February 2004 - mathfun@mailman.xmission.com

[math-fun] sum phi(n)/2^n
by Chris Landauer 13 Feb '04

13 Feb '04

hihi, all - i transformed rkg's sum first, since (if i'm using the right phi function = euler totient) phi(n) = sum(d|n) mu(n/d) d sum(n>=1) phi(n)/2^n = sum(n>=1, d>=1, d|n) mu(n/d) d / 2^n (here we take m / n/d) = sum(m>=1, d>=1) mu(m) d / 2^(m d) (here we use sum(d>=1) d x^d = x/(1-x)^2, with x = 2^(-m)) = sum(m>=1) mu(m) 2^(-m)/(1 - 2^(-m))^2 which converges very quickly (after m>=53, the values don't change in simple double precision arithmetic) M sum to M 51 1.367630801985022293 53 1.367630801985022070 55 1.367630801985022070 which agrees well enough with rkg's number (to 15 places or so) 1.367630801985022350790508146 well, to check the rest of the digits, i then wrote out some completely ad hoc multiple-precision functions and got (at varying precisions) 1. 3676 3079 9663 1. 3676 3080 1985 0223 0400 1. 3676 3080 1985 0223 5079 0509 2651 1. 3676 3080 1985 0223 5079 0508 1462 1311 4322 1. 3676 3080 1985 0223 5079 0508 1462 1308 8139 0748 9199 8962 7951 8188 1. 3676 3080 1985 0223 5079 0508 1462 1308 8139 0748 9199 8962 7948 5295 6598 4637 6215 6711 9392 which agrees with rkg's numbers as far as they go more later, cal

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[math-fun] sum phi(n)/2^n
by Richard Guy 13 Feb '04

13 Feb '04

David Angell asked about this at the Western Number Theory Conference. Would someone check my hand calculation, 1.367630801985022350790508146 and try it out on Simon Plouffe's whatever-it-is? Thanks! R.

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Re: [math-fun] Two questions in geometric combinatorics
by Michael D Beeler 12 Feb '04

12 Feb '04

Richard Guy writes: > The case n = 7 mod 8 is left to the reader. I think odd-sized pyramids are impossible, by a 4-coloring: layer 1: A layer 2: B C D layer 3: A D C B A B layer 4: B C D A B A D C D C layer 5: A D C B A B C D C D A B A B A layer 6: B C D A B A D C D C B A B A B C D C D C D layer 7: A D C B A B C D C D A B A B A D C D C D C B A B A B A B layer 8: B C D A B A D C D C B A B A B C D C D C D A B A B A B A D C D C D C D C and so on. A 4-sphere "tetrad" will have one of each color regardless of where it is in this grid. To be solvable, any size of pyramid must have an equal number of each color. >From Richard's packing, we know that all even-order pyramids are solvable, thus must contain an equal number of all colors. Consider an odd-numbered layer. Parse it into rhombi, as in this example of layer 7: A ------- /D C/ /B A/ B ------- ------- ------- /C D/ /C D/ /A B/ /A B/ A ------- ------- ------- ------- ------- /D C/ /D C/ /D C/ /B A/ /B A/ /B A/ B ------- ------- ------- All the rhombi are color-balanced, but the right edge contains only A's and B's, so the layer as a whole cannot be color-balanced. So, adding an odd-numbered layer to an even-sided (solvable) pyramid makes it un-solvable. -- Mike Richard Guy <rkg @cpsc.ucalgary.ca> Sent by: math-fun-bounces+mbeeler=csc.com 02/09/2004 05:38 PM Please respond to math-fun To: Daniel Asimov <dasimov(a)earthlink.net>, math-fun <math-fun(a)mailman.xmission.com> cc: Subject: Re: [math-fun] Two questions in geometric combinatorics A sufficient condition is n even. For n = 2 it's trivial. For larger even n, partition the tetrahedron into pairs of layers, of edge 2k and 2k-1, 0 < k < (n+2)/2. E.g., k = 3 with the b-layer sitting on the a-layer. /a\ / b \ /a a\ ------- \b b/ /a\ \ a / /a\ / b \ \b/ / b \ /a a\ /a a\ ------- ------- \b b/ \b b/ /a\ \ a / /a\ \ a / /a\ / b \ \b/ / b \ \b/ / b \ /a a\ /a a\ /a a\ ------- ------- ------- The case n = 7 mod 8 is left to the reader. R. On Sun, 1 Feb 2004, Daniel Asimov wrote: > ----- Original Message ----- > From > To: <math-fun(a)mailman.xmission.com> > Sent: Sunday, February 01, 2004 2:24 PM > Subject: Re: [math-fun] Two questions in geometric combinatorics > > ------------------------------------------------------------------------- > Dylan asked: > << > On Fri, Jan 30, 2004 at 05:17:01PM -0500, asimovd(a)aol.com > wrote: > << > Let A_n denote a regular tetrahedral arrangement of points in > 3-space, n to an edge. (The layers successively contain > 1, 3, 6,... points.) > >> > > Can you be more explicit what this arrangement is? I am > completely baffled. > >> > --------------------------------------------------------------------------- > > The arrangement A_n of points in 3-space is just a regular tetrahedron's > worth of the face-centered cubic lattice (from the triangular viewpoint). > I.e., starting from the top, the (k)th layer is an k-sided equilateral > triangle > of k(k+1)/2 points arranged as in the triangular lattice. The layers are > spaced so each point's nearest neighbors are (let's say) exactly one unit > apart from it. > > The problem (to repeat) is, For which n can A_n be partitioned into disjoint > "tetrads", where a tetrad here means 4 mutually nearest points) ? An > obvious > necessary condition is that 4 divides #(A_n), which happens exactly when > n = 0,2,4,6, or 7 (mod 8). > > --Dan > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] First player win?
by Michael D Beeler 12 Feb '04

12 Feb '04

Unless I'm missing something, five field kono should have been cooked long ago. Because only diagonal moves are allowed, the board consists of two independent fields: "odd" field (13 squares) with 3 R pieces, 3 B pieces, and 7 empty squares, so 13! / 3! 3! 7! = 34230 possible states "even" field (12 squares) with 4 R pieces, 4 B pieces, and 4 empty squares, so 12! / 4! 4! 4! = 34650 possible states The full game is the composite of the two fields. Even if all possible combinations had to be analyzed, that's only 1,189,188,000. But each field can be analyzed separately, and then an analysis of the full game can be made much more simply than by examining all possible combinations, no? (Hand waving goes here... :-) -- Mike ps. there is a nice five field kono player at http://www.darkfish.com/kono/FiveFieldKono.html

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[math-fun] First player win?
by Richard Guy 12 Feb '04

12 Feb '04

Five Field Kono, {\it Symmetry plus}, {\bf22}(Autumn 2003) 16. Traditional game from Korea. Played on a 5 by 5 board. Moves are diagonal as in checkers, but there is no jumping or taking. Object is to get your 7 R (or B) pieces into the opponent's starting position before she does. In the diagram below, `o' represents an initially empty square. Presumably the moves can, should it be necessary or desirable, be backwards, as with a checkers King, as well as forwards. R R R R R . . . . . . . . . . . . . . . . . . . . R o o o R . . . . . . . . . . . . . . . . . . . . o o o o o . . . . . . . . . . . . . . . . . . . . B o o o B . . . . . . . . . . . . . . . . . . . . B B B B B This is small enough to yield to brute force with modern computer expertise, but munsters may come up with something more intellectually pleasing. R.

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[math-fun] Re: Apollonian gaskets and Dionysian fillings
by James Propp 10 Feb '04

10 Feb '04

>>Jim: Since there are various versions of Apollonian gaskets, could you >>tell us exactly the one you are using to get started here? > >Start with three mutually externally tangent circles, with a big circle >that contains and is tangent to the other three. This gives a foursome >of mutually tangent circles. My original problem concerned situations with an additional constraint: all four circles should have radii that are reciprocals of integers. Then it can be shown that ALL of the infinitely many circles in the gasket have this property too, and I believe the same is true of the space-filling family of circles derived from the gasket by iterated inversion. I was trying to find combinatorial interpretations of the numbers that turn up, analogous to the fact that numerators and denominators of fractions in the Stern-Brocot tree count perfect matchings of "snake graphs" like o---o---o | | | o---o---o | | o---o---o---o---o | | | | | o---o---o---o---o These numbers are related to the Ford circles, which are an especially nice subset of the most fundamental of Apollonian gaskets, namely the one in which the four original "circles" are two parallel lines and two mutually tangent circles in between them. So, although my long-range goal is to understand this and other specific gaskets from a combinatorial point of view, I'd also like to find out what's known about the space-fillers derived from them from a purely geometrical point of view. Jim Propp

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Re: [math-fun] limit question
by MCKAY＠vax2.concordia.ca 10 Feb '04

10 Feb '04

qgfvedgydhgghsdghbsdghbghFrom: IN%"math-fun(a)mailman.xmission.com" "math-fun" 9-FEB-2004 16:49:25.65 To: IN%"math-fun(a)mailsdhbjsdhjdshjdshjdshjdhjman.xmission.com", IN%"rwg(a)tc.spnet.com" CC: IN%"jdb(a)math.arizona.edu" Subj: RE: [math-fun] limit question hjsdhjsdh Return-path: <math-fun-bounces+mckay=vax2.concordia.ca(a)mailman.xmission.com> Received: from newnewsflash.concordia.ca (newnewsflash.Concordia.CA [132.205.7.80]) by vax2.concordia.ca (PMDF V6.2 #30759) with ESMTP id <01L6EL9KOZOU008ODP(a)vax2.concordia.ca> for mckay(a)vax2.concordia.ca (ORCPT mckay(a)vax2.concordia.ca) Mon, 09 Feb 2004 16:49:23 -0400 (EDT) Received: from mailman.xmission.com (mailman.xmission.com [198.60.22.29]) by newnewsflash.concordia.ca (8.12.10/8.12.10) with ESMTP id i19LoWWP015972 for <mckay(a)vax2.concordia.ca>; Mon, 09 Feb 2004 16:50:32 -0500 Received: from localhost ([127.0.0.1] helo=mailman.xmission.com) by mailman.xmission.com with esmtp (Exim 3.35 #1 (Debian)) id 1AqIx3-0002H7-0C for <mckay(a)vax2.concordia.ca>; Mon, 09 Feb 2004 14:28:21 -0700 Received: from smtp224.math.arizona.edu ([128.196.224.202] ident=mail) by mailman.xmission.com with esmtp (Exim 3.35 #1 (Debian)) id 1AoSYU-00042C-00 for <math-fun(a)mailman.xmission.com>; Wed, 04 Feb 2004 12:19:22 -0700 Received: from puma.math.arizona.edu ([128.196.226.62] helo=localhost.localdomain) by smtp224.math.arizona.edu with smtp (Exim 3.36 #1 (Debian)) id 1AoSYo-0002SN-00; Wed, 04 Feb 2004 12:19:42 -0700 Received: (nullmailer pid 30336 invoked by uid 20079); Wed, 04 Feb 2004 19:19:11 +0000 Date: Wed, 04 Feb 2004 12:19:11 -0700 From: jdb(a)math.arizona.edu Subject: Re: [math-fun] limit question Sender: math-fun-bounces+mckay=vax2.concordia.ca(a)mailman.xmission.com To: math-fun(a)mailman.xmission.com, rwg(a)tc.spnet.com Cc: jdb(a)math.arizona.edu Errors-to: math-fun-bounces+mckay=vax2.concordia.ca(a)mailman.xmission.com Reply-to: math-fun <math-fun(a)mailman.xmission.com> Message-id: <1075922351.270925.30335.nullmailer(a)localhost.localdomain> Precedence: list X-Mailman-Approved-At: Mon, 09 Feb 2004 14:27:43 -0700 X-BeenThere: math-fun(a)mailman.xmission.com X-Mailman-Version: 2.1.1 X-Scanned-By: MIMEDefang 2.35 List-Post: <mailto:math-fun@mailman.xmission.com> List-Subscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=subscribe> List-Unsubscribe: <http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>, <mailto:math-fun-request@mailman.xmission.com?subject=unsubscribe> List-Archive: <http://mailman.xmission.com/cgi-bin/mailman/private/math-fun> List-Help: <mailto:math-fun-request@mailman.xmission.com?subject=help> List-Id: math-fun <math-fun.mailman.xmission.com> Original-recipient: rfc822;mckay(a)vax2.concordia.ca Dear Bill, I thought I might mention a paper by D. H. Lehmer on results concerning the central binomial coefficent which may be of interest to you in case you don't already know it: Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92(1985), 449-457. Best, John _______________________________________________ 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] Re: Apollonian gaskets and Dionysian fillings
by James Propp 10 Feb '04

10 Feb '04

>Jim: Since there are various versions of Apollonian gaskets, could you >tell us exactly the one you are using to get started here? Start with three mutually externally tangent circles, with a big circle that contains and is tangent to the other three. This gives a foursome of mutually tangent circles. Now, anytime you have four mutually tangent circles, you can replace any of the four by a different circle that is tangent to the other three. Repeat this sort of replacement operation ad infinitum, and you've got the kind of gasket I mean. Check out http://arxiv.org/abs/math.MG/0101066 for some pictures of this sort of gasket, and http://ups.physics.wisc.edu/~hal/SSL/apollonian.pdf for pictures that start down the road from gaskets to space-fillers. Jim

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[math-fun] Re: Help with a sequence needed
by Dean Hickerson 10 Feb '04

10 Feb '04

Mostly to Jon Perry: > Consider a tower of height n on the left hand side of a square, and we wish > to create a horizontal line along the bottom of the square. > > We can push any block at the top of a column 1 space to the right, and then > gravity takes over. > > How many ways can we do this? > > So far I have for n=1, 0,1,2,5, and I guess n=5 is 15 (Catalan numbers? > A000108) > > If we label the blocks for convience, then the 5 options for n=4 are: > > 444332 > 444322 > 443222 > 433222 > 433322 You missed some for n=4. How many depends on exactly what you want to count. If you're just interested in the number of possible arrangements at the end, then there are 6; the missing one is given by 433342. But there are other sequences of moves that lead to the same final arrangements as some of the above. If I haven't made any mistakes, there are 10 ways to move the blocks; each is shown here along with the final arrangement on the bottom row: 433222 1432 433232 1423 433322 1423 433342 1243 443222 1342 443242 1324 443422 1324 443432 1234 444322 1324 444332 1234 In general, with n blocks, we can get any permutation along the bottom row that has 1 in the leftmost position, so the number of final arrangements is (n-1)!. (Use induction on n: First move the top n-2 blocks to the bottom row, in the desired order. Then push some of them 1 space to the right, leaving a gap that block 2 can be pushed into.) I don't know how many possible sequences of moves there are. Dean Hickerson dean(a)math.ucdavis.edu

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Re: [math-fun] Two questions in geometric combinatorics
by Mike Stay 09 Feb '04

09 Feb '04

The triangular numbers are 1,3,6,10,... * * * * * * * * * * If you stack these triangles, you get a tetrahedron. ----- Original Message Follows ----- > > On Fri, Jan 30, 2004 at 05:17:01PM -0500, asimovd(a)aol.com > > wrote: Let A_n denote a regular tetrahedral arrangement > > of points in 3-space, n to an edge. (The layers > > successively contain 1, 3, 6,... points.) > > Can you be more explicit what this arrangement is? I am > completely baffled. > > Peace, > Dylan > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > [Attachment: dpt(a)lotus.bostoncoop.net.sig] -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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