February 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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[math-fun] New Tetrads, New orchard solution (A006065)
by ed pegg 28 Mar '20

28 Mar '20

I've put in an update at http://www.mathpuzzle.com/ --Ed Pegg Jr

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[math-fun] Heighway triple points redux
by Bill Gosper 21 Mar '20

21 Mar '20

I thought OEIS A260482, A260747..A260750 completely described the (initial) Dragon triple points, all of which have preimages a(n)/15/2^k. But I just noticed an editor's remark that there is another infinite family of the form a(n)/7/2^k. These are much less "photogenic" than the old set (e.g. gosper.org/dragtrip!.png) <http://gosper.org/dragtrip!.png> because two preimages always(?) differ by 2/7/2^k, and ~75% of them have the 3rd preimage within 4/7/2^k. But they can get "arbitrarily" lopsided: In[508]:= undrag[(69 + 27 I)/160] Out[508]= {1539/3584, 2551/3584, 2553/3584} (Differences presumably limited to 40%.) Also, Julian has privately remarked that there are no irrational triple points. --rwg

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Re: [math-fun] sum of 2 triangular numbers is a triangular number
by Brad Klee 10 Mar '20

10 Mar '20

Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate: Pythagorean is to Trithagorean as {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} is to {n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}. — Brad > On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane(a)gmail.com> wrote: > > PS Brad, I was not trying to "diss"you, I simply did not understand what > you were saying. > I proposed a formula that seemed to give all the trythagorean triples. > I'm quite willing to believe either of the following: > You had already given that formula two days ago > My formula is wrong, and you had given the correct formula two days ago! > Best regards > Neil > > Neil J. A. Sloane, President, OEIS Foundation. > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. > Phone: 732 828 6098; home page: http://NeilSloane.com > Email: njasloane(a)gmail.com > > > >>> On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane <njasloane(a)gmail.com> wrote: >> Dear Grad, >> Trouble is, I don't speak Mathematica. Could you possibly translate what >> you said in that last email into human? >> Best regards >> Neil >> Neil J. A. Sloane, President, OEIS Foundation. >> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. >> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. >> Phone: 732 828 6098; home page: http://NeilSloane.com >> Email: njasloane(a)gmail.com >>> On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee(a)gmail.com> wrote: >>> Hi Neil, >>> I think you may have missed this little piece of code: >>> PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ >>> MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + >>> k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], >>> And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] >>> (*https://oeis.org/A046079*) >>> Length[PythagoreanTriples[#]] & /@ Range[2, 20] >>> Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} >>> If you hack out the part that says Solve[...] and maybe put a "d" in >>> for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], >>> with "d" for divisor. Just to double-check for possible typo, let's try: >>> Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 >>> - z^2] >>> Out[]=0 >>> Then, we can refine the idea of an "analog", >>> Pythagorean : Trithagorean :: >>> {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, >>> n*(n + 1)/d + (d - 2)/4}. >>> The other relevant sequence is https://oeis.org/A309507, see >>> my message from Sat. Feb 22. --Brad >>> On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane <njasloane(a)gmail.com> wrote: >>>> I mentioned this on Feb 19 2020 and several people sent very helpful >>>> replies. >>>> For squares, we have the theorem that all solutions to x^2+y^2=z^2 in >>>> integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. >>>> Expanding on the replies from Michael Collins and Rich, it looks like >>> the >>>> following is the analog for triangular numbers T_n = n(n+1)/2. >>>> ALL solutions to T_n+T_k=T_m are given by the list of what one might >>> call >>>> Trithagorean triples: these are the triples >>>> [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q >>> is >>>> any odd divisor of T less than n, >>>> plus these triples with the first two coordinates swapped. >>>> I haven't found this in the literature, but it can hardly be new. >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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