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[math-fun] The eraser game
by Éric Angelini 28 May '20

28 May '20

Hello Math-Fun, (I'm sorry to propose this here – but SeqFan didn't publish my last three posts, dunno why) The idea is to always extend the sequence E with the sum of the last two terms – and to apply the "eraser rule" when needed: E = 1, 2, 3, 5, 8, 13, 2(1), 15, (1)7, ... The eraser rule says that the last term of E and the coming sum cannot share any digit. The duplicate digit (parenthesized) are thus erased from the sum. (When needed, the remaining digits are concatenated to form a new integer). Example: We see that after 8 + 13 = 21 we must erase the 1 from 21 and proceed with 2; this 2 will be added to 13 to produce 15; but as 2 + 15 = 17 we will erase 1 again from 17 and extend E with 7; etc. The above start stops quickly: E = 1, 2, 3, 5, 8, 13, 2(1), 15, (1)7, 22, (2)9, 31, 40, 71, (111) stop. What about leading zeros in the "erased" sum? Let's see: E = 598, 414, (1)0(1)2,... Well, let's erase any leading zero – we will thus proceed here with: E = 598, 414, 2, 416, (41)8, 424, etc. Questions: What is the lexicographically earliest start pushing the sequence into a loop? Is there a start leading to an infinite sequence? Best, É. As usual, the same text as above is here -- with colors and pictures: http://cinquantesignes.blogspot.com/2020/05/the-eraser-game.html

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Re: [math-fun] Fibred knots
by Dan Asimov 27 May '20

27 May '20

So I wanted to know what is this glueing map h : T^2 - D^2 —> T^2 - D^2 which for the trefoil knot K determines its complement S^3 - K via S^3 - K = (T^2 - D^2) x [0,1] / (x,1) ~ (h(x),1) ??? I finally figured it out, and it's kind of pretty. (No doubt this was known long, long ago to knot theorists.) Consider the hexagonal torus: a regular hexagon with each pair of its opposite sides identified by a translation across the hexagon. Now remove a small disk D^2 centered at the center of the hexagon. Finally, the glueing map h just takes this holey hexagonal torus and rotates it 120º into itself. That's it. That's S^3 - K when K is the trefoil. If you're partial to knots in R^3 instead, just remove another point from that. (Something similar happens with all (p,q) torus knot complements. —Dan ----- There's an interesting feature of many knots called "fibring", and such knots are called "fibred knots". A knot can be thought of as living in R^3 but it can be more useful to think of it as living in the 3-sphere S^3 (which is topologically just R^3 with a point added at infinity). To distinguish knots — since all knots are topologically circles — people have studied not the knot itself but what becomes of S^3 after the knot is removed. S^3 - K is called the knot complement. For many knots K in S^3, it turns out that there is a lovely surjective map π : S^3 - K —> S^1 to the circle, such that the inverse image of each point of S^1 is always a certain M (with boundary) in a continuous manner. If we removed one of these surfaces, say π^(-1)(1), what remained of S^3 - K would be topologically M x (0, 1), the product of the surface with an open interval. If we like we can glue on the missing end-surfaces to get M x [0, 1]. Then knot complement is completely determined by the way the end-surfaces are glued back together: S^3 - K = M x [0,1] / (x,1) ~ (h(x),0) where h is some self-homeomorphism h : M —> M of M. ----

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Re: [math-fun] Knot Annihilator
by Dan Asimov 27 May '20

27 May '20

We need three constraints to accomplish what? —Dan ----- Version 1.0 in Mathematica: https://0x0.st/ip7E.txt Analyzing outputs, I found the following for case (3,4): {X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], 2 Sin[4 t]} 0 = -81 + 117 X^2 - 40 X^4 + 4 X^6 + 117 Y^2 - 112 X^2 Y^2 + 12 X^4 Y^2 - 40 Y^4 + 12 X^2 Y^4 + 4 Y^6; 0 = -8 X^3 Y + 8 X Y^3 + 27 Z - 24 X^2 Z + 4 X^4 Z - 24 Y^2 Z + 8 X^2 Y^2 Z + 4 Y^4 Z; 0 = 81 - 81 X^2 - 81 Y^2 + 32 X^2 Y^2 + 16 X^2 Z^2 + 16 Y^2 Z^2 Contrary to expectation, we need not two but three constraints. The first two equations define the knot + 8 vertical lines at crossing points. The third constraint filters out vertical lines. -----

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[math-fun] Fibred knots
by Dan Asimov 27 May '20

27 May '20

There's an interesting feature of many knots called "fibring", and such knots are called "fibred knots". A knot can be thought of as living in R^3 but it can be more useful to think of it as living in the 3-sphere S^3 (which is topologically just R^3 with a point added at infinity). To distinguish knots — since all knots are topologically circles — people have studied not the knot itself but what becomes of S^3 after the knot is removed. S^3 - K is called the knot complement. For many knots K in S^3, it turns out that there is a lovely surjective map π : S^3 - K —> S^1 to the circle, such that the inverse image of each point of S^1 is always a certain M (with boundary) in a continuous manner. If we removed one of these surfaces, say π^(-1)(1), what remained of S^3 - K would be topologically M x (0, 1), the product of the surface with an open interval. If we like we can glue on the missing end-surfaces to get M x [0, 1]. Then knot complement is completely determined by the way the end-surfaces are glued back together: S^3 - K = M x [0,1] / (x,1) ~ (h(x),0) where h is some self-homeomorphism h : M —> M of M. The trefoil can be thought of a the intersection of the zero-locus of the polynomial P : C^2 —> C (C = complex numbers) defined by P(z,w) = z^3 + w^2 with a small 3-sphere S^3 about the origin (0,0) of C^2. Then on S^3 - K, that polynomial is always non-zero. So we can define a mapping π : S^3 - K —> S^1 via π(z,w) = P(z,w) / |P(z,w)| and that defines the fibring as above. For the trefoil, it turns out that the fibre M is just a torus with a disk removed. Even the unknot (a geometric circle) has a fibring: The fibre is just a disk. But some knots have no such fibring. —Dan

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Re: [math-fun] Knot Annihilator
by Dan Asimov 27 May '20

27 May '20

Actually, I'm just trying to understand the comment "we need three constraints". I'm not asking about the details of your calculation. I'm asking about what is the final goal that requires three constraints? I ask because, if the final goal is to represent the (3,4) torus knot in 3-space as the intersection of the zero-loci of each of several real polynomials, I would have expected that only two are required. —Dan ----- Again, I think it my last message was fine as stated, but I will explain a few finer points. Call the three polynomials S(X,Y), H(X,Y,Z), and F(X,Y,Z). They all equal zero for the parametric curve {X(t),Y(t),Z(t)} after reducing terms. The linear height function, H, can be factored as H = L(X,Y)+C(X,Y)*Z with L the product of four linear factors, and C the product of two circular factors. Eight solutions of L=0 & C=0, are a solutions of H=0 & S=0 regardless of Z, see: https://0x0.st/iph1.png So the algebraic varietry { (X,Y,Z) : H=0 & S=0 } accidentally contains 8 vertical lines. If (X_0,Y_0) is any one of the eight singular points, then F(X_0,Y_0,Z)=0 implies either: 48 Z^2-45 = 0, or 144 Z^2 - 567 = 0. On the set of singularities--which corresponds to the set of knot crossings--the quadratic filter function F chooses just two points from the singular, vertical line, one on the overpass, and one on the underpass. Thus the variety: { (X,Y,Z) : H=0 & S=0 & F=0 } Is the (4,2) torus knot in the strict sense, where it contains no extra points or mirror images. -----

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[math-fun] Why R0 is a useless concept
by Henry Baker 27 May '20

27 May '20

I think that we are all in violent agreement that these epidemiological models are 'ill-conditioned', hence *any* noise in the input can be dramatically *amplified* in such a way that it can often overwhelm any 'answer'. Analogy: those screeching noises that are often heard from audio public address systems that have positive feedback; the screeches often overwhelm the person speaking. Re network-simulation Monte Carlo models, e.g., the Imperial model: Monte Carlo models require enough iterations/runs in order to *average out* the sampling noise, *and to fully "explore" the nether/tail regions of the particular probability density function*. The most trivial Monte Carlo model is that of estimating the *mean* of a distribution by computing statistics from N samples. How many samples are required in order to assure a reasonable estimate of the mean? Answer: N ~ O(distribution variance). OK. Let's take an oversimplified 'superspreader' model for R0: 99% of the time, R0=2, and 1% of the time, R0=98. The mathematical mean of this bimodal distribution is 2.96, and the mathematical variance of this distribution is ~91. But I just ran this experimental model and it takes at least 15,000 random samples of this distribution just to get a decent approximation to one number -- its mean! The reason why so many samples are required is that the relatively rare event where R0=98 has to occur often enough to average out against the vastly more probable R0=2 events. But we're only getting started. R0 appears as the *base* of an exponential in various epidemic models -- e.g., (R0)^(a*t), for some constant a. The most elementary statistics classes don't deal with *products* of random variables, much less *exponentials* of random variables. One simple way to understand such products and exponentials uses *lognormal* distributions. If X=L(m,v) is a lognormal distribution with parameters m,v, then the distribution for the exponential X^n is L(n*m,n*v). The mean of L(n*m,n*v) is (exp(m+v/2))^n; the variance of L(n*m,n*v) is exp(2*m+v)^n*(exp(v)^n-1). If we choose m,v to match the mean and variance of the bimodal distribution above, then m~-0.1322 and v~2.4348, so the mean of X^n is (2.96)^n and the variance of X^n is (2.96)^(2n)*(11.414^n-1) ~ 100^n. So what if we have to sample, e.g., (R0)^10, i.e., a*t=10 -- to compute its mean ? How many samples will we need to get a decent approximation ? (Note that this is the 10-fold product of independently chosen R0's, so we can't simply average numbers like sample^(1/10).) Since variance of (R0)^10 is ~100^10 = 10 *billion*, it could take O(10 billion) random samples to get a decent approximation to the mean of (R0)^10. I'd be willing to bet that the Imperial model was not run 10 billion times! But this is merely one positive feedback loop in such a Monte Carlo network simulation. What happens when there are multiple feedback loops ? How many runs might then be required ? The problem here is that our samples have to explore an incredibly wide and incredibly shallow distribution, and then accumulate enough weight for each sample to guarantee some reasonable accuracy for our result. But even if we performed such a computation, what would it mean when the *variance* of the distribution is so wide -- hence the weight of any particular value is so tiny -- of what practical use is *any* particular value -- e.g., the "mean" ? This is the reason why "R0" models make no sense in the presence of superspreaders -- there is no single 'R0' that captures any useful aspect of the behavior.

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[math-fun] Knot Annihilator
by Brad Klee 27 May '20

27 May '20

Version 1.0 in Mathematica: https://0x0.st/ip7E.txt Analyzing outputs, I found the following for case (3,4): {X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], 2 Sin[4 t]} 0 = -81 + 117 X^2 - 40 X^4 + 4 X^6 + 117 Y^2 - 112 X^2 Y^2 + 12 X^4 Y^2 - 40 Y^4 + 12 X^2 Y^4 + 4 Y^6; 0 = -8 X^3 Y + 8 X Y^3 + 27 Z - 24 X^2 Z + 4 X^4 Z - 24 Y^2 Z + 8 X^2 Y^2 Z + 4 Y^4 Z; 0 = 81 - 81 X^2 - 81 Y^2 + 32 X^2 Y^2 + 16 X^2 Z^2 + 16 Y^2 Z^2 Contrary to expectation, we need not two but three constraints. The first two equations define the knot + 8 vertical lines at crossing points. The third constraint filters out vertical lines. --Brad

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[math-fun] A Conway-ish approach to the dyadic Eisenstein rationals?
by James Propp 27 May '20

27 May '20

I'm trying to see what sorts of algebraic structures can be built up in a Conway-style "genetic" way. I think I may have an approach to constructing the surcomplex numbers that's different from Conway's, but I'm stuck; maybe one of you can help me see what the next step is (or help me see that the enterprise is doomed). Some background may be helpful here. We say that one dyadic rational number is simpler than another if the first has smaller denominator, or, in the case where they have the same denominator, the first has smaller absolute value. Thus 1/4 is simpler than 3/4 but 1/2 is simpler than both of them. This gives a partial ordering on dyadic rationals. The simplest dyadic rational is 0; the next simplest are +1 and -1; the next simplest are +2, -2, +1/2, and -1/2; etc. Conway approached simplicity via birthday ordinals but this is not necessary. Every open interval I in the dyadic rationals contains a unique simplest number I*, and we have the addition rule I*+J* = ((I*+J) \cap (I+J*))* where the sum of a number and an interval is defined in the obvious way (shift the interval by the number, or if you prefer, take the Minkowski sum of the interval with the singleton set associated with the number). This formula is just Conway's definition of addition in disguise. To rip away the disguise, write I* as {x_L | x_R} where x_L and x_R are the left- and right-endpoints of I respectively. If we write I = (x_L, x_R) and I* = x and J = (y_L, y_R) and J* = y, then Conway's definition of addition tells us that x + y is the simplest dyadic rational that is bigger than both x + y_L and x_L + y and smaller than both x + y_R and x_R + y; that is, it is the simplest dyadic rational in the intersection of the interval (x + y_L, x + y_R) = x + (y_L, y_R) = I* + J and the interval (x_L + y, x_R + y) = (x_L, x_R) + y = I + J*. (Has anyone seen this way of reformulating surreal numbers before?) Now on to my real question. Define the dyadic Eisenstein rationals as the complex numbers of the form a + bw where a and b are dyadic rationals and w is a primitive cube root of 1; define its denominator as the maximum of the denominators of a and b. Every dyadic Eisenstein rational z can be written as a + bw, a + bw^2, or aw + bw^2 with a,b non-negative; accordingly define the "triangle norm" of z as a+b (so that the level sets of the triangle norm are triangles centered at 0 in the complex plane; it's not a true norm since the level sets aren't centrally symmetric). We say that one dyadic Eisenstein rational is simpler than another if the first has smaller denominator, or, in the case where they have the same denominator, the first has smaller triangle norm. The role played by intervals in the dyadic rationals will be played by open equilateral triangles that point to the right (like the "play" icons on many devices); hereafter when I write "triangle" I mean that kind of triangle. I can prove that every triangle T in the dyadic Eisenstein rationals contains a unique simplest number T*. What I haven't been able to prove, but believe to be true, is that for all triangles T,U we have (*) T*+U* = ((T*+U) \cap (T+U*))* If this is true, then there's probably a way to "genetically" build up the Eisenstein rationals (perhaps extending beyond the complex numbers into the surcomplex realm) starting from the empty set, much as Conway built up the surreals. (This leaves out the problem of defining multiplication genetically, but hey, you gotta start somewhere.) I've proved the conjecture for triangles that are symmetric about the real axis; in this case (*) follows from facts about surreal numbers. (I should mention that there's an alternative, simpler way to build up the surcomplex numbers genetically, using rectangles instead of triangles. But then all the theorems one needs to prove are trivialities, as all we're doing is looking at ordered pairs of surreal numbers and doing the obvious things with them. This is pretty close to Conway's construction of the surcomplex numbers by adjoining a formal square root of -1. But what interests me is the idea of genetic construction itself. The only things I've seen genetic constructions for are the surreals and the nimbers.) Jim Propp

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[math-fun] Amazing(?) Binomial Identity For A304126.
by Brad Klee 26 May '20

26 May '20

a[n_] := Sum[Binomial[6 n, n + k] Binomial[4 n, k] (-1)^k, {k, 0, 4 n}] zero[n_] := Binomial[3 n, n] Binomial[6 n, 3 n] - Binomial[5 n, n] a[n] zero /@ Range[0, 10] See also: https://oeis.org/A304126 ( nice! ) In terms of OEIS numbering, A304126 neither divides A005809 nor A066802, yet it does divide their product A113424. This can be proven using Zeilberger's algorithm, or via ExpToODE according to the integral definition, a_n = 1/(2*pi) Int_{0..2*Pi} (2^10*cos(x)^6*sin(x)^4)^n *dx . More on this soon... --Brad

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[math-fun] I just "Julianized" the base 2-i 'four-around-one" spacefill.
by Bill Gosper 26 May '20

26 May '20

It's pretty boring <http://gosper.org/2-ifills.png>. No quadruple points. All triple point inverse image denominators seem to be 5^(n+1) or twice that. —rwg You can find Julian's magic piecewiserecursivefractal in OEIS under Dragon triple points. If you are sufficiently hardcore to want "Eisenstein rationals", ask me for Julian's latest version with extra optional simplification. If you're feeling adventuresome, you could try to capture the definitions of bas2mi and im2sab with FindImageText and TextRecognize on the linked big png.

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