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[math-fun] Fun with shift register sequences (connecting A000031 and A000016)
by Joerg Arndt 01 Apr '19

01 Apr '19

To generate the necklaces for length n... Let N = 2^n and f(j) = 2*j if j < N/2, otherwise (2*j+1) (mod N). This induces a permutation of the elements [0...N-1]. Now write down the cycles, and for a cycle (r, s, t, ..., z), map its elements to {0,1} using m(x) = 0 if x < N/2, otherwise 1. An example for n=5. The cycles are (0) (1 2 4 8 16) (3 6 12 24 17) (5 10 20 9 18) (7 14 28 25 19) (11 22 13 26 21) (15 30 29 27 23) (31) The map m gives (dots for zeros) . ....1 ...11 ..1.1 ..111 .1.11 .1111 1 These are the primitive parts of the necklaces of length n. In other words: the Lyndon words whose length divide n. (Btw. for n prime this gives a neat algorithm to generate a random Lyndon word of length n with just n operations). This is implied in section "Example using inverse Burrows-Wheeler transform" in https://en.wikipedia.org/wiki/De_Bruijn_sequence All this is likely known but I have not seen it in this particular form. (Btw. the generalization from binary to k-ary is easy). Now here is a twist: Using the map f(j) = 2*j if j >= N/2, otherwise (2*j+1) (mod N). (note "j >= N/2", the condition is "swapped") and otherwise the procedure above I get n = 0: . # = 1 n = 1: .1 # = 1 n = 2: ..11 # = 1 n = 3: ...111 .1 # = 2 n = 4: ....1111 ..1.11.1 # = 2 n = 5: .....11111 ...1.111.1 ..1..11.11 .1 # = 4 n = 6: ......111111 ....1.1111.1 ...1..111.11 ...11.111..1 ..1.1.11.1.1 ..11 # = 6 n = 7: .......1111111 .....1.11111.1 ....1..1111.11 ....11.1111..1 ...1...111.111 ...1.1.111.1.1 ...11..111..11 ..1..1.11.11.1 ..1.1..11.1.11 .1 # = 10 These do seem to be the shift register sequences counted in A000016. IS THIS CORRECT? If it is, I'll enter comments to both sequences. Best regards, jj

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Re: [math-fun] Glass moon
by Keith F. Lynch 27 Mar '19

27 Mar '19

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > If your first thought is that is that the moon would be a burning > glass, leaving a trail of destruction during every solar eclipse, > think again. In a week I will explain why that wouldn't happen > unless someone beats me to it, which I hope they will. Nobody did, so here goes. In any purely optical system, "etendue" (look it up) can never decrease, due to the second law of thermodynamics. That means any refraction, reflection, diffraction, etc., of all or part of the sun can never be any brighter per unit angular area than the sun itself. As every child chasing ants with a burning glass knows (it's what my generation had instead of video games), to roast an ant you need to hit it with light from multiple directions at once. The lens has to look much larger than the sun from the ant's perspective. This lesson was reinforced a couple decades later when I worked with what was then the world's largest solar furnace, at Kirtland Air Force Base. A medium-sized desert was lined with large mirrors, and anyone at the focal point when it was sunny would have a very bad day. At least one person suffered from moonburn (!) while aligning it all night using the full moon. He should have remembered to wear moonscreen. As seen from Earth's surface, the sun and the moon are almost exactly the same angular size. So if the moon is replaced with a lens that focuses the sun, the result will look exactly the same as the sun, no brighter or dimmer. You'll actually be viewing just one point on the sun, but that won't make a visible difference unless you're using a telescope. That's if the sun is centered exactly behind the moon, as with totality during a total solar eclipse. If the sun is not at all behind the moon, then the moon will be dark (or, with a very low probability, will show some planet for a few seconds). The brightest the sky can get in such a setup is if the edge of the sun is just barely behind the center of the moon. In that case you get two overlapping bright circles. That should be 1.609 times brighter than the sun alone. Not enough to set anything on fire. > How would the refractive index have to vary with depth for the moon > to have a focal length equal to the distance to Earth's surface? Nobody has answered this yet. A closed-form solution is beyond me. I guess I'll write a simulation. I can drop a dimension and use a circle instead of a sphere. And I can drop half the circle, due to symmetry. I'll approximate the smoothly varying refractive index with perhaps a hundred concentric semi-circles, each with a different fixed refractive index, and do ray tracing. That will give me a hundred different interacting knobs to twiddle. Perhaps a hill-climbing algorithm will get it to converge on the optimal solution. First I'll try focusing on a point on the moon's surface rather than Earth's, and see if my program rediscovers the Luneburg lens. Any thoughts on such a program? Thanks. It's hard to believe nobody has solved this already.

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[math-fun] This is a test
by Henry Baker 27 Mar '19

27 Mar '19

Testing to see if math-fun is working for me.

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Re: [math-fun] Siddarth Ghoshal has written an amazing identity
by Dan Asimov 27 Mar '19

27 Mar '19

Hey, who do you think will write the software that writes the software that writes the software that will prove the Riemann Hypothesis? —Dan Fred Lunnon a écrit: ----- Good to know that we haven't managed to make ourselves completely redundant yet, then! -----

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[math-fun] Siddarth Ghoshal has written an amazing identity
by Bill Gosper 27 Mar '19

27 Mar '19

https://www.wolframcloud.com/objects/d0cafba7-0635-40d3-83af-c71cd08ca033 and possibly seminal paper <http://gosper.org/total_term_computations_case2.pdf> . "While this work stands alone as a very interesting identity, it suggests more deeply that the lacunary series above CAN be extended to a function on the entire complex plane, through some advanced generalization of analytic continuation." —rwg

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[math-fun] Multiplication is O(n log n)
by Mike Stay 27 Mar '19

27 Mar '19

https://hal.archives-ouvertes.fr/hal-02070778/document -- Mike Stay - metaweta(a)gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com

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[math-fun] Y-squared puzzle
by Dan Asimov 25 Mar '19

25 Mar '19

Let Y denote the union of 3 unit intervals sharing one common endpoint. Consider the *cartesian square* Y x Y. Since Y is the union of 3 intervals I_j, j=1,2,3, Y^2 must be the union of 9 squares Q_jk = I_j x I_k that may intersect each other, but only along a common edge or vertex.) Let G be the graph consisting of all the edges of Y x Y. PUZZLE: ------- How many distinct simple closed curves occur in G ??? Extra credit section: ------------------------------------------------------------------- We can also look at the space that is the nth cartesian power of Y: Y^n = Y x Y x ... x Y (n times), and ask: ----- What is the "embedding dimension" E(n) of Y^n, defined as the least d in Z+ such that Y^n is homeomorphic to a subset of Euclidean space R^d ??? ---- Since Y is (homeomorphic to) a subset of R^2, clearly Y^n is a subset of R^(2n). Question: ----- For instance, can Y^2 be embedded in R^3 ??? ----- ***> A proof or a counterexample is sought. ------------------------------------------------------------------- —Dan

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Re: [math-fun] Pi s ~= nanocentury
by Keith F. Lynch 24 Mar '19

24 Mar '19

That's roughly equivalent to what I noticed long ago, that the number of seconds in a year is almost exactly equal to 10^7.5. It's convenient to memorize the common logs of common things, so I can do mental arithmetic on very large and very small quantities when I'm out walking, which is when I do my best thinking. Other weird units include the microfortnight, used in the VMS operating system, the hubble-barn unit of volume (which combines a very small area with a very long distance), and the Planck units, which define the speed of light, the gravitational constant, and Planck's constant as being 1, and derive all other units from those. In the Planck system the units of mass and energy are of reasonable size, but pretty much all other units are either absurdly large or ridiculously small. Also see https://en.wikipedia.org/wiki/List_of_humorous_units_of_measurement

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[math-fun] Pi s ~= nanocentury
by Georg Dr. Fischer 23 Mar '19

23 Mar '19

It may be an old joke for you, but I did not know that pi seconds are almost a nanocentury. - cf. < https://en.wikipedia.org/wiki/List_of_humorous_units_of_measurement>.

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[math-fun] Continued fractions of Gaussian integers for complex numbers
by Dan Asimov 23 Mar '19

23 Mar '19

Let z be any complex number whose real and imaginary parts are not both rational. Set z_0 = floor(Re(z)) + i*floor(Im(z)) (the lower left corner of the integer grid square that z lies in). Thereafter set z_(n+1) = floor(Re(1/z_n)) + i*floor(Im(1/z_n)) Then z_0, z_1, ..., z_n, ... in Z[i] form a series of Gaussian integers representing z, generalizing continued fraction expansions of irrational real numbers. * Have such things been studied? * For almost all real numbers x, the geometric mean of all the "digits"* gm(x) = lim (K_0 K_1 ... K_n)^(1/n) n —> oo of the cfe of x is equal to the same number. What about the corresponding geometric mean of the Gaussian integers gm(z) = (z_0 z_1 ... z_n)^(1/n) ??? —Dan ————— * Should we call them "tigids"?

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