April 2008 - mathfun@mailman.xmission.com

[math-fun] Collatz/Gosper shirts
by Jason 13 Apr '08

13 Apr '08

(Apologies in advance if this is excessively commercial; in my defense, I'm not out to make any money, and it all derives directly from threads on this list) A few years ago I plotted out the iterations to reach 1 for the Collatz conjecture for n \in{1,1e9} (and posted the results here). The graph was so interesting that I made a T-shirt out of it, and today I finally got around to adding it to one of their "marketplace" accounts: http://www.cafepress.com/lunkwill.254254793 I included a $1 markup on the items just for the heck of it, but let me know if you'd like it at cost. I also did a version of the shirt for my friend doing a postdoc at MIT, changing the text from "COLLATZ" to "Increment, Halve, Triple. Forever, Please.": http://www.cafepress.com/lunkwill.254237008 The preview isn't intelligible, so here's the image: http://lunkwill.org/ihtfp-black.png (If you want this one, you may want to wait until my friend gets it and lets me know if it looks okay) I also became enamored of Bill's space-filling function, and made some sweatshirts of it: http://www.cafepress.com/lunkwill.254257188

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Re: [math-fun] 24 puzzle
by Dan Asimov 13 Apr '08

13 Apr '08

Andy wrote: << [I wrote]: << Find the set of prime numbers p such that p^2 == 1 (mod 24). If (n, 24) = 1, then n^2 == 1 (mod 24). So that would be all primes > 3. >> This statement "If (n, 24) = 1, then n^2 == 1 (mod 24)" is true and easy to verify, but it is also profound. There is no number > 24 for which the corresponding statement is true. (This is also not hard to prove, but I wonder if there is a deep underlying reason it is true.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Prouhet-Tarry-Escott in Magic Square
by Richard Guy 13 Apr '08

13 Apr '08

Paul, Never seen it before, but it surely can't be new???? Let's try it on the math-fun network. R. On Fri, 11 Apr 2008, Paul Muljadi wrote: > Professor Guy, > > Have you seen a Prouhet-Tarry-Escott solution in the > Lo Shu square? > > 4 9 2 > 3 5 7 > 8 1 6 > > The two number sets from the magic square are obtained > by combining numbers of last rows and last columns in > the clockwise and anticlockwise directions: {492, 276, > 618, 834} and {294, 438, 816, 672}. > > Tarry-Escott solution: > 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + > 816^1 + 672^1 > 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + > 816^2 + 672^2 > 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + > 816^3 + 672^3. > > If you have, who discovered it first? > > Paul Muljadi > > >

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[math-fun] crossword puzzle clue in today's (Saturday) New York Times (Byron Walden)
by Thane Plambeck 13 Apr '08

13 Apr '08

(already yesterday's NYT, I guess, it's after midnight), but I just finished it. Here's the clue: Setting numbered in multiples of the square root of 2 (5 letters) I agonized, but eventually got it by working around it with other clues. -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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Re: [math-fun] Prouhet-Tarry-Escott in Magic Square (fwd)
by Paul Muljadi 12 Apr '08

12 Apr '08

Here's what we have so far: ---------------------------------------------------------------------------- 6 1 8 4 9 2 7 5 3 <-> 3 5 7 2 9 4 8 1 6 ---------------------------------------------------------------------------- 618^n + 753^n + 294^n = 816^n + 357^n + 492^n (rows) 672^n + 159^n + 834^n = 276^n + 951^n + 438^n (columns) 654^n + 132^n + 879^n = 456^n + 231^n + 978^n (diagonals \) 852^n + 174^n + 639^n = 258^n + 471^n + 936^n (diagonals /) True for n = 1 and 2. True again if you delete the first digit in each number. True again if you delete the second digit in each number. True again if you delete the last digit in each number.[1] ----------------------------------------------------------------------------- 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.[2] ----------------------------------------------------------------------------- References: [1] R. Holmes, The Magic Magic Square, The Mathematical Gazette, December 1970, p.376 [2] ? ---------------------------------------------------------------------------- The Lo Shu is more "magical" than most of us have known for thousands of years! Paul --- Richard Guy <rkg(a)cpsc.ucalgary.ca> wrote: > Ditto. R. > > ---------- Forwarded message ---------- > Date: Sat, 12 Apr 2008 09:23:06 +0200 > From: Christian Boyer <cboyer(a)club-internet.fr> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > To: 'math-fun' <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Prouhet-Tarry-Escott in > Magic Square > > Same properties on: > > > > 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 > + > > > 816^1 + 672^1 > > > 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 > + > > > 816^2 + 672^2 > > > 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 > + > > > 816^3 + 672^3. > > True again if you delete the first digit in each > number. > True again if you delete the second digit in each > number. > True again if you delete the last digit in each > number. > > Christian. > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > >

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[math-fun] spacefilling "curves"
by Bill Gosper 12 Apr '08

12 Apr '08

Joerg, Jason, et al>I'd much like to see a generalization of the string substitution rules for the 2D curve, I hate this substitution rule picture of "spacefilling curves". I even hate the term "spacefilling curve". A curve is a range set, the image of some function of the unit interval. If it's really a spacefilling function, the "curve" is a two (or more) dimensional blob. There is no way to define this blob as the "limit" of some sequence of squiggly lines, nor to prove that the blob is not only completely filled, but has countably many points triply covered. To spacefill a blob, find one with a self similar dissection and order the n subblobs in a Hamiltonian path (the path suggested by your polygonal squiggles). Chop the unit interval into n pieces of which the subblobs are the images. Recurse. Now you can *exactly* evaluate your spacefilling *function* (and its up to three-valued inverse). Choose t in the unit interval. Write it in base n. Construct a finite state machine that turns the digits base n^2 into pairs of digits base n. So rationals map to rationals, which you can give exactly via loop detection. You can even play with irrationals. I seem to recall proving that if P(t) = (x,y) is the Peano squarefilling function, then the only solutions of P(y) = (1,y) are y = theta and 1-theta, where theta := the Thue parity constant .0110100110010110... With some more work, you can write the *exact* fourier coefficients that trace out a time-smoothed approximation (a curve!) with rotating vectors. (Difficult paper: http://gosper.org/fst.dvi . Crude pix: http://gosper.org/fst1.png,http://gosper.org/fst2.png, http://gosper.org/fst3.png,http://gosper.org/fst4.png .) But the squiggly lines are pretty, you say. Yes! And the way to get them, in abundance, is to choose some rational(?) step size and offset, evaluate the function at those equally spaced t, and connect the dots. See www.tweedledum.com/rwg/. Click Sampled Peano squarefiller, Sampled trianglefilling function, Sampled Sierpinski halfsquare filler . The latter is the two-dimensional limit of the Snowflake construction-- a really nice spacefilling function, but you'd never know it from the recursive polygon view, which completely retraces itself and makes a simple grid from a bunch of crosses. >P.S.: is there any paper dealing with three- or higher dimensional Hilbert curves that is not totally obscure? Foo, just self-similarly dissect a 3D blob and make an FSM to crank out 3-vectors. --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [math-fun] a perplexing asymmetry
by Steve Witham 12 Apr '08

12 Apr '08

[I get the digest so forgive me if Jim has reported this already.] The smallest n where f(n) < n/2 is 5495. Here are the record-breakers for f(n)-n/2 that I've computed so far: n f(n)-n/2 T(n) (see below) 3 +0.5 2 4 +1.0 2 13 +1.5 4 40 +2.0 8 43 +2.5 8 124 +3.0 14 181 +3.5 16 622 +4.0 26 1435 +4.5 34 1882 +5.0 36 4225 +5.5 44 5495 -0.5 48 <-- 8992 +6.0 52 12343 +6.5 56 19108 +7.0 62 24632 -1.0 66 24713 -1.5 66 31891 +7.5 68 71968 +8.0 78 71971 +8.5 78 152990 -2.0 88 153017 -2.5 88 288568 +9.0 96 288577 +9.5 96 549512 -3.0 106 549513 -3.5 106 665470 +10.0 108 1575895 +10.5 120 1575898 +11.0 120 2850236 -4.0 128 2851713 -4.5 128 9440910 -5.0 146 10034029 +11.5 146 10038334 +12.0 146 16214491 +12.5 154 >From: James Propp <jpropp(a)cs.uml.edu> > >(1) How big can |f(n) - n/2| be? I suspect it's O(log n). (For >n up through 500, f(n) - n/2 ranges between 0 and 7/2.) It looks sort of log like, but if you graph those numbers above, there are very piecewise-linear features, which doesn't make me confident I know what it's doing. f(n)-n/2 is like a moving slice on the far tail of a growing, skewed bell curve. >(3) What's the best way to compute f(n)? It would be good to >know ahead of time two numbers L(n) and T(n) such that if you >replace the infinite one-dimensional lattice {1,2,3,...} by the >finite interval {1,2,3,...,L(n)} and run the arithmetic rule >for T(n) steps, with particles that leave 1 going leftward being >absorbed at 0 and particles that leave L(n) going rightward being >ignored, then f(n) is just n minus the number of particles that >have been absorbed at 0 by time T(n). T(n), the time step where you should stop, is easy to see as you compute. It's the point where the population at lattice point 1 reaches 1. Once initially colonized (& skipping every other time step), (I believe but haven't proven) a lattice point's population never increases again. So once point 1 reaches 1, it will never drop particles to the left again. The third column of numbers above shows T(n). It's log-like starting at n ~= 3000, but it might be line segments approximating log^2. L(n) is how wide to make the rows. Once you've cut back the number of rows drastically, making L(n) < T(n) doesn't help much, maybe a factor of two. But the best speedup is to keep the 2D table between calculations. The change between calculating f(n) and f(n+1) is to increment one entry on each row. Here's the inner loop in Python: i = 1 # Column 0 absorbs particles. for t in range( len(A) ): # for each row A[ t ][ i ] += 1 if i > 0: # Once in column 0, stay there. if A[ t ][ i ] % 3 == 2: i -= 1 else: i += 1 # Once at the bottom you occasionally need to add two rows. So successive visits to a 2D point go down to the right, next time left, next time right; right, left, right, etc. It's a rounder-router! Guess you can do the whole calculation by keeping each count mod 3. Might be nice to plot in color. --Steve

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[math-fun] peano staircase
by Thane Plambeck 12 Apr '08

12 Apr '08

seb przd strikes again http://flickr.com/photos/sbprzd/2405205641/ -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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[math-fun] 24 puzzle
by Dan Asimov 11 Apr '08

11 Apr '08

Find the set of prime numbers p such that p^2 == 1 (mod 24). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] a perplexing asymmetry
by James Propp 09 Apr '08

09 Apr '08

Rich Schroeppel writes: >Have you looked at whether varying the split ratio also gives >the odd lower bound results? What about 1:3 split (with various >rounding rules), or 1:4 split, or 1:2:2; or 1:1:1, or fiddling >with the reach 1:*:1:1? Nice idea. As you noticed, the 1:3 split requires a tie-breaking rule for rounding numbers halfway between two integers, so I'll skip over that (since I don't know what the right tie-breaking rule should be --- though I have a strong conviction that the sweetest general theory will be one in which number of particles is conserved over time). The 1:4 split gives the sequence 1, 2, 2, 3, 4, 5, 6, 6, 7, ... and I notice right off the bat that, although the asymptotic value of the nth term is 3n/4, some of the terms fall short of this value; for instance, the 3rd term is less than 9/4. There may still be an asymmetry here, but it's not as dramatic as in the 1:2 split case. I'm not sure what you mean by 1:2:2. I'm guessing the idea is that nearly 1/5 go left, nearly 2/5 go right, and nearly 2/5 stay put, but I don't know how to do the rounding, since the integers closest to n/5, 2n/5, and 2n/5 don't always add up to n (e.g., take n=1), and I'm extremely prejudiced against non-conservative rules. Jim

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