September 2016 - mathfun@mailman.xmission.com

Re: [math-fun] Bar bet
by Don Reble 23 Sep '16

23 Sep '16

> "The volume of a regular tetrahedron (triangular pyramid with unit edges) > is exactly half the volume of a square pyramid with unit edges." > True or false? --rwg Trick puzzle: prove that the volume of _every_ such pyramid (unit edges, regular polygon base) is a quadratic surd. -- Don Reble djr(a)nk.ca

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[math-fun] Dario Alpern's QUAD
by Zak Seidov 22 Sep '16

22 Sep '16

Dear SF & MF gurus, plz HELP! Something is wrong with Dario Alpern's QUAD: https://www.alpertron.com.ar/QUAD.HTM They say that Java is obsolete(?) and I should do something that I don't understand:( I use Win XP and Win10 if it matters. Thanks, Zak Seidov

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Re: [math-fun] Dario Alpern's QUAD
by Axel Vogt 21 Sep '16

21 Sep '16

> From: Zak Seidov <zakseidov(a)mail.ru> > Subject: [math-fun] Dario Alpern's QUAD ... > https://www.alpertron.com.ar/QUAD.HTM > > They say that Java is obsolete(?) and I should do something > that I don't understand:( > I use Win XP and Win10 if it matters. The site says "This page uses JavaScript, ..." and after allowing that it works for me (Firefox 45), no need for Java.

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Re: [math-fun] Dario Alpern's QUAD
by Dave Dyer 21 Sep '16

21 Sep '16

Oracle has been making it harder and harder to deploy java applets, requiring real (not self-signed) certificates and changes to the tags associated with applets. Without recent maintenance, there's no chance applets will work. Also finding a browser which still supports java applets is getting harder - chrome and edge are completely out.

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Re: [math-fun] Dario Alpern's QUAD
by Dave Dyer 21 Sep '16

21 Sep '16

Oracle has been making it harder and harder to deploy java applets, requiring real (not self-signed) certificates and changes to the tags associated with applets. Without recent maintenance, there's no chance applets will work. Also finding a browser which still supports java applets is getting harder - chrome and edge are completely out.

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Re: [math-fun] The largest twin-primes known today...
by Bill Gosper 21 Sep '16

21 Sep '16

I think Allan means 3^2, not 3^12. This would have been an extremely easy factorization even for Lehmer's optical contraption <https://en.wikipedia.org/wiki/Lehmer_sieve>. Or Macsyma back when it trial-divided by consecutive odd numbers. Which is pretty much like GCDing with their product, now that we can store big numbers. I just wrote a tiny (and very untested) recursive program in a similar vein: In[832]:= Clear[dumfac]; dumfac@1 = {}; dumfac[n_Integer] := (dumfac@n = {n}; dumfac@n = Join[dumfac[#], dumfac[n/#]] &@ GCD[n, Binomial[#, Floor[#/2]] &@Floor[Sqrt[n]]]) In[833]:= dumfac[2996863034895] // tim During evaluation of In[833]:= 0.188517 sec, 5 Out[833]= {5, 3, 3, 18583, 3583757} In[834]:= PrimeQ /@ % Out[834]= {True, True, True, True, True} --rwg On 2016-09-20 16:05, Allan Wechsler wrote: > Factoris says the coefficient factors to 3^12.5.18583.3583757. > > On Tue, Sep 20, 2016 at 6:55 PM, Dan Asimov <asimov(a)msri.org> wrote: > >> P.S. I'm puzzled by this statement about these twin primes at that link: >> >> ----- >> They will enter Chris Caldwell's “The Largest Known Primes Database” >> (http://primes.utm.edu/primes) ranked 1st for twins, and each entered >> individually ranked 4180th overall. >> ----- >> >> Why do they have the same ranking overall? Does it not matter that >> one is bigger than the other? >> >> —Dan >> >> >> > On Sep 20, 2016, at 3:35 PM, Dan Asimov <asimov(a)msri.org> wrote: >> > >> > Is anything known about the factorization of that coefficient >> > >> > (other than its divisibility by 5 and 9)? >> > >> > Was it chosen as part of some algorithm that singled out relatively >> > likely candidates? Or was it just stumbled on at random from a >> > massive search? >> > >> > >> > Or maybe one of our factoring experts can factor it? (You know who you >> are.) >> > >> > —Dan >> > >> > >> >> On Sep 20, 2016, at 3:18 PM, Eric Angelini <Eric.Angelini(a)kntv.be> >> wrote: >> >> >> >> >> >> ... are: >> >> 2996863034895*2^1290000-1 >> >> and >> >> 2996863034895*2^1290000+1 >> >> >> >> http://www.primegrid.com/download/twin-1290000.pdf >>

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[math-fun] The largest twin-primes known today...
by Eric Angelini 21 Sep '16

21 Sep '16

... are: 2996863034895*2^1290000-1 and 2996863034895*2^1290000+1 http://www.primegrid.com/download/twin-1290000.pdf Best, É.

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[math-fun] Something's rotten about Common Core math
by Henry Baker 20 Sep '16

20 Sep '16

FYI -- Parts of this article are pretty funny! https://www.salon.com/2015/11/28/youre_wrong_about_common_core_math_sorry_p… Youre wrong about Common Core math: Sorry, parents, but it makes more sense than you think

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[math-fun] gyroelongated bifastigium
by Bill Gosper 17 Sep '16

17 Sep '16

On 2016-09-14 17:45, James Buddenhagen wrote: > Haha -- that is an endearing name :) . > Do you happen to have an off file for that? > I'm still trying to figure out what it is! > You mean like Graphics3D@Polygon@{{{-(1/2), -(1/2), -(1/(2 2^(1/4)))}, {-(1/2), 1/ 2, -(1/(2 2^(1/4)))}, {-(1/2), 0, 1/4 (-2^(3/4) - 2 Sqrt[3])}}, {{1/2, 0, 1/4 (-2^(3/4) - 2 Sqrt[3])}, {1/2, 1/2, -(1/(2 2^(1/4)))}, {1/ 2, -(1/2), -(1/(2 2^(1/4)))}}, {{-(1/2), 0, 1/4 (-2^(3/4) - 2 Sqrt[3])}, {-(1/2), 1/2, -(1/(2 2^(1/4)))}, {1/2, 1/2, -(1/(2 2^(1/4)))}, {1/2, 0, 1/4 (-2^(3/4) - 2 Sqrt[3])}}, {{-(1/2), -(1/2), -(1/( 2 2^(1/4)))}, {-(1/2), 0, 1/4 (-2^(3/4) - 2 Sqrt[3])}, {1/2, 0, 1/4 (-2^(3/4) - 2 Sqrt[3])}, {1/2, -(1/2), -(1/(2 2^(1/4)))}}, {{1/ 2, 1/2, -(1/(2 2^(1/4)))}, {0, 1/Sqrt[2], 1/(2 2^(1/4))}, {1/Sqrt[ 2], 0, 1/(2 2^(1/4))}}, {{-(1/2), 1/ 2, -(1/(2 2^(1/4)))}, {-(1/Sqrt[2]), 0, 1/(2 2^(1/4))}, {0, 1/Sqrt[ 2], 1/(2 2^(1/4))}}, {{-(1/2), -(1/2), -(1/(2 2^(1/4)))}, {0, -(1/ Sqrt[2]), 1/(2 2^(1/4))}, {-(1/Sqrt[2]), 0, 1/(2 2^(1/4))}}, {{1/ 2, -(1/2), -(1/(2 2^(1/4)))}, {1/Sqrt[2], 0, 1/( 2 2^(1/4))}, {0, -(1/Sqrt[2]), 1/(2 2^(1/4))}}, {{1/2, 1/ 2, -(1/(2 2^(1/4)))}, {-(1/2), 1/2, -(1/(2 2^(1/4)))}, {0, 1/Sqrt[ 2], 1/(2 2^(1/4))}}, {{-(1/2), 1/ 2, -(1/(2 2^(1/4)))}, {-(1/2), -(1/2), -(1/(2 2^(1/4)))}, {-(1/ Sqrt[2]), 0, 1/( 2 2^(1/4))}}, {{-(1/2), -(1/2), -(1/(2 2^(1/4)))}, {1/ 2, -(1/2), -(1/(2 2^(1/4)))}, {0, -(1/Sqrt[2]), 1/( 2 2^(1/4))}}, {{1/2, -(1/2), -(1/(2 2^(1/4)))}, {1/2, 1/ 2, -(1/(2 2^(1/4)))}, {1/Sqrt[2], 0, 1/( 2 2^(1/4))}}, {{-(1/(2 Sqrt[2])), -(1/(2 Sqrt[2])), 1/4 (2^(3/4) + 2 Sqrt[3])}, {-(1/Sqrt[2]), 0, 1/( 2 2^(1/4))}, {0, -(1/Sqrt[2]), 1/(2 2^(1/4))}}, {{1/Sqrt[2], 0, 1/( 2 2^(1/4))}, {0, 1/Sqrt[2], 1/(2 2^(1/4))}, {1/(2 Sqrt[2]), 1/( 2 Sqrt[2]), 1/4 (2^(3/4) + 2 Sqrt[3])}}, {{1/(2 Sqrt[2]), 1/( 2 Sqrt[2]), 1/4 (2^(3/4) + 2 Sqrt[3])}, {0, 1/Sqrt[2], 1/( 2 2^(1/4))}, {-(1/Sqrt[2]), 0, 1/( 2 2^(1/4))}, {-(1/(2 Sqrt[2])), -(1/(2 Sqrt[2])), 1/4 (2^(3/4) + 2 Sqrt[3])}}, {{1/Sqrt[2], 0, 1/(2 2^(1/4))}, {1/( 2 Sqrt[2]), 1/(2 Sqrt[2]), 1/4 (2^(3/4) + 2 Sqrt[3])}, {-(1/(2 Sqrt[2])), -(1/(2 Sqrt[2])), 1/4 (2^(3/4) + 2 Sqrt[3])}, {0, -(1/Sqrt[2]), 1/(2 2^(1/4))}}} ? A fastigium is apparently just a triangular prism with square sides. This is two of them endcapping a square antiprism. --rwg Johnson had to eschew nonconvex solids because they're infinitudinous, narrowly excluding such gems as bilunagyrobicupola. > On Wed, Sep 14, 2016 at 4:10 AM, Bill Gosper <billgosper(a)gmail.com> wrote: >> is nonconvex, but the name is irresistible. >> gosper.org/gyroelongatedbifastigium.png >> Resistince is futile. --rwg

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Re: [math-fun] A powerful Zero Torque Square (Ed Pegg Jr)
by Walter Trump 16 Sep '16

16 Sep '16

Hi Ed, your square is a candidate for the most interesting magic square of order 4. It additionally has the longest Anderson Graph (a closed path from entry 1 to 2 to 3 to ... to 16 to 1) of all classical magic 4x4 squares (together with the square Frenicle # 361.) The length of the graph is 43.942968 units. For general orders we can say that all magic and semi-magic squares are balanced. (Obvious) And all centrally symmetric magic squares remain balanced if their entries were squared. (Not difficult to prove) Question: Are there non-symmetric magic squares of orders greater than 4 with this property? In this context Peter Lolys work should be mentioned. He showed that the moment of inertia is equal for all classical magic squares of the same order. See: http://home.cc.umanitoba.ca/~loly/MathGaz.pdf And: https://oeis.org/A126275 Walter

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