> > >> > > >> > [image: t-15.png] >> > > >> > >> > > >> > The construction we show here is only one possible polynomial; >> there >> > > are >> > > >> > many more. For instance, the values in u can be permuted >> > > arbitrarily, as >> > > >> > long as the same value of u is used throughout the sequence. >> > > >> > >> > > >> > We finish with a solution for n=121. >> > > >> > >> > > >> > [image: t-121.png] >> > > >> > >> > > >> > >> > > >> > On Sat, Jul 30, 2016 at 2:21 PM, James Propp < >> jamespropp(a)gmail.com <javascript:;>> >> > > >> wrote: >> > > >> > >> > > >> > > Tom, >> > > >> > > >> > > >> > > So far it looks like, for prime m, n has a >> > > >> > > > rotationally symmetrical solution of order m >> > > >> > > > if n is divisible by m^2. >> > > >> > > >> > > >> > > >> > > >> > > And a rotationally symmetrical solution of order m^r if n is >> > > divisible >> > > >> by >> > > >> > > m^(r+1) (still assuming that m is prime), right? >> > > >> > > >> > > >> > > If we're looking at symmetrical solutions, we should probably >> > > include >> > > >> > even >> > > >> > > values of n back in the game. For instance, when n=8 is there a >> > > >> solution >> > > >> > > with fourfold symmetry? >> > > >> > > >> > > >> > > Jim Propp >> > > >> > > _______________________________________________

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[math-fun] Reentrant polygons with area zero
by James Propp 16 Aug '16

16 Aug '16

For which n>2 does there exist a reentrant n-gon whose vertices are evenly spaced on a circle and whose signed area (aka algebraic area) is zero? When n is even, there is an easy solution: draw a circuit from 1 to 2 to 3 to ... to n/2-1 to n/2 to n to n-1 to n-2 to ... to n/2+1 to 1 (here I'm numbering the n vertices cyclically). For n=3, it's trivially impossible to find such an n-gon, and I've checked by brute force (using Mathematica) that there's no such n-gon when n=5 or n=7. For n=9, there is such an n-gon (which I found through brute-force search): http://mathenchant.org/nonagon.pdf Mathematica assures me that 1/2 (-1/2 Sqrt[3] sin(\[Pi]/18)+1/2 sin(\[Pi]/9)-sin((2 \[Pi])/9)-2 sin(\[Pi]/18) sin((2 \[Pi])/9)-1/2 cos(\[Pi]/18)-1/2 Sqrt[3] cos(\[Pi]/9)+2 cos(\[Pi]/18) cos((2 \[Pi])/9)+2 sin(\[Pi]/9) cos(\[Pi]/9)) simplifies to 0, which implies the claim. Is there a nice (preferably trig-free) way to prove it? That is: is there a nice way to show that the total area of the three scalene black triangles equals the total area of the four nonscalene black triangles, maybe using dissection and area-preserving linear transformations, or maybe using algebraic tricks involving roots of unity? I suspect that when n is prime, there's no such polygon. Can any of you find a proof? Also, can any of you construct such a polygon with n=15? It'd be nice if the answer to my original question were "Precisely when n is composite". Jim Propp

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[math-fun] Algebra 2 exercise
by Bill Gosper 15 Aug '16

15 Aug '16

If a regular *n*-gon with unit circumradius has side s, the 2*n*-gon has side Sqrt[2 - Sqrt[4 - s^2]]. Shew that this denests to (-2 + s + Sqrt[4 - s^2])/Sqrt[4 - 2 s] when 0 ≤ s ≤ 2. --rwg

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