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[math-fun] Polyomino question
by franktaw＠netscape.net 27 Feb '06

27 Feb '06

What is the smallest polyomino that contains all polyominos of size n? For n = 1 or 2, the value is clearly n. For n = 3, it's 4; either of the following will work: OOO OOO O O For n = 4, there is a unique size 6 polyomino that works: OOOO OO For n = 5, the best I can do is 9: O OOOOO OOO I don't know if this is unique; I would be very surprised if 8 is possible. For n = 6, I can get 12: OO OOOOOO OOOO Again, this may not be unique, and might possibly be improved. For n = 7, the best I've found is 17: OOOOO OOOOOOO OOO OO I would not be at all surprised if there is a solution with 16, but I didn't find one. If 17 is the best, it is not unique; some simple variants of this also work. For n = 8 and beyond, I have no real idea. ---- We can establish a general upper bound from the consideration that a size n polyomino always fits in a k x (n+1-k) box for some k. It follows that sum_{k=1}^{ceiling(n/2)) n+1-k = ceiling(n/2)*(n+floor(n/2)+1)/2 is an upper bound. (This can be improved to ceiling(n/2)*(n+floor(n/2)-1)/2 + 1 by noting that, for k>1, the polyomino cannot occupy all four corners of the box.) This is asymptotically 3/8 n^2. From the very limited data above, it looks more like 1/4 n^2. Even that the sequence grows quadratically is not certain, however. 2n-2 is a simple lower bound, since fitting the general "line" polyomino requires n in a single row (or column), while the general "snake" polyomino has only 2 squares in any given row. Franklin T. Adams-Watters 16 W. Michigan Ave. Palatine, IL 60067 847-776-7645 ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com

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RE: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
by Schroeppel, Richard 27 Feb '06

27 Feb '06

Fascinating! I had a different objective in mind when I wrote the problem. My goal was to reach the fairly difficult pi^2/6 result using series manipulation, starting from a definition of pi based on 1 - 1/3 + 1/5 - ... or the like. Ideally, one would avoid calculus and complex integrations and the theory of dilogarithms, or Euler's trick with the power series for sin x. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Robert Baillie Sent: Wednesday, February 22, 2006 8:39 AM To: math-fun Subject: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8 HAKMEM 239 has this interesting question: "119 (Schroeppel): Can someone square some series for Pi to give the series Pi^2/6 = 1 + 1/2^2 + 1/3^2 + ... ?" Would you settle for squaring a series for Pi/Sqrt(8) to get a series for Pi^2/8? Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 --++ ... Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9+2 + 1/11^2 + ... Also, notice that you can square the first series by adding the squares of the individual terms! I first noticed this by comparing exercises 6a and 6b on page 398 in Wilfred Kaplan's "Advanced Calculus". The series for Pi/Sqrt(8) comes from the Fourier series for f1(x) = 0 for -Pi <= x < 0, f(x) = 1 for 0 <= x <= Pi. The series is f[x_] := (1/2) + (2/Pi) Sum[ Sin[(2n-1)x]/(2n-1), {n, 1, Infinity}] f[Pi/4] gives the series for Pi/Sqrt[8]. The series for Pi^2/8 comes from the Fourier series for g1(x)= Abs[x], -Pi <= x <= Pi. g[x_] := (Pi/2) - (4/Pi) Sum[ Cos[(2n-1)x]/((2n-1)^2), {n, 1, Infinity}] g[0] gives the series for Pi^2/8. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Smallest bimagic square: unknown!
by Christian Boyer 27 Feb '06

27 Feb '06

A magic square is bimagic if it remains magic after squaring each of its integers. The smallest bimagic squares using consecutive integers are known: 8x8. BUT... the smallest bimagic squares using distinct integers (= not forced to be consecutive) are still unknown! 5x5, 6x6, 7x7??? If 5x5 is impossible, a proof??? Edouard Lucas was the first to work on the subject, in 1891, easily proving that 3x3 is impossible. For more information, go to www.multimagie.com/indexengl.htm Click on "Magic squares of squares" in the left menu, then click on "Open problems". And, from this table, you will find the current status of the research. Any help or idea on the subject is welcome! Christian.

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RE: [math-fun] Problem Puzzler 1999
by Eric Angelini 27 Feb '06

27 Feb '06

First pangrammatic self-describing french crossword, by Frédéric Martin, into a 22x22 grid : http://www.cetteadressecomportecinquantesignes.com/22x22.htm Both an English and a German version can be seen there: http://carremagique.site.voila.fr/ Best, É.

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Re: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
by dasimov＠earthlink.net 25 Feb '06

25 Feb '06

(*) (a_1 + a_2 + . . . )^2 = a_1^2 + a_2^2 + . . . is indeed a fascinating equation to ponder. Assume hereon that a_i -> 0 as i -> oo. Note that since sum (a_i) may not be absolutely convergent (or perhaps *must* not?), the LHS may (must?) depend on the order in which its terms are summed. Maybe the intelligentest order is to do (**) sum {n = 2 to oo} of (sum {i+j=n} of a_i*a_j). The truth of (*) <=> the cross terms (sum {i <>j} a_i*a_j) = 0. Of course, if the cross terms are NOT absolutely convergent, then there are always some (inf'ly many) orderings of them that sum to 0. (Or any other chosen sum.) So unless we make some assumption about the ordering of (a_1 + a_2 + . . .)^2, (*) is not that interesting; I propose (**). What would be really cool to see would be a *geometrical* proof of sum (1/n^2) = pi^2 / 6. That is, some object of volume pi^2 / 6 that can be partitioned into pieces of volume 1/n^2. P.S. Also note that 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 . . . . = integral {0 to 1} (1+x^2) / (1+x^4) dx --Dan

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[math-fun] where can I buy a flexing hyperboloid?
by Thane Plambeck 25 Feb '06

25 Feb '06

This is from Courant and Robbins, "What is Mathematics", pg 214 * * * One of the most remarkable properties of the hyperboloid is that although it contains two families of intersecting straight lines, these lines do not make the surface rigid. If a model of the surface is constructed from wire rods, free to rotate at each intersection, then the whole figure may be continuously deformed into a variety of shapes **** So, I'd like to play with such a model. Can I buy one somewhere? Does anything come to mind? I doubt I have patience to make one. I'm also a bit unclear on what "free to rotate at each intersection" means in the quoted passage. Can you fold it into nonhyperboloids? Google was unhelpful. Here's a photo from the book http://static.flickr.com/37/104048548_2327835121_o.gif -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
by Robert Baillie 25 Feb '06

25 Feb '06

HAKMEM 239 has this interesting question: "119 (Schroeppel): Can someone square some series for Pi to give the series Pi^2/6 = 1 + 1/2^2 + 1/3^2 + ... ?" Would you settle for squaring a series for Pi/Sqrt(8) to get a series for Pi^2/8? Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 --++ ... Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9+2 + 1/11^2 + ... Also, notice that you can square the first series by adding the squares of the individual terms! I first noticed this by comparing exercises 6a and 6b on page 398 in Wilfred Kaplan's "Advanced Calculus". The series for Pi/Sqrt(8) comes from the Fourier series for f1(x) = 0 for -Pi <= x < 0, f(x) = 1 for 0 <= x <= Pi. The series is f[x_] := (1/2) + (2/Pi) Sum[ Sin[(2n-1)x]/(2n-1), {n, 1, Infinity}] f[Pi/4] gives the series for Pi/Sqrt[8]. The series for Pi^2/8 comes from the Fourier series for g1(x)= Abs[x], -Pi <= x <= Pi. g[x_] := (Pi/2) - (4/Pi) Sum[ Cos[(2n-1)x]/((2n-1)^2), {n, 1, Infinity}] g[0] gives the series for Pi^2/8.

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[math-fun] A generalization of the Gamma function reflection formula
by Eugene Salamin 24 Feb '06

24 Feb '06

It is known that 1/(z! (-z)!) = sin(pi z)/(pi z). Does there exist a similar simplification formula for 1/(z! (wz)! (wwz)! ...), w an n-th root of 1, and the product is over all n roots? When n is even, we can use the reflection formula, but what happens when n is odd? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Tiling a square
by Kerry Mitchell 23 Feb '06

23 Feb '06

Hi all, I've found out, through sheer playing around, that a 15x15 square can be tiled exactly using 4 5x5 squares, 9 3x3 squares, and 11 2x2 squares. No big deal. What I'd like to do is determine whether or not a certain combination of smaller squares will tile a larger square, without actually finding the tiling. For example, will 2 5x5s, 11 3x3s and 19 2x2s also tile a 15x15? I've already figured out that the areas need to match. :-) Any pointers or pearls of wisdom? Thanks, Kerry Mitchell -- lkmitch(a)gmail.com www.fractalus.com/kerry

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[math-fun] 7-color torus animation ("Szilassi 42-hedron")
by R. William Gosper 22 Feb '06

22 Feb '06

Those of you with PCMacsyma 2.4 might enjoy www.tweedledum.com/rwg/tor7.mfe . Click on the last image, then the rocket icon. To tumble and zoom the moving image, click off the rocket, click the camera mount icon, Continuous Draw, then reclick the rocket. Control the speed with the movie camera icon. If you get sporadic gray frames, try reducing the section height or notebook width. If you wish to abuse it, the code is all there in the notebook. --rwg PS, rwg>>PS, I like Klarner's distinction: perfect packing -> tiling. JPropp>Can someone say what Klarner's distinction is? (I have used the >two phrases interchangeably in my own work.) Sorry, probably everybody but me had it right all along. I was misusing packing to mean tiling.

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