March 2020 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

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[math-fun] Chess variant
by David Wilson 13 Nov '20

13 Nov '20

In Paul Morphy's "Opera House" chess game, Morphy never moves a piece backward. I propose Opera House Chess, with initial position and moves as in chess, except pieces cannot move backward. Is it playable? With victory conditions modified?

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[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

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Re: [math-fun] FactorInteger
by Bill Gosper 20 Aug '20

20 Aug '20

Slight improvement: Mathematica's front end by default displays a centerdot for implicit multiplication. Strangely, this doesn't work in old the FI function below, which displays blanks. Blanks also (by default) separate digits in blocks of 3. So the display is sometimes ambiguous. But Wolfram CE Harry Calkins showed me this workaround: FI⨉[n : _Rational | _Integer] := Defer @@ {FactorInteger@n} /. {{a_Integer, 1} -> a, {a_Integer, b_} -> a^b} /. {a_} -> a /. List -> Times E.g. In[144]:= FI⨉@BernoulliB@50]@// tim During evaluation of In[144]:= 0.006522,1 (seconds) Out[144]= (5^2 417202699 47464429777438199)/(2 3 11) (The first Bernoulli number with a square factor!) Dirty trick. I used a tiny ⨉ symbol instead of x, so you need Front End completion to input FI⨉. You may want to edit the definition. Instead of little centerdots between the factors, there's a Preferences variable Appearances|Numbers|Multiplication|Automatically display multiplication symbol between multiplied numbers Use symbol: which you can bash to be a mini ⨉. Sorry, I don't know how to bash it under program control, as in an init file. Unfactor is way nicer than with FactorInteger: UnFI⨉@Defer[x_] := x In[153]:= UnFI⨉@%144 Out[153]= 495057205241079648212477525/66 —rwg On Wed, Jan 4, 2017 at 12:44 AM Bill Gosper <billgosper(a)gmail.com> wrote: > There are > > In[14]:= 52! > Out[14]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > ways to shuffle a deck. > That's just a "little" over a googol ... > > In[366]:= Log[10.^100, %] > Out[366]= 0.679066483922048 > > to the 2/3 power. The square of the cube root. And vice versa. So if > anybody asks how many > ways to shuffle a deck, "Oh, about the cube root of a googol squared." > But this is actually low > by a fairly substantial factor of > In[372]:= (10.^100)^(.679 - 2/3) > > Out[372]= 17.1132830416181 > > The actual number of shufflings is enormously composite: > > In[15]:= FactorInteger@%14 > Out[15]= > {{2,49},{3,23},{5,12},{7,8},{11,4},{13,4},{17,3},{19,2},{23,2},{29,1},{31,1},{37,1},{41,1},{43,1},{47,1}} > > But this is ugly and hard to read. The doc suggests > > In[16]:= CenterDot@@(Superscript@@@%) > Out[16]= > 2^49\[CenterDot]3^23\[CenterDot]5^12\[CenterDot]7^8\[CenterDot]11^4\[CenterDot]13^4\[CenterDot]17^3\[CenterDot]19^2\[CenterDot]23^2\[CenterDot]29^1\[CenterDot]31^1\[CenterDot]37^1\[CenterDot]41^1\[CenterDot]43^1\[CenterDot]47^1 > > This actually looks neat. Except for the ^1s. Try it. But it's unclear > (actually, a nice little Mathematica exercise) how to get your original > number back. > So I wrote something called FI > > In[364]:= FI[52!] > Out[364]= 2^49 3^23 5^12 7^8 11^4 13^4 17^3 19^2 23^2 29 31 37 41 43 47 > > In[365]:= ReleaseHold@% > Out[365]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Here is the definition: > FI[n : _Rational | _Integer] := > Times @@ ((HoldForm @@ {#1[[1]]})^#1[[2]] &) /@ FactorInteger[n] > > And here's a spoiler for how to invert FactorInteger: > > In[367]:= Times @@ (Power @@@ FactorInteger[73 137]) > Out[367]= 10001 > > which also works on that CenterDot expression: > > In[368]:= Times @@ (Power @@@ %16) > > Out[368]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Finally, you might enjoy finding a fairly nice formula for how many 0s are > on the end of n!, and, > more generally, what are exponents of the various prime factors. > --Bill >

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[math-fun] How does a dull spoon manage to cut a stringy piece of meat
by Bill Gosper 14 Jul '20

14 Jul '20

while sliding back an forth against the inside of a slippery bowl? --rwg

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[math-fun] Existence from parity
by James Propp 20 Apr '20

20 Apr '20

What are people’s favorite examples of existence proofs that show that a set is not empty by showing that its cardinality is odd? Jim Propp

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[math-fun] Another pic of Hilbert's spacefill
by Bill Gosper 19 Apr '20

19 Apr '20

to accompany should've been "sampHilbert" <http://www.tweedledum.com/rwg/sampeano.htm> Hilbert's [spacefilling] function is dense with quadruple points, i.e. points in [0,*i*] with four distinct preimage points in [0,1]: gosper.org/hilbquad.png . E.g., the preimage of 1/2 + *i*/4 is {5/48, 7/48, 41/48, 43/48}. This picture connects, in the order they were swept, all the points in [0,*i*] with preimages having denominators ≤ 3⨉4096. (All quadruple point preimages have denominators 3⨉2ⁿ.) E.g., the lower-leftmost vertex is 1/32 + *i*/64, the image of 5/12288, 7/12288, 41/12288, and 43/12288. The white segments are retraced boundary, retroflexed in the middle, (making quadruple points). (Spacefilling functions map closed intervals to closed sets.) The Heighway Dragon is dense with mere triple points. Here's one: Hi res Heighway <http://gosper.org/dragtrip!.png> . —rwg

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Re: [math-fun] Pandigital Trivia Spoiler
by Keith F. Lynch 01 Apr '20

01 Apr '20

Hans Havermann <gladhobo(a)bell.net> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> What do 3076521984, 3718250496, and 6398410752, and no other >> numbers, have in common?" > They are obviously pandigital numbers and thus share the divisor > 3^2. The GCD of the three numbers divided by 9 is a large power of > 2. A quick check shows that no other pandigital has that large a > power-of-two divisor (the next largest being for 9805234176). Correct. My three numbers are all divisible by 2^21. No 10-digit base-10 pandigital numbers are divisible by 2^22. I've looked into the highest powers of small primes among the n-digit base-n pandigital numbers for small n. Here's a table. As always with tables, it should be viewed in a fixed-width font. base: 2 3 4 5 6 7 8 9 10 11 12 13 2: 1 0 3 1 11 0 5 2 21 0 29 1 3: 0 1 3 5 7 9 10 3 15 17 18 21 5: 0 1 2 1 5 5 6 9 9 11 12 13 7: 0 1 2 2 3 1 7 6 8 9 9 13 11: 0 1 1 2 2 3 5 7 5 1 8 9 13: 0 0 1 2 2 3 3 5 5 6 8 1 I can easily push it to higher primes, but not to higher bases. Computing the entries for base n will take roughly n times longer than computing them for all smaller bases put together. I may extend this table to powers of composite numbers. It should be obvious why the entry equals 1 whenever the row and column numbers are equal. The "opposite" problem, finding n-digit base-n pandigital primes, isn't very interesting. There are only three of them. Can you find them? (I haven't looked into non-standard bases, such as negabinary, Fibonacci, or balanced ternary.) Which of these rows and columns would be of interest to OEIS? None of them appear to be there yet.

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[math-fun] An array with all the negative and positive integers, plus zero
by Éric Angelini 01 Apr '20

01 Apr '20

Hello Math-Fun, have a look here, on my personal page: https://bit.ly/2wGyzpv You will see how to build this array; it starts (fixed font): > A new array for the OEIS? 0 +1 +3 +7 +11 +5 +14 ... -1 +2 -4 +4 -6 +9 ... -3 +6 -8 -10 -15 ... -9 -14 -2 -25 ... -5 -12 -23 ... -7 -35 ... -28 ... ... The idea is to insert in an infinite “triangular” array, all integers from -∞ to +∞, with no duplicates. We start, as usual, in the upper left corner with 0. The array now is filled with those two rules: >>Rule 1) we always try to extend the upper row with the smallest positive integer a(n) not present so far in A (A is the array), integer that doesn’t lead to a contradiction immediately, or later in the rows below it; >>Rule 2) a(n) and a(n-1), the term at the left of a(n), produce another integer in the row below them; this term is the difference (smallest term – biggest term). If this leads to a contradiction (immediately or later in the antidiagonal), we try (biggest term – smallest term). If both operations lead to a contradiction, we try a(n)+1 instead of a(n) and repeat the whole process, filling A by antidiagonals. Example: After 0, we write 1 because this 1 doesn’t lead to a contradiction: we will have -1 below the pair (as 0 – 1 = -1): (...) Best, É.

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Re: [math-fun] math-fun Digest, Vol 205, Issue 35
by Steve Witham 31 Mar '20

31 Mar '20

From: Mike Stay <metaweta(a)gmail.com> I do get the "not secure" complaint, but the server sends an EarthLink 404 page, not a page on hyperbolic pairing. On Mon, Mar 30, 2020 at 12:39 AM Steve Witham <sw(a)tiac.net> wrote: > > http://www.rong.address Oops, habit. http://www.mac-guyver.com/switham/2020/03/HyperbolicPairing/index.html --Steve

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