May 2020 - mathfun@mailman.xmission.com

[math-fun] Only eight
by Hans Havermann 04 May '20

04 May '20

https://oeis.org/A305942 I recently extended this conjectured sequence (the number of decimal expansions of powers of two that contain exactly n digits 0). There would of course be analog sequences for the other nine digits. I wondered if any of these ten sequences might have 0 as a term. In other words, is there an n for which the decimal digit d does not appear in any power of two exactly n times? The digit 7 appears exactly 275923 times for *only eight* powers of two, the smallest number of solutions that I found (for any d, n up to 295000). It would take a gargantuan effort to bring this down to seven or six solutions, so no, I don't think zero is ever going to be a known. Against a backdrop of all counts of the digit 7 in powers of two from 2^9100000 to 2^9240000, here are my eight solutions in graphical form: http://chesswanks.com/num/eight.png

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[math-fun] 40 numbers under 40
by Cris Moore 03 May '20

03 May '20

https://www.theonion.com/40-numbers-under-40-1843198348 <https://www.theonion.com/40-numbers-under-40-1843198348> Cris Moore moore(a)santafe.edu

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Re: [math-fun] Moebius-like torus name?
by Keith F. Lynch 03 May '20

03 May '20

Steve Witham <sw(a)tiac.net> wrote: > What does one call a tube looped around as if to make a torus, but > connected clockwise-to-counterclockwise, as in this sewing pattern? > z y>--C-->z y > +---------+ > x|w>--A-->x|w > v|^ v|^ > ||| ||| > B|D B|D > ||| ||| > v|^ v|^ > y|z<--C--<y|z > +---------+ > w x<--A--<w x If I'm understanding your diagram correctly, that's a Purse of Fortunatus. If you start with a square, and attach the top to the bottom without any twists and attach the left side to the right side without any twists, you get a torus. If, instead, you attach either the top to the bottom or the left to the right with a twist, i.e. reverse the direction, you get a Klein Bottle. If you reverse the directions of both connections, you get a Purse of Fortunatus. None of these surfaces have any intrinsic curvature (i.e. triangles have the same sum of angles, and circles have the same ratio of circumference to diameter, as in a flat plane). Their embeddings in 3-space, however, have extrinsic curvature. And, except for the torus, are self-intersecting unless embedded in 4-space or higher. The Klein Bottle and the Purse of Fortunatus are non-orientable, i.e. the chirality of a figure can change by moving the figure. I don't know what the analogous results would be in higher dimensions, i.e. gluing opposite sides of a cube together, flipping none, some, or all of them. It would be interesting if one of those was the topology of our universe. I've read that some astronomers have done searches for repeating patterns of galactic clusters in an attempt to test this. If our space is non-orientable, then if you travel far enough in a straight line, you'll return to Earth swapped left-to-right. That might mean that you'd be converted to anti-matter. Or it might just mean that pine trees would smell like lemons and you couldn't digest the food. It would be a good defense against viruses, however.

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[math-fun] Handy solution manuals for math textbooks
by Eugene Salamin 03 May '20

03 May '20

https://waxworksmath.com/solutions.aspx -- Gene

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Re: [math-fun] Couple generator (The Erdos-Bacon isomorphism)
by Steve Witham 03 May '20

03 May '20

> From: James Propp <jamespropp(a)gmail.com> > Date: 5/3/20, 9:11 AM > > Is there a database I can query that will give me a pair of famously > linked individuals whose first names start with the respective letters? There are three I can think of: https://oakland.edu/enp/Â (Erdos Number Project Data Files link on the left). http://www.rhydlewis.eu/movies/ There are links to the data and Python code Lewis used, at the bottom of the page. Maybe https://csxstatic.ist.psu.edu/downloads/data CiteSeer data gives the most easily mineable "famousness" information. Funny, Bogart and Bacall are excluded from this sort of thing because they're on the diagonal. Â --Steve

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[math-fun] Fwd: this lifted my spirits: hexaflexasafety
by Cris Moore 03 May '20

03 May '20

from the inimitable Vi Hart: https://www.youtube.com/watch?v=AmN0YyaTD60 <https://www.youtube.com/watch?v=AmN0YyaTD60> - Cris

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[math-fun] Couple generator
by James Propp 03 May '20

03 May '20

Say I want to describe a two-player game, and for mnemonic reasons, I want the names of the players to start with two specific letters (such as the letters I and D in my previous email). Is there a database I can query that will give me a pair of famously linked individuals whose first names start with the respective letters? (They could be siblings or spouses or friends or rivals, as long as the relationship between them is common knowledge.) This is related to a party game (I believe I invented it but I could be mistaken) in which two letters are drawn from a bag, and within some limited time each player, working on their own, has to generate pairs of words starting with those respective letters. For instance, S and P could give "salt / pepper", "strict / permissive", "Standard" / "Poor", etc. People go around the room saying what they found, and anything that was found by just one person gets crossed out. Then there's a scoring stage, whose details I forget, that rewards people for finding pairs that were found by a small but still positive number of other people. Jim Propp

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[math-fun] We love commas
by Éric Angelini 03 May '20

03 May '20

Hello Math-Fun, S = 2,5,26,159,1447,10274,45206,280278,2298281,... If we frame any comma above with its two closest digits we get 2.5 for the first comma, 5.2 for the second comma, 6.1 for the third one, etc. Look now: the division of 5 by 2 starts with 2.5 the division of 26 by 5 starts with 5.2 the division of 159 by 26 starts with 6.1 etc. [In other words, a(n+1)/a(n) starts with [x.y] with x = the rightmost digit of a(n) and y = the leftmost digit of a(n+1)]. As usual, we want S to be the lexico- graphically earliest seq of distinct terms with this property... but! Here comes the (Math-)fun part. I thought that S would increase monoto- nically for ever when I bumped into a(9) = 2298280 (and not, as above, = 2298281). I realized that S could decrease at some point! Indeed, this example works: 120,24,... as the result of 24/120 starts with 0.2). But this "zero comma" trick doesn't systematically work with integers ending in 0 (it works rarely, in reality). If it doesn't, we have to backtrack -- and increase by 1 the terms ending in "0" that would stop the sequence; this is what I did above. The question remains: will the "+1" trick always work? Or will S stop at some point? Enter into a loop? Extend for ever? [I've noticed that the ratio a(n+1)/a(n) gets closer at every stage to [x.y], (the divisor), but I don't know if it means something for the future of S]. Best, É. à+ É. Catapulté de mon aPhone

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[math-fun] An Erdos-Szekeres game
by James Propp 03 May '20

03 May '20

The Erdos-Szekeres theorem tells us that every list of n^2+1 distinct numbers must contain either an increasing sequence of length n+1 or a decreasing sequence of length n+1. This suggests a two-player game (call the players I and D) whose state after k moves is a list of k distinct numbers, where the players take turns generating a new number via some random mechanism and inserting it somewhere in the list, such that player I (resp. player D) wins when the sequence contains an increasing (resp. decreasing) subsequence of length n+1. (I was thinking of giving the players actual names, but all I could come up with is Ivana and Donald, and I think all of us can use a break from politics.) The E-S theorem guarantees that someone will win within n^2+1 moves. One could play this game with a bag of tiles (Scrabble tiles would be a nice choice, since the number of letters in the alphabet is conveniently of the form n^2+1). Or one could roll a die repeatedly, sampling with replacement rather than without, and bound the length of the game using a variant of the Erdos-Szekeres theorem in which repeats are allowed and the subsequences are permitted to be weakly increasing/decreasing. Has anyone studied such a game? I just did some searching and found other Ramsey theory games but not ones involving an element of chance. Jim Propp

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Re: [math-fun] "Dedecision"
by Steve Witham 01 May '20

01 May '20

(A) "Dedecision" is a Dedekind cut or the process of "making" one. The nice thing is that -cision means (making of) a cut. As in, well, decision, and also precision, concision, incision, excision, etc. I hope Dedekind doesn't mind. So, I wanted to get the *exact* value of round((j + 1/2) sqrt(2)) where j is an int. Um in Python. I have a floor-sqrt for arbitrarily big ints. And I know that I'll never have an exact halfway case to round because sqrt(2) is irrational. "All I have to do is" guess low by less than one, then see whether that's too low. The code is... well the derivation is ugly but what falls out looks sort of nice... and it has intermediate values like 8 j^2, but the thing I retained from what I read about Dedekind is that as long as you have a test for all the rationals, you've "captured" the real number. --Steve def round_j_plus_half_sqrt_2(j): """ Given int j, return *the exact* round((j + 1/2) * sqrt(2)). """ # round((j + 1/2) * sqrt(2)) # = round(sqrt((j + 1/2)**2 * 2)) # = round(sqrt((j**2+j+1/4))*2) # = round(sqrt((2*j*(j+1)+1/2))) # STEP 1: Take the floor rather than round. # k = floor(sqrt( 2*j*(j+1)+1/2) ) # = floor(sqrt(floor(2*j*(j+1)+1/2))) # = floor(sqrt( 2*j*(j+1) ) ) k = isqrt(2 * j * (j + 1)) # STEP 2: If k + 1/2 is too small, round it up. # test whether (k + 1/2)**2 < 2 * j * (j + 1) + 1/2 too_small = bool(4 * k * (k + 1) + 1 < 8 * j * (j + 1) + 2 ) return k + too_small -fin-

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