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[math-fun] Cutting up a strip of squares
by Neil Sloane 11 Apr '20

11 Apr '20

Here's a simple question I've been struggling with for weeks. Draw a row of n adjacent squares (so 2(n+1) boundary points). Join every pair of boundary points by a line segment. The resulting graph G_n is a K_{n+1, n+1} with two extra boundary lines. Here is a colored illustration of G_3: https://oeis.org/A331452/a331452_8.png (A331452 has many other pictures.) Thanks to the work of Legendre (2009), Griffiths (2010), and Max Alekseyev (2010, 2015),we know the numbers of regions (or cells) in G_n, which is A306302: 4, 16, 46, 104, 214, 380, 648, ... (and has an elegant formula), and also the number of vertices (A331755) and edges (A331757). My question is, why are all the cells in G_n triangles or quadrilaterals? Why are there no pentagons or higher-order cells? Max gives an indirect proof of this fact referring back to an old Russian theorem about training sets for threshold functions. But there ought to be a proof just from geometry. Once we know this fact, we can deduce the numbers of triangular cells (A324042) and quadrilaterals (A324043).

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[math-fun] More on Occam's Razor
by Brad Klee 11 Apr '20

11 Apr '20

The other day Keith Lynch was saying that COVID-19 has been a good reminder to the public about exponential functions, and I agree. The next test is whether or not the public can understand what we could loosely call "logistic peaks". These are even more simple than the output of the COVID calculator mentioned earlier. For example: [1] https://demonstrations.wolfram.com/SIREpidemicDynamics/ [2] https://demonstrations.wolfram.com/ReducingFatalitiesFromCoronavirusEpidemi… The second, latest effort [2] is a rehash of [1], and both of them have unnecessary parameters, as does mathworld: [3] https://mathworld.wolfram.com/Kermack-McKendrickModel.html Change variabless by dividing both S & I by beta, and this reduces to one parameter, gamma. It is a shape parameter, where low values of gamma indicate asymmetry and relatively quick emergence. Once you understand this, the question is: how well do you think you can estimate shape and width (and possibly amplitude)? When yr "stepping, walking, cutting, flicking, jumping, chopping, walking" you just say: "I don't know what the shape will be", choose a fixed shape that looks enough like the _one_ complete data set from China, try to match the S-shape from emergence to peak on other data sets, and get something like: https://0x0.st/iSIz.png I am hoping that "Insect Pandemics" makes it through the editorial stack at Wolfram soon, so that we can get a hyperlink to another perspective on how the analysis can be done, and a reminder that other more enjoyable pandemics will be started shortly. The other question is what should you do if you were part of the March purge? I don't know, but will say that I have been enjoying the opportunity to do some more hiking and field work, see also: https://www.inaturalist.org/observations/41879763 (digged out of a dung pat and cleaned yesterday) Hooray for social distancing! "Treat me good", Brad

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[math-fun] 1, 2, 3, drink (my numbers)
by Éric Angelini 10 Apr '20

10 Apr '20

Hello Math-Fun, we want to spell some numbers with other numbers, like here: TWENTYONEsevenFOURTEEN This spells ONE because the letters in position 21, 7 and 14 are precisely E, O, and N: T W E N T Y O N E s e v e n F O U R T E E N 1 2 3 4 5 6 7 8 9 o 1 2 3 4 5 6 7 8 9 o 1 The above example is not unique – there are many ways to spell such numbers: ONEtwoTHREE (spells 1 with 1, 2, 3) TWENTYONEsevenFOURTEEN (spells 1 with 21, 7, 14 – opening example)* (...) The other part of the mail is on my webpage, with the question: > Is it possible to spell all numbers from 1 to 20 in the *star mode* (this is, without using in its spelling the number we want to produce)? https://bit.ly/3c9X7pI Best, É.

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Re: [math-fun] 5883800000000000 and other pandigital numbers
by Keith F. Lynch 10 Apr '20

10 Apr '20

Michael Kleber <michael.kleber(a)gmail.com> wrote: > Nice table, Keith! Thanks. > I have little to offer, but at least my copy of Mathematica can > answer one of your questions by brute force: The first prime that > doesn't divide any of the 10! base-10 pandigitals is apparently > 111119. Thanks, though that wasn't actually one of my questions. > (Computing the 10! prime factorizations took under a minute.) Likewise. But it gets much slower for each higher base. Also, if I were ever to go beyond base 16, I could no longer use 64-bit integers, which would create an additional slowdown. > (I guess doing the composite-numbered rows would involve some > factor/round/max logic that I haven't written but isn't hard.) Since I only bothered to go through powers of 16 (to make the table square, given that I stop at base 16), I hand-coded that (in C): if (p[i] == 2) { n[4] = n[2]/2; n[6] = min(n[2],n[3]); n[8] = n[2]/3; n[10] = min(n[2],n[5]); n[12] = min(n[3],n[4]); n[14] = min(n[2],n[7]); n[16] = n[2]/4; } if (p[i] == 3) { n[6] = min(n[2],n[3]); n[9] = n[3]/2; n[12] = min(n[3],n[4]); n[15] = min(n[3],n[5]); } if (p[i] == 5) { n[10] = min(n[2],n[5]); n[15] = min(n[3],n[5]); } if (p[i] == 7) { n[14] = min(n[2],n[7]); } Had I been lazy, I could have simply tried repeatedly dividing the pandigital number by each of the composite numbers. > To find the first 0 in the column, i.e. the first missing prime: > For[n=1, maxpower[Prime[n]] > 0, n++, Null] > {n, Prime[n]} > (* Answer: {10545, 111119} *) I see that Mathematica starts counting with 1, not 0. (If 111119 is prime 10545, then 2 is prime 1, not prime 0.) > The first zero in the base-6 column is 229, and in base-8 is 4663. Confirmed. Starting with base-2, I get 3,2,17,67,229,2,4663,.... Since 111119 is an interesting-looking number in base 10, I looked at those other numbers in the bases they go with: 11,2,101,232,1021,2,11067. Not very interesting. But your 3257447 is 111111B in base 12. That can't be a coincidence. (Well, it can, but it's very unlikely.) What prime do you get for base 9, and what does it look like in that base? > None of this counts as "efficient", certainly, since the work grows > as worse than the factorial of the base. Indeed. > For base 12, I had to change my Mathematica thing to not materialize > all those lists of 12! things all at once, since they wouldn't fit > in memory. It would never even occur to me to try to put them in memory. I deal with each pandigital number one at a time. My actual question was: >> For each column, especially base 10, where's the first zero, i.e. >> the smallest number that no N-digit base-N pandigital number is >> divisible by? That's a much easier question, as the answer can't exceed N^2. The answers are 3,2,16,4,36,2,64,8,100,2,144,4,... I wonder if the answer is always N^2 for even values of N other than 2. And if it's ever N^2 for odd values of N. For odd N does it ever exceed N itself?

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[math-fun] 5883800000000000 and other pandigital numbers
by Keith F. Lynch 09 Apr '20

09 Apr '20

Yes, 5883800000000000 is pandigital -- in base 14, where it's 75AD3621CB9408. No other pandigital number, in that or any smaller base, is divisible by a higher power of 10. Here's the latest version of my table, showing, for each small base, what's the highest power of each small number that divides any N-digit base-N pandigital number. (As always, this and every table should be viewed in a fixed-width font.) Base: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 2: 1 0 3 1 11 0 5 2 21 0 29 1 36 0 7 3: 0 1 3 5 7 9 10 3 15 17 18 21 23 4: 0 0 1 0 5 0 2 1 10 0 14 0 18 0 3 5: 0 1 2 1 5 5 6 9 9 11 12 13 15 6: 0 0 3 1 1 0 5 2 8 0 3 1 13 0 7: 0 1 2 2 3 1 7 6 8 9 9 13 13 8: 0 0 1 0 3 0 1 0 7 0 9 0 12 0 2 9: 0 0 1 2 3 4 5 1 7 8 9 10 11 10: 0 0 1 1 3 0 5 2 1 0 8 1 11 0 11: 0 1 1 2 2 3 5 7 5 1 8 9 11 12: 0 0 1 0 1 0 2 1 6 0 1 0 10 0 13: 0 0 1 2 2 3 3 5 5 6 8 1 11 14: 0 0 1 0 1 0 2 1 6 0 1 9 1 0 15: 0 1 2 1 2 3 3 3 6 7 6 9 9 1 16: 0 0 0 0 2 0 1 0 5 0 7 0 9 0 1 Note that: For powers of any even number, bases that are congruent to 3 mod 4 have no solutions That's because in any odd base, a number is odd if and only if it has an odd number of odd digits. Any N-digit pandigital number of a length congruent to 3 mod 4 will have an odd number of odd digits, hence every permutation of those digits will represent an odd number. The diagonal values are all 1. That's because in an N-digit pandigital number, there's only one 0, and for a base-N number that digit has to go on the end for the number to be divisible by N. For divisors that are a power of 2, the values are the corresponding values for base 2, divided by which power of 2 it is, rounded down. For instance since the highest power of 2 in any 10-digit base-10 number is 21, the highest power of 16 (2^4) in any 10-digit base-10 number is 21/4 = 5. Analogously with powers of other divisors, e.g. powers of 3 vs. powers of 9 or of 27. That's because base 2^N numbers can be viewed as base-2 numbers chunked in sets of N base-2 digits. Analogously with powers of other divisors. For base N=2^M, the highest power of 2 that will divide an N-digit pandigital number is 2M-1. For instance in base 16 (2^4), it's 7. Again, that's because of chunking. In base 16 (hexadecimal), if the number ends with 80, that's 1000 0000 in base 2 (binary), as every programmer of my generation knows. That's seven 0s. There's no way to get more without a second hexadecimal 0. Questions that remain: Is there an efficient way to fill in the table for higher bases? Currently, I'm testing every N-digit base-N pandigital number for divisibility by every small prime. For each column, especially base 10, where's the first zero, i.e. the smallest number that no N-digit base-N pandigital number is divisible by? Which, if any of these rows and columns belong in OEIS? I don't think any of them are there yet. Where's the error in this post that I will notice immediately after I send it?

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[math-fun] Hyperbolic pairing simpler
by Steve Witham 09 Apr '20

09 Apr '20

Simpler description of "hyperbolic" pairing. Look at all the hyperbolas like y = n/x, where n is an int >= 1. Start with y = 1/x, which goes through the point (1, 1); and declare 1 <=> (1, 1). For each successive n, start at (1, n) and slide down y=n/x to (n, 1). As the curve passes through integer points, give each one you pass the next available int label. At the bottom, take the lift up to (n+1, 1). Where Cantor's pairing rides up escalators, this is an infinite day at the infinite ski resort. So, (n = 1) 1 <=> (1, 1), (n = 2) 2<=> (1, 2), 3<=> (2, 1), (n = 3) 4<=> (1, 3), 5<=> (3, 1), (n = 4) 6<=> (1, 4), 7<=> (2, 2), 8<=> (4, 1), (n=5) 9<=>(1, 5), 10<=>(5, 1), (n=6) 11<=>(1, 6), 12<=>(2, 3), 13<=>(3, 2), 14<=>(6, 1)... It can be remembered and programmed from that much. It must have been thought of before but I haven't successfully googled it. (There are hyperbolic pair interactions in physics, and side-pairing transformations in hyperbolic geometry.) I've been trying to optimize speed when I'm more interested how this pairing is optimal for the goal that space_cost( (x, y) ) = ~(space_cost(x) + space_cost(y)). That makes it a point of comparison for more reasonable-to-implement pairings. Also I *think* it relates the law of large numbers to data compression in a cool way. I like that it gets its optimality by just being a list of pairs in order of their products, without being defined in terms of digits, bits, or trees. 30 28 24 21 17 15 11 31 9 25 6 18 32 4 12 22 33 2 7 13 19 26 34 1 3 5 8 10 14 16 20 23 27 29 35 --Steve P.S. d(n) is the number of divisors of n; a conventional name for the sum of d(n) is D(n), the "divisor summatory function." "We are riding the escalator in our dream department store." --Bill Nelson

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[math-fun] Two topological visualization questions.
by Andy Latto 06 Apr '20

06 Apr '20

Since there's only one embedding of a circle in R^3 up to homotopy, there's an embedding of a mobius strip in R^3 where the edge is a geometric perfect circle. But I find myself unable to visualize such a thing. Has anyone seen a 3-d model of this surface? Second-best thing would be a graphic of such a thing, preferably one that you could rotate in 3 dimensions. I'd also like to better visualize Boy's surface, or any other immersion of RP^2 in R^3. It would also be interesting to have an insight into why immersing a Klein bottle in R^3 is easy, while immersing RP2 is "hard". I don't know of any formal sense in which this is true, but apparently Boy came up with this surface when challenged by Hilbert to prove that immersing RP^2 in R^3 was impossible. Also, are these two questions related? That is, can you immerse a mobius strip in R^3 in such a way that the boundary is a geometric circle, and that the union of this mobius strip and a disk with the same boundary is still an immersion (of RP^2 in R^3)? Andy Latto andy.latto(a)pobox.com -- Andy.Latto(a)pobox.com

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[math-fun] Etymology of "Sudanese Möbius Band"
by James Buddenhagen 04 Apr '20

04 Apr '20

The Möbius band with circular boundary discussed here recently is apparently (I learned the day before yesterday) called the Sudanese Möbius band with "Sudanese" being a composite of the first names of Sue Goodman and Daniel Asimov (see https://en.wikipedia.org/wiki/M%C3%B6bius_strip#M%C3%B6bius_band_with_round… ) I was just wondering if this is the same Dan Asimov who used to post to math-fun quite often?

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[math-fun] Shouldn't 'social distance' be relativistic?
by Henry Baker 04 Apr '20

04 Apr '20

Wouldn't the appropriate metric be distance in space-time, where c = speed of cough? Jez askin'...

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[math-fun] Every breath you take
by James Propp 03 Apr '20

03 Apr '20

There's a popular science writing trope about how every breath you take contains atoms that were once breathed by X [or, for extra piquancy, in X's dying breath]. I think the one I saw in the late 1960s had X = Archimedes. My first question is, do any of you remember a specific source from the 1900s? (At https://www.newscientist.com/article/mg15020308-500-the-last-word/ someone writes "Is it true that every time we take a breath of air or swallow a mouthful of water, we consume some of the atoms breathed or swallowed by Leonardo da Vinci (as I read in a children’s science book in 1960)?" but doesn't mention the source.) My second question is, what's the best writeup of this that you know? I just borrowed (in e-book form) Sam Kean's recent book "Caesar's Last Breath", which I'm guessing will do a decent job of explaining the science. My third question is, how far back does this sort of observation go? Does it have an attributable original source? (Maybe Kean addresses this; it's a pretty thick book.) If the history of science has a subfield treating history of science popularization, and people do PhD's in it, maybe someone has tracked this trope in all its incarnations, showing how the choice of who gets to play the role of X reflects underlying cultural assumptions about significance and worth. Jim Propp

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