May 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

4 10

[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?

4 4

[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

1 3

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 >

1 1

[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

5 9

[math-fun] triadic dragon reminiscence
by Bill Gosper 11 Jul '20

11 Jul '20

Half a triadic dragon <http://gosper.org/semitriadic.png> self-similarly trisects into knobby canes (blue, orange+green, red+purple). But the whole triadic dragon self-similarly trisects, so two thirds of this half-dragon (blue+orange+green) is itself a triadic dragon. Deleting the blue cane leaves a (not self similar) congruently bisected backwards Z, orange+green, red+purple. But it is also in fact self-similarly trisected! (orange, green+red, purple). The Triadic Dragon joining 0 to 1 has bounding box West: -1/16, East: 17/16 (with index 239(!)/240), and N|S bounds ±9*i*/(16√3) (less certain here). Is there a plane figure that can be dissected into two, as well as three, but not six congruent pieces? —rwg

1 1

[math-fun] MacOS Mail
by Henry Baker 09 Jun '20

09 Jun '20

I realize that this is off-topic, so feel free to delete this if you wish. I personally use Linux & Windows, but my wife uses MacOS. Over the past 2 years, the MacOS "Mail" program has become impossibly slow -- all the while use the same hardware! I can't believe that Mac users would put up with this. Note that everything else works fine; it's just the Mail program. Do any of you use Macs? Do you find the Mac Mail program intolerably slow?

7 10

[math-fun] A generalization of an old Conway Problem
by Victor Miller 06 Jun '20

06 Jun '20

John Conway once posed the following problem: The Ten Divisibilities I have a ten digit number, abcdefghij. Each of the digits is different, and a is divisible by 1 ab is divisible by 2 abc is divisible by 3 abcd is divisible by 4 abcde is divisible by 5 abcdef is divisible by 6 abcdefg is divisible by 7 abcdefgh is divisible by 8 abcdefghi is divisible by 9 abcdefghij is divisible by 10 Find the number. One can do this computationally in a number of ways, such as exhausting over all permutations of [1,10], or branch and bound. I decided to do this via integer programming to investigate the generalization to other bases than 10. Interestingly enough, I found that there were only solutions for bases 2,4,6,8,10,14 (tested through base 27). I don't know if they've petered out completely. Any ideas? Victor

4 7

Re: [math-fun] Some math puzzles
by Dan Asimov 04 Jun '20

04 Jun '20

MSRI is the Mathematical Sciences Research Institute in Berkeley, California. —Dan Bernie Cosell wrote ----- >> Here are four problems from the puzzle column in the latest newsletter >> from MSRI, faintly edited. > >What's MSRL? The puzzles are fun. > /b\ -----

5 6

[math-fun] Chiral Genesis
by Brad Klee 31 May '20

31 May '20

See also: https://www.inaturalist.org/observations/47995301 . --Brad

1 0