February 2012 - mathfun@mailman.xmission.com

Re: [math-fun] Next generation sorting algorithm
by Dave Dyer 20 Feb '12

20 Feb '12

Is this an open competition for slowest sort? How about trying and testing all possible permutations.

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Re: [math-fun] Next generation sorting algorithm
by Dave Dyer 20 Feb '12

20 Feb '12

Is this an open competition for slowest sort? How about trying and testing all possible permutations.

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Re: [math-fun] isosceles triangle dissections
by Stuart Anderson 20 Feb '12

20 Feb '12

>> On Wed, Feb 15, 2012 at 4:11 PM, Bill Gosper <billgosper(a)gmail.com> wrote: >> > Neil has just uncovered fq.math.ca/Scanned/6-6/hoggatt.pdf >> > predating my http://mathworld.wolfram.com/SquareDissection.html >> > of a square into ten acute isosceles triangles. >> > >> > Neil also found in a Gardner book a reference to a Monthly paper >> > (June-July 1962, pp550-552) claiming that *any* obtuse triangle can >> > be cut into eight acute isosceles triangles, implying at most nine for >> > dissecting a right isosceles, in contradiction of my round(tan(69)) >> > solutions assertion. Has anyone a picture of this dissection? >> > --rwg Martin Gardner gave the 9 triangle dissection in "The Last Recreations" p253 (but Google books and Amazon wont show me that page) also found the following; Dissecting Triangles into Isosceles Triangles - A high school student paper http://math.ca/crux/v22/n3/page97-100.pdf Jim Loys webpage with some dissections of a square into isosceles triangles with 7 and 6 triangles; http://www.jimloy.com/puzz/iso.htm "Can a square be dissected into a finite number of obtuse, isosceles triangles? (1) Show that a square can be dissected into a finite number of OBTUSE, ISOSCELES triangles. (2) What is the minimum number of OBTUSE, ISOSCELES triangles needed to accomplish this? http://answers.yahoo.com/question/index?qid=20090224164301AAtkOqy - A square dissection with 10 obtuse, isosceles triangles is given. http://i524.photobucket.com/albums/cc321/Krejakovic/10tr.jpg

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[math-fun] Hirschhorn's modular identity
by Bill Gosper 20 Feb '12

20 Feb '12

rwg> http://reference.wolfram.com/mathematica/ref/QPochhammer.html (Neat examples) gives "Hirschhorn's modular identity ": [I just pasted the above from my Sent Mail. It ends with ((q;q)_oo)^5=(q^5,q^5)_oo, pasted from the Mma doc, but apparently vaporized on transmission.] rwg> QPochhammer[q, q]^3 = QPochhammer[q^3, q^3] *except* at q^binomial(n,2), n>1. It seems that QPochhammer[q, q]^3 - QPochhammer[q^3, q^3] == -3*Sum[(-1)^n*q^Binomial[n, 2]*Floor[2*n/3], {n, 0, Infinity}] which ought to be some kind of theta derivative, but I can't place it. --rwg

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Re: [math-fun] Curses, anticipated again.
by Bill Gosper 19 Feb '12

19 Feb '12

Here <http://gosper.org/isos10.png>'s a 10 triangle solution (like with the square), while we await the wondrous 9. Easy one: Everybody knows that you get a cube when you start with a regular tetrahedron and perform the Snowflake recursion in 3D. But what do you get if you start with an octahedron instead, and add tetrahedra? Answer <http://gosper.org/cuboct.png>. (Simply typed in by Julian.) What percentage of true/false quiz takers would bite on "This correct name for this solid is "Cuboctahedron"." ? In another, overlong thread I said >>>> Trying the g.f. sum, I hit this weirdy: >>>> Sum[(c^n*QPochhammer[b, q, n])/(b^n*QPochhammer[c*q, q, n]), {n, 0, >>>> Infinity}] == >>>> (b*(-1 + c))/(-b + c) >>>> >>>> independent of q. Plotting for b=2/3,c=1/3, it holds somewhat past >>>> q=1, and then >>>> develops a very interesting collection of poles. >And fake poles? ( http://gosper.org/poles.png) All but one of the poles are completely spurious artifacts of insufficient WorkingPrecision. c=1,b=2 near q=1 is about the worst floating point noise I've ever seen. (Amplitude >> 10^4.) --rwg On Wed, Feb 15, 2012 at 4:11 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > Neil has just uncovered fq.math.ca/Scanned/6-6/hoggatt.pdf > predating my http://mathworld.wolfram.com/SquareDissection.html > of a square into ten acute isosceles triangles. > > Neil also found in a Gardner book a reference to a Monthly paper > (June-July 1962, pp550-552) claiming that *any* obtuse triangle can > be cut into eight acute isosceles triangles, implying at most nine for > dissecting a right isosceles, in contradiction of my round(tan(69)) > solutions assertion. Has anyone a picture of this dissection? > --rwg > This is weird: Calculating with pen and paper for a change, it > seems that sunlight falling on the diagrams and equations actually > makes them *easier* to read.

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[math-fun] Mma sum
by Veit Elser 19 Feb '12

19 Feb '12

How does one report bugs to Wolfram? In:=Sum[(-1)^k Binomial[m, k] k, {k, 0, m}] Out:=0 It gets the correct answer when m is given explicitly. I'm guessing it knows Sum[z^k Binomial[m, k] k, {k, 0, m}] and then chokes when it evaluates for z=-1. Veit

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[math-fun] Waring-type question
by Dan Asimov 19 Feb '12

19 Feb '12

In what follows, "cube" means the cube of a positive integer. It is known that every sufficiently large integer is the sum of distinct cubes. See < http://oeis.org/A001476 >. QUESTION: Is every sufficiently large integer the sum of *more than one* distinct cubes? --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Curses, anticipated again.
by Stuart Anderson 19 Feb '12

19 Feb '12

Date: Wed, 15 Feb 2012 17:38:06 -0800 From: Bill Gosper <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Subject: [math-fun] Curses, anticipated again. Neil has just uncovered fq.math.ca/Scanned/6-6/hoggatt.pdf predating my http://mathworld.wolfram.com/SquareDissection.html of a square into ten acute isosceles triangles. Neil also found in a Gardner book a reference to a Monthly paper (June-July 1962, pp550-552) claiming that *any* obtuse triangle can be cut into eight acute isosceles triangles, implying at most nine for dissecting a right isosceles, in contradiction of my round(tan(69)) solutions assertion. Has anyone a picture of this dissection? --rwg This is weird: Calculating with pen and paper for a change, it seems that sunlight falling on the diagrams and equations actually makes them *easier* to read. http://mathworld.wolfram.com/SquareTiling.html M. Laczkovich has shown that there are exactly three shapes of non-right triangles that tile the square with similar copies, corresponding to angles (pi/8,pi/4,5pi/8), (pi/4,pi/3,5pi/12), and (pi/12,pi/4,2pi/3) (Stein and Szabó 1994). in degrees that is (22.5, 45, 112.5), (45,60,75), and (15, 45, 120) The Gosper-Hoggatt tilings use these angles. Stuart

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Re: [math-fun] cubed squares
by Stuart Anderson 19 Feb '12

19 Feb '12

> Message: 3 > Date: Sat, 18 Feb 2012 18:06:05 -0500 > From: Warren Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] cubed squares > Message-ID: > > The cubed square with the most water-volume is > just to use the entire square without any tiles. > [Maximizes sum_j (s_j)^3.] > Course, it is not so easy if you demand #tiles=27 (or whatever). Exactly, and tiles are always >21 for perfect (all squares in the dissection are different sizes) squared squares Squared squares produced using Tuttes rotor-stator symmetry method seem to make big deep ponds. See the Federicos (PJF) order 50 squares in http://www.squaring.net/sq/ss/spss/o31+/SPSSo31-75.pdf > > More interesting in my view is > to MINIMIZE sum_j (s_j)^3. The minimum size possible on the boundary of a perfect squared square is 5, Gambini proved this and produced some with 5 on the boundary using his packing program. Such squares would hold very little water, as small boundary squares need to be surrounded by other small squares creating an area with very low elevation and volume, and the water would just drain out from the boundary. There are many examples in the pdf above also in order 40s, 50's and 60's; I am trying to imagine other ways to minimise the water but cant think of another method of constructing perfect squares that would have this outcome. Then theres also the Imperfect squared squares, which exist in much greater numbers, which I havnt considered yet > Stuart

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Re: [math-fun] mu question
by Michael Reid 18 Feb '12

18 Feb '12

> Letting mu be the Moebius function, and nu be the Mertens function > defined as > > nu(n) = SUM(k = 1..n, mu(k)) > > Then I notice > > [1] SUM(k = 1..n, mu(k)*[n/k]) = 1 > > [2] SUM(k = 1..n, nu([n/k])) = 1 > > where [x] is the floor of x. I wondered if there was any simple > relationship linking [1] and [2], or perhaps some relationship to the Sure, write the second sum as a double sum, using the definition of nu(-) , and then switch the order of summation. > Moebius transform > > a(n) = SUM(d|n; b(d)) <=> b(n) = SUM(d|n; mu(n/d)*a(d)). >

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