November 2004 - mathfun@mailman.xmission.com

[math-fun] (Fwd) [NOBNET 4625] Mathematical Games
by Bernie Cosell 09 May '05

09 May '05

I know that in addition to your likely being interested in this, I know that several of you have contributed to Mathematical Games over the years and so might want to get in touch with Ed... /Bernie\ ------- Forwarded message follows ------- Date: Mon, 8 Nov 2004 06:30:37 -0800 (PST) From: ed pegg <ed(a)mathpuzzle.com> Subject: [NOBNET 4625] Mathematical Games To: nobnet(a)iijnet.or.jp Martin Gardner just recently celebrated his 90th birthday. Also, maa.org is *finally* releasing the entire collection of Mathematical Games on CD rom next month. There were frightening amounts of copyright issues to work out, but they have been resolved. Catalog Code: TDG/FW04 ISBN 0-88385-545-3 List Price: $54.95 This is all 15 Mathematical Games books on a CD-Rom. I'd like to do a "behind the scenes" look at Mathematical Games for an article. Martin did lots, but he avidly used a lot of help from hundreds of people. I'd like to write an article about the background help. If you ever assisted in one of Martin's columns, I'd like to hear from you -- a note about what you did. I'll use these notes in my article. Ed Pegg Jr http://www.maa.org columnist http://www.mathpuzzle.com ------- End of forwarded message ------- -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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[math-fun] A missing sequence from group theory?
by N. J. A. Sloane 04 Feb '05

04 Feb '05

To MathFun and SeqFans: A possibly new sequence? Could someone calculate it? Let t(G) = no. of unitary factors of the abelian group G, and let T(n) = Sum t(G) taken over all abelian groups of order <= n. There are several papers giving estimates for T(n), e.g. T(x) = c_1 x (log x + 2 gamma - 1) + c_2 x + ... My question is, how does the sequence T(n) begin? References and definitions follow, from MSN. NJAS MR1850168 Zhai, Wen Guang On the average number of unitary factors of finite abelian groups. II [MR1813450 (2001j:11092)]. (Chinese) Acta Math. Sinica 44 (2001), no. 4, 667--672. 11N45 MR1813450 (2001j:11092) Zhai, Wenguang On the average number of unitary factors of finite abelian groups. II. Acta Math. Sin. (Engl. Ser.) 16 (2000), no. 4, 549--554. (Reviewer: R. C. Baker) 11N45 (11L07 20K01) MR1641042 (99e:11124) Zhai, Wenguang; Cao, Xiaodong On the average number of unitary factors of finite abelian groups. Acta Arith. 85 (1998), no. 4, 293--300. (Reviewer: R. C. Baker) 11N45 MR1613290 (99f:11124) Wu, J. On the average number of unitary factors of finite abelian groups. Acta Arith. 84 (1998), no. 1, 17--29. (Reviewer: R. C. Baker) 11N45 (11L07 20K01) 1225427 (94k:11108) Schmidt, Peter Georg(D-MRBG) Zur Anzahl unitrer Faktoren abelscher Gruppen. (German) [On the number of unitary factors in abelian groups] Acta Arith. 64 (1993), no. 3, 237--248. Part of this review follows: Let $\scr A$ denote the set of all abelian groups. Under the operation of direct product, $\scr A$ is a semigroup with identity element $E$, the group with one element. $G_1$ and $G_2$ are relatively prime (German: teilerfremd) if the only common direct factor of $G_1$ and $G_2$ is $E$. We say that $G_1$ and $G_2$ are unitary factors of $G$ if $G=G_1\times G_2$ and $G_1,G_2$ are relatively prime. Let $t(G)$ denote the number of unitary factors of $G$, and let $T(x)=\sum_{G\in {\scr A}, |G|\le x} t(G).$ There are constants $A,B,$ and $C$ such that $\Delta(x)\colon =H(x)-Ax\log x - Bx -C\sqrt x=o(\sqrt x).$ In this paper, the author shows that $\Delta(x)\ll x^{3/8} \log^{7/2}x.$ The previous best estimate for $\Delta(x)$ had been $\Delta(x)\ll x^{31/82} \log^2 x$; this was given by H. Menzer \ref["Exponentialsummen und verallgemeinerte Teilerprobleme", Habilitationsschrift, Friedrich-Schiller-Univ., Jena, 1992; per bibl.]. Further references from the Wu (1998) paper: Many accented letters have been dropped! # R. C. Baker and G. Harman, Numbers with a large prime factor, Acta Arith. 73 (1995), 119--145. MR1358192 (97a:11138) # E. Cohen, On the average number of direct factors of a finite abelian group, ibid. 6 (1960), 159--173. MR0118764 (22 #9535) # E. Fouvry and H. Iwaniec, Exponential sums with monomials, J. Number Theory 33 (1989), 311--333. MR1027058 (91b:11097) # S. W. Graham and G. Kolesnik, Van der Corput's Method of Exponential Sums, Cambridge Univ. Press, Cambridge, 1991. MR1145488 (92k:11082) # C. H. Jia, The distribution of square-free numbers, Sci. China Ser. A 36 (1993), 154--169. MR1223084 (94f:11095) # G. Kolesnik, On the number of abelian groups of a given order, J. Reine Angew. Math. 329 (1981), 164--175. MR0636451 (83b:10055) # E. Krtzel, On the average number of direct factors of a finite abelian group, Acta Arith. 51 (1988), 369--379. MR0971087 (89m:11091) # H.-Q. Liu, The greatest prime factor of the integers in an interval, ibid. 65 (1993), 302--328. MR1259341 (95d:11117) # H.-Q. Liu, On some divisor problems, ibid. 68 (1994), 193--200. MR1305200 (96a:11089) # H.-Q. Liu, Divisor problems of 4 and 3 dimensions, ibid. 73 (1995), 249--269. MR1364462 (96k:11118) # P. G. Schmidt, Zur Anzahl unit\"arer Faktoren abelscher Gruppen, ibid. 64 (1993), 237--248. MR1225427 (94k:11108) # P. Sargos and J. Wu, Multiple exponential sums with monomials and their applications in number theory, Prpublications 97/ n37 de l'Institut lie Cartan, Universit Henri Poincar (Nancy 1). # J. Wu, On the distribution of square-full and cube-full integers, Monatsh. Math., to appear. cf. MR1657818 (2000a:11125)

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[math-fun] 'Cogwheels of the Mind' ...
by Guy Haworth 30 Nov '04

30 Nov '04

May I ask for any views or reviews of Anthony Edwards' 'Cogwheels of the Mind' ... in addition to those via Google on Amazon, MAA, Sc. Amer. Guy Haworth

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[math-fun] Re: continued fractions and surreals
by Mike Stay 30 Nov '04

30 Nov '04

More results: The function f that maps the stern-brocot tree onto the tree of surreals is closely related to the Minkowski questionmark function. For x in [1,2], f(x) = ?(x-1). Define the pseudo-lg as plg(x) = 2^k x - k - 1, 2^(-k) <= x < 2^(-k+1), k>=0 a linear interpolation between powers of 1/2. Let f' = inverse(f). Then f'(plg(f(x))) = x-1 for x in [1,2]. For x>=2, f'(f(x)-1) = x-1. And of course, for x>0, -f(1/x) = x. Using f, we can compute homomorphisms of continued fractions. For example, what is the point corresponding to the golden ratio? Since plg(x) = 2x-2 for all the convergents, we solve x=-plg(x)=-(2x-2) and find that f(phi) = 2/3. What about if we use the binary logarithm lg instead of plg? f'(lg(f(x))) >= x-1 on [1,2]. It's a bumpy fractal with equality at powers of 1/2 and some extra points I haven't characterized yet. The golden ratio in this case is x=-lg(x)=-log(x)/log(2) x=lambertw(log(2))/log(2) What kind of tree has the homomorphism g(x), where g satisfies g'(lg(g(x))) = x-1 on [1,2]? lg(g(x)) = g(x-1) g(x) = exp(g(x-1)) Looks like a job for continuous tetration (flow stuff) to me. -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] Re: continued fractions and surreals
by Mike Stay 24 Nov '04

24 Nov '04

Moving from the Stern-Brocot tree to the tree of surreals is done by a function f closely related to the Minkowski Question Mark function. In fact, when mapping rationals in the range (1,2) to their corresponding entries in the surreal tree in (0,1), we get f(x) = ?(x-1). The answer to the question "What function plays the role of lg in the Stern-Brocot tree?" where lg is log_2, is g(x)=f^-1(lg(f(x))) and doesn't seem to have a simpler form. ------- Original message -------- The section [of ONAG] on the game "contorted fractions" explores the area in great depth, but there's another way of looking at continued fractions with surreal numbers that I haven't seen mentioned anywhere. The shift operator acting on the Stern-Brocot tree deletes the root (x=1) and the left branch (x<1). This is equivalent to subtracting 1 until the number is less than or equal to 1. When we find ourselves at the root, we halt; if we're in the left branch, we jump to the mirror position in the right branch (1/x) and continue. We can do the same operation on the tree of surreals. Subtracting 1 will get you to the LR branch--that is, first go right, then left (0<x<1). To get to the root or left branch (x<=0), we have to apply the "pseudo-lg" function: p-lg(x) = { x * 2^k - k - 1 where 1/2^k <= x < 1/2^{k-1} This is just a linear interpolation between powers of 1/2. Finally, if we're at the root (x=0) we halt; if we're in the left branch, we jump to the mirror image in the right branch (-x). If we use lg instead of p-lg, cf's typically involve the Lambert W-function, and the omega constant plays the role of the golden mean. What function plays the role of lg in the Stern-Brocot tree? -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] More cool election maps (non-partisan, on-topic)
by Henry Baker 12 Nov '04

12 Nov '04

More cool election maps that distort the US map to account for voting amounts instead of land area: http://www-personal.umich.edu/~mejn/election/

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[math-fun] continued fractions and surreals
by Mike Stay 12 Nov '04

12 Nov '04

The section on the game "contorted fractions" explores the area in great depth, but there's another way of looking at continued fractions with surreal numbers that I haven't seen mentioned anywhere. The shift operator acting on the Stern-Brocot tree deletes the root (x=1) and the left branch (x<1). This is equivalent to subtracting 1 until the number is less than or equal to 1. When we find ourselves at the root, we halt; if we're in the left branch, we jump to the mirror position in the right branch (1/x) and continue. We can do the same operation on the tree of surreals. Subtracting 1 will get you to the LR branch--that is, first go right, then left (0<x<1). To get to the root or left branch (x<=0), we have to apply the "pseudo-lg" function: p-lg(x) = { x * 2^k - k - 1 where 1/2^k <= x < 1/2^{k-1} This is just a linear interpolation between powers of 1/2. Finally, if we're at the root (x=0) we halt; if we're in the left branch, we jump to the mirror image in the right branch (-x). If we use lg instead of p-lg, cf's typically involve the Lambert W-function, and the omega constant plays the role of the golden mean. What function plays the role of lg in the Stern-Brocot tree? -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] Permutation Readings
by Jon Perry 08 Nov '04

08 Nov '04

If p is a permutation of n digits, we define it's 'pattern' as follows: Begin at the left most digit, and make a list of the first numbers higher than this one, including this one. Move to the right-most digit, and make a list of the first numbers lower than this one. Move to the next available left digit, and repeat until the permutation is exhausted. For example, take 358926471. We begin 3589 - no number is higher than 9, so we can stop. Then 1. 2 is the next available left digit, so we write 26, and finally 74. So this permutation has the pattern: 3589 1 26 74. Another example; 87214536 becomes 8 631 7 542 This example is easy to see as not unique - e.g. 87245136 has the same pattern. Now for the questions: 1) Is there a combinatorial interpretation for a pattern? 2) Given a pattern, determine the number of permutations that fit it. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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