January 2005 - mathfun@mailman.xmission.com

[math-fun] patterns in steel, damascus steel, interesting patterns
by Simon Plouffe 10 Jan '05

10 Jan '05

Hello, I just contacted an old friend who lives in France and specialize in the fabrication of knives and 'eggs of Damascus', the interesting thing is that when they make those egss, patterns appears on the surface which are intriguing. http://damascus.free.fr/f_damas/quest.htm and see the eggs here : http://oeufs-damas.com/index.php http://oeufs-damas.com/oeufs/oeuf_08.jpg They are considered decorative and they sell it. (I am not trying to sell those things, just pointing to an interesting phenomena). Simon Plouffe

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[math-fun] Query into pseudoprimes
by David Wilson 08 Jan '05

08 Jan '05

It is no longer surprising that there are even pseudoprimes 2n for which 2^(2n) == 2 (mod 2n) specifically, 2n = 161038 (see A006935). This got me to wondering if there were always multiples of This got me to wondering if for any m, there always exists k >= 2 satisfying [1] m^(km) == m (mod km) and if so, what would the least such k(m) be? Empirically, it looks as if k(m) exists for all m, and programmatically I got the following k(m) for 2 <= m <= 100. 80519,2,3,2,5,2,7,2,3,2,11,2,13,2,3,2,17,2,19,2,3,2,23,2,5,2,3,2,29,2,31, 2,3,2,5,2,37,2,3,2,41,2,43,2,3,2,47,2,7,2,3,2,53,2,5,2,3,2,59,2,61,2,3,2,5, 2,67,2,3,2,71,2,73,2,3,2,7,2,79,2,3,2,83,2,5,2,3,2,89,2,7,2,3,2,5,2,97,2,3 Superficially, this appears to be k(m) = 80519, if m = 2 lpf(m-1), if m > 2. where lpf(x) is the least prime factor of x (A020639). I assume we can prove that k = lpf(m-1) satisfies [1], any takers? But maybe things aren't so simple. According to my program, there are some m for which k(m) < lpf(m-1). In particular: k(152) = 143 < 151 = lpf(152-1) k(158) = 77 < 157 = lpf(158-1) k(338) = 161 < 337 = lpf(338-1) k(368) = 77 < 367 = lpf(368-1) k(380) = 299 < 379 = lpf(380-1) k(524) = 437 < 523 = lpf(524-1) k(542) = 143 < 541 = lpf(542-1) k(548) = 77 < 547 = lpf(548-1) k(608) = 407 < 607 = lpf(608-1) k(614) = 611 < 613 = lpf(614-1) k(662) = 581 < 661 = lpf(662-1) k(674) = 319 < 673 = lpf(674-1) k(692) = 377 < 691 = lpf(692-1) k(998) = 161 < 997 = lpf(998-1) ... These may actually be discrepancies, however, there is also the distinct possibility of computational error, so maybe someone could verify or refute these values?

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[math-fun] Book recommendation
by Daniel Asimov 08 Jan '05

08 Jan '05

Recently I've been browsing with great delight a recent book called Mathematical Puzzles: A Conoisseur's Collection, by Peter Winkler. These puzzles have been chosen because they're typically much more interesting than, say, the puzzles in the Monthly. They're aimed at people with a math background, but not necessarily an advanced one. Some of the solutions are fairly complex, but almost none require math beyond calculus (and very few go even that far). Besides a sections on probability, geometry, games, combinatorics, etc., there's also a non-mathy fun section on geography puzzles, a section of especially tough puzzles, and finally, a section of easy-to-state unsolved problems. Most math-funners have seen some of these problems before, but I'd guess at least 2/3 of them were new to me, and well worth the price ($18 on Amazon (paperback)). --Dan

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Re: [math-fun] Modular square root
by Mike Stay 06 Jan '05

06 Jan '05

Rabin's public key algorithm is based on the difficulty of computing square roots mod n=p*q, p,q prime. ----- Original Message Follows ----- > Except for semicolons, Michael Kleber writes: > > << > [T]here will be at most two [square roots of an integer > mod p]. > > Working mod p, you can still do a^2=b^2; a^2-b^2=0; > (a+b)(a-b)=0; a=b or a=-b, since there are no zero > divisors mod p. >> > > There's something different and interesting going on with > square roots in the integers mod n for n not prime, > and especially if n|24. > > --Dan > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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[math-fun] A060839
by David Wilson 06 Jan '05

06 Jan '05

A060839 is multiplicative.

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[math-fun] New sequence
by David Wilson 06 Jan '05

06 Jan '05

With all this discussion of square roots I engendered, I decided to invesitigate the number of square roots of 1 mod n. This turned out to be given by A060594, which is a multiplicative sequence with a(2^n) = 1 if n = 0; 1 if n = 1; 2 if n = 2; 4 if n >= 3. a(p^n) = 1 if n = 0; 2 if n >= 1. This shows that every modulus m has 2^k square roots of 1 for some k. With a little thought, I was able to convince myself that the least modulus m with 2^k square roots of 1 is 1 if k = 0; 3 if k = 1; 4*p[n-1]# if k >= 2. where p[n] is the nth prime and p# is the primorial of p.

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Re: [math-fun] Modular square root
by Jpr2718＠aol.com 06 Jan '05

06 Jan '05

David Wilson asks: Is there a fairly fast algorithm to find the square root of a number to a given prime modulus p? Henri Cohen, ``A Course in Computational Algebraic Number Theory,'' gives several methods. John Robertson

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[math-fun] 13yrold programmers; dodacahedron sequence (2)
by Richard Schroeppel 06 Jan '05

06 Jan '05

Some non-member messages were dropped over the holidays. -- Rich --------------- Date: Mon, 27 Dec 2004 18:05:29 -0800 From: Richard Fateman <fateman(a)cs.berkeley.edu> To: "R. William Gosper" <rwg(a)osots.com> CC: math-fun(a)mailman.xmission.com Subject: Re: [math-fun] Learning computer programming for a bright 13-year old ? You might be surprised to learn that speaking fractions is ambiguous. 30/400 and 34/100 are quite different numbers, but each can be pronounced thirty four hundredths Best wishes for the new year! RJF R. William Gosper wrote: >>My very bright 13-year-old nephew would like to learn computer >>programming (from scratch). He's quite familiar with using software, > > > Dan, there are some kids where the absolutely best place to start would > be the console toggle switches and indicator lamps on an ancient > mainframe, if you can find a live one in a computer museum, e.g. There > are other kids who take best to the opposite extreme, say generic > functions (object-oriented LISP). For a certain concrete thinker, I'm > planning the following first program: Combine Macsyma's string > manipulation functions, especially CARDINAL_STRING (12 -> "twelve") and > ORDINAL_STRING (12 -> "twelfth") to make a FRACTION_STRING function > (11/12 -> "eleven twelfths"). There are nice opportunities for bugs > ("one twelfths", 1/2 -> "one second", 3 -> "three firsts") and features > (improper vs proper format, "three quarters"), generalization > (RATIONAL_STRING, NUMBER_STRING even taking floats), and finally > Robinson Crusoeing one's own CARDINAL_ and ORDINAL_ STRINGs. (Captious: > "third", "thirteenth", "twenty third".) --rwg ------------------ From: "Rainer Rosenthal" <r.rosenthal(a)web.de> To: "wouter meeussen" <wouter.meeussen(a)pandora.be>, <njas(a)research.att.com>, <seqfan(a)ext.jussieu.fr>, <math-fun(a)mailman.xmission.com> Date: Tue, 28 Dec 2004 20:15:36 +0100 Subject: Re: Possibly new sequence related to dodecahedron > more detail is in a 70kb Excel spreadsheet : > http://users.pandora.be/Wouter.Meeussen/DodecahedralVectorSum.xls > Dear Wouter, I couldn't open your interesting file, because I did not have access to the following: http://users.pandora.be/Program Files\ Wolfram Research\MathematicaLinkForExcel\Excel97\MLX.xla Please help, thanks, Rainer Rosenthal r.rosenthal(a)web.de ------------------------- Date: Thu, 30 Dec 2004 20:16:36 -0500 From: franktaw(a)netscape.net (Franklin T. Adams-Watters) To: seqfan(a)ext.jussieu.fr, math-fun(a)mailman.xmission.com Subject: RE: Possibly new sequence related to dodecahedron A pair of related, also finite, sequences: the number of "polypents" (connected sets of pentagons, up to symmetry) on an dodecahedron, and "polyiamonds" on an icosohedron. These sequences can also be done for smaller Platonic solids, but are too simple to be interesting. This is especially so for the tetrahedron, where the sequence is: 1,1,1,1,1 For the square, we have: 1,1,1,2,2,1,1. For the octahedron, the sequence is (I think): 1,1,1,1,3,3,3,1,1 "N. J. A. Sloane" <njas(a)research.att.com> wrote: >The following is an interesting question raised by >Wouter Meeussen <wouter.meeussen(a)pandora.be> on Dec 27 2004: > >He said: >>> Based on the symmetry point-group, how many 'different' sets of 1, 2, .., 10 vertices >>> exist on a dodecahedron? >>> >>> If I count different norms of sums of 1, 2, .., 10 unit vectors, I find: >>> 1,5,12,22,34,50,65,78,78,86 >>> but there might be 'different' sets that accidentaly sum to the same resultant's norm. >>> So the given integers are a lower bound. > >We can rephrase the question as follows. > >Let G be the full icosahedral group, of order 120. >Let v_1, ..., v_20 be the vertices of the dodecahedron. >Let S(n) be the set of vectors > > v_{i_1} + v_{i_2} + ... + v_{i_n} > >where 1 <= i_1 <= i_2 <= ... <= i_n <= 20. >Then what is s(n), the number of orbits of G on S(n)? >(so s(1) = 1, s(2) = 6, ...) > >Presumably this is different from A039742: >... >%S A039742 1,6,50,475,4881,52835,593382,6849415,80757819,968400940,11773656517, >%T A039742 144791296055,1797935761182 >%N A039742 Lattice animals in the fcc lattice (12 nearest > neighbors), connected rhombic dodecahedra, edge-connected cubes. >... > > >Repeat for icosahedron, etc.! > >NJAS > -- Franklin T. Adams-Watters 16 W. Michigan Ave. Palatine, IL 60067 847-776-7645

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Re: [math-fun] Modular square root
by Daniel Asimov 06 Jan '05

06 Jan '05

Except for semicolons, Michael Kleber writes: << [T]here will be at most two [square roots of an integer mod p]. Working mod p, you can still do a^2=b^2; a^2-b^2=0; (a+b)(a-b)=0; a=b or a=-b, since there are no zero divisors mod p. >> There's something different and interesting going on with square roots in the integers mod n for n not prime, and especially if n|24. --Dan

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Re: [math-fun] Modular square root
by Jeffrey Shallit 06 Jan '05

06 Jan '05

See my book with Bach, Algorithmic Number Theory, sections 7.1 and 7.2 Jeffrey Shallit

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