March 2020 - mathfun@mailman.xmission.com

[math-fun] Mathologer cubic messup?
by Bill Gosper 14 Mar '20

14 Mar '20

https://www.youtube.com/watch?v=N-KXStupwsc offers an attractive expression for (one) solution of the general cubic, but I can't make it work. Have I entered this incorrectly? In[488]:= Power[#1 - Sqrt[#1^2 + #2^3], (3)^-1] + Power[#1 + Sqrt[#1^2 + #2^3], (3)^-1] - b/3/a &[-b^3/27/a^3 + b c/6/a/a - d/2/a, c/3/a - b^2/9/a/a] Out[488]= -(b/(3 a)) + (-(b^3/(27 a^3)) + (b c)/(6 a^2) - d/(2 a) - Sqrt[(-(b^2/(9 a^2)) + c/(3 a))^3 + (-(b^3/(27 a^3)) + (b c)/( 6 a^2) - d/(2 a))^2])^( 1/3) + (-(b^3/(27 a^3)) + (b c)/(6 a^2) - d/(2 a) + Sqrt[(-(b^2/(9 a^2)) + c/(3 a))^3 + (-(b^3/(27 a^3)) + (b c)/( 6 a^2) - d/(2 a))^2])^(1/3) —rwg

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[math-fun] Sudoku
by Bernie Cosell 14 Mar '20

14 Mar '20

I've been pondering Sudoku recently. It is easy [I think] to generate a sudoku grid. But beyond that there are a bunch of questions about turning it into a puzzle that I haven't seen much talked about. Most of the articles I've found fret about numbers, rather than algorithms [minimum number of clues necessary, how many grids are there with the, the maximum number of clues and still have an ambiguous grid, etc] I'm wondering about such things like: after you generate a grid, how do you figure out which cells to give as clues, recalling that the clues should be optimal for that grid [that is, yield a unique solution and have the solution not be unique if any clue is removed]? How do they determine the "difficulty" of a puzzle? /Bernie\ Bernie Cosell bernie(a)fantasyfarm.com -- Too many people; too few sheep --

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[math-fun] Pi Day video
by James Propp 14 Mar '20

14 Mar '20

I just posted a Pi Day video called "The Truth About Pi". Am I serious? Well, I named my alter ego Beau Gus Bogusian; that should give you a clue. https://youtu.be/_0xqNNd6RpE <https://t.co/f1BFCBtxpT?amp=1> Jim

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[math-fun] Your favorite elliptic curve?
by Brad Klee 12 Mar '20

12 Mar '20

My Favorite Elliptic Curve: A Tale of Two Types of Triangles Author(s): Richard K. Guy Source: The American Mathematical Monthly, Vol. 102, No. 9 (Nov., 1995), pp. 771-781 Sometimes it’s fun to appreciate the special rather than the general, and that is well conveyed in the aforementioned article. Does anyone else have a favorite elliptic curve they want to mention? —Brad

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[math-fun] Richard Guy, 1916-2020
by James Propp 10 Mar '20

10 Mar '20

Not sure if someone else already mentioned this (though I see that Wikipedia already knows): *From:* "Kharaghani, Hadi" <kharaghani(a)uleth.ca> *Date:* 9 March 2020 at 17:52:16 GMT It is with extreme sadness and sorrow to let you know of passing of Professor Richard Guy at 8:35 in the morning today. Hadi Jim Propp

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[math-fun] What is "too difficult"?
by Brad Klee 10 Mar '20

10 Mar '20

https://0x0.st/iAau.pdf Undergrads in Arkansas apparently are not rated for thinking in terms of analogy, so this problem has been judged inappropriate for a class on physics. The other difficulties are that they don't know how to fit data, work with error bars, calculate standard deviations, integrate functions, etc. In this case, adding, subtracting, and function evaluation is too much to ask for, because students (mostly from engineering) expect cookies to be cut in a particular way. Too bad because I was interested to hear what they think about the Nature article, see also: https://media.nature.com/original/magazine-assets/d41586-019-03241-9/d41586… Anyone here have an opinion about the local extinction data? Or suggestions what to do as a teacher in this situation? --Brad

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Re: [math-fun] A simple naive question
by Henry Baker 10 Mar '20

10 Mar '20

Take the *decimal* expansion of A,B, and then reinterpret each decimal digit as a *hexadecimal* digit. Call these new numbers A', B'. Finally, concatenate A':a:B', where 'a' is the hex digit for 5+5. The receiver inverts the process after scanning hex digits looking for 'a'. Of course, you can get a more compact representation by using base-eleven instead of base-16. -- Google 'BCD'='Binary Coded Decimal'; compliments to IBM 1620, 1401, system/360, Intel 4004, ... At 10:30 AM 3/10/2020, Ãric Angelini wrote: >Hello Math-Fun, > >I would like to transmit two _composite_ numbers A and B in the form of single number C. > >How can I do that? > >For instance, if Alice concatenates A et B, Bob will have a problem in reconstructing those numbers out of C: > >A = 865445 B = 12377007 C = 86544512377007 > >Is there a clever, economic way to say where C must be split? > >Or do you know another technique (dealing perhaps with prime numbers)? > >This has nothing to do with cryptography (I guess) -- just a (silly) question. > >Best, Ã.

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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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[math-fun] quick gif I was going to show at G13
by Bill Gosper 07 Mar '20

07 Mar '20

How does this happen? <http://gosper.org/snoflsnofl.gif> —Bill

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[math-fun] Oddity: What is the average value of the Dragon Curve?
by Bill Gosper 06 Mar '20

06 Mar '20

Which we can now define as the restriction of our Dragon Function to dyadic rationals. Successively higher averages look a little crazy: In[370]:= Table[Mean[First@dragun@# & /@ Range[0, 1, 2^-n]], {n, 11}] Out[370]= {1/2 + I/6, 2/5 + I/5, 7/18 + I/6, 7/17 + (3 I)/17, 9/22 + (13 I)/66, 2/5 + I/5, 103/258 + (17 I)/86, 103/257 + (51 I)/257, 137/342 + (205 I)/1026, 2/5 + I/5, 1639/4098 + (273 I)/1366} Note the three occurrences of 2/5+i/5. Remarkably, In[393]:= FindSequenceFunction[%370, n] Out[393]= ((1/5 + I/10) ((2 - I) - I I^n + 2^(1 + n)))/(1 + 2^n) In[373]:= Limit[%, n -> ∞] Out[373]= 2/5 + I/5 How often do we see sequences tipping off their limit like this? In fact, every fourth value is 2/5 + I/5. Perhaps more remarkably, 2/5+i/5 is an endpoint of the common boundary where two half-size Dragon images self-similarly unite to form the Dragon joining 0+0i to 1+0i via 1/2+i/2. This is not obvious without the Dragon Function. (Hey, wait a minute. Why aren't the averages in Out[370] dyadic rationals, like the actual Dragon values?-) —rwg

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