June 2005 - mathfun@mailman.xmission.com

[math-fun] Self-junk
by Eric Angelini 21 Jun '05

21 Jun '05

Hello SeqFan and Math-Fun, Can someone find the last term of this (finite) sequence? 1 21 2 11 3 4 111 1122 5 6 7 8 9 22 23 22222 222222 2222222 22222222 222222222 24 25 26 222 27 28 29 32 33 34 35 36 37 38 39 42 43 44 45 46 47 48 49 52 53 54 55 56 57 58 59 62 63 64 65 66 667 68 69 72 2222 73 22223 74 Building rules: 1) The length of each integer is given by the succession of the digits of the sequence itself (first integers have length 1, 2, 1, 2, 1, 1, 3, ...) 2) No two identical integers in the sequence 3) When adding a new integer to the sequence, put always the smallest unused so far 4) Always have in mind that the final sequence must be as long as possible -- this explains why the above 8th integer is "1122" and not "1111", for instance: "1111" would block the sequence very fast: 1 21 2 11 3 4 111 1111 5 6 7 8 9 -- stop, no more available integer of length "1"... Best, É.

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[math-fun] squarefree string draws Sierpinski triangle
by R. William Gosper 21 Jun '05

21 Jun '05

As in my 5 Dec 96 mail (Thanks Rich!), let d(n) be the alternating sum of the nonzero digits of the balanced ternary representation of n. E.g., d(-69) -> -1,0,1,1 -> -1-1+1 = -1. Then d is an odd function, and mod 3 is a squarefree (stutter-free) sequence. For n>=0, 0 1 2 1 0 1 2 0 2 1 0 2 0 1 2 1 0 1 2 0 1 0 2 1 2 0 2 1 0 2 0 1 0 2 1 2 0 1 2 1 0 1 2 0 2 1 0 2 0 1 2 1 0 1 2 0 1 0 2 0 1 2 1 0 2 1 2 0 2 1 0 1 2 0 1 0 2 . . .. Claim: the series of increments (-)^k exp(2 i pi d(k)/3) from k = -(3^n+1)/2 to (3^n-1)/2 draws a self-avoiding polygonal approximation to a curve which sweeps out the triangular Sierpinski "sponge" (or "gasket"). (Try it!) (Spongebob Trianglepants?) For k>=0, it draws a diagonal chain of consecutively larger ones (= the right half of an infinitely large one). There's a lot of interesting math here. I believe n (3 - 1) t ---------- 2 2 i pi d(k) ==== ----------- \ k 3 > (- 1) e / ==== n 3 + 1 k = - ------ 2 f(t) := limit ----------------------------------- n -> oo n 2 is a continuous map of [0,1] onto the closure of the Sierpinski triangle with the unit interval of the real axis as base. This gives us explicit valuations of f where t has a simple ternary expansion, e.g., the rationals. (f(.012101202..._3) = ?) Other claims: n 3 - 1 ------ 2 2 i pi d(k) ==== ----------- \ k 3 > k (- 1) e / ==== n 3 + 1 k = - ------ i sqrt(3) 2 - --------- = limit -------------------------------------, 6 n -> oo n n 2 3 n 3 - 1 ------ 2 2 i pi d(k) ==== ----------- \ 2 k 3 > k (- 1) e / ==== n 3 + 1 k = - ------ 1 2 - = limit --------------------------------------, 8 n -> oo 2 n n 3 2 n 3 - 1 ------ 2 2 i pi d(k) ==== ----------- \ 3 k 3 > k (- 1) e / ==== n 3 + 1 k = - ------ 7 i sqrt 3 2 - ---------- = limit --------------------------------------, ... 432 n -> oo 3 n n 3 2 --rwg PS, Weisstein gives (Thue-Morse sequence) the base 3 squarefree string 0 2 1 0 1 2 0 2 1 2 0 1 0 2 1 0 1 2 0 1 0 2 1 2 0 2 1 0 1 2 0 2 1 2 0 1 0 ... based on renaming the base 4 digits of twice the parity constant. Simpler: Parity(n+1) - Parity(n), where Parity(n) is the usual sum mod 2 of the binary digits of n. (Add 2 mod 3 for the exact match.)

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Re: [math-fun] Ethnomathematics
by Henry Baker 20 Jun '05

20 Jun '05

Interesting, thank you. I think that today's WSJ editorial is significantly more negative than the article you pointed to below. Perhaps Diane's views have changed/evolved since the article below. Perhaps people substantially less talented than the teacher in "Stand & Deliver" have "captured" ethnomathematics and transformed it into another "P.C." initiative. At 08:16 AM 6/20/2005, Kerry Mitchell wrote: >I don't know how similar this is, but it's by the same author and it's on the same subject, and it's freely accessible: > >http://college.hmco.com/currentconflict/students/multicultural/ravitch_arti… > >Kerry >-------------- Original message from Henry Baker <hbaker1(a)pipeline.com>: -------------- > >> In this morning's Wall Street Journal, Diane Ravitch has an editorial on >> "Ethnomathematics". I don't have access to the WSJ Online, so I can't >> include this short editorial, but you can get the gist of "ethnomathematics" >> from this website: >> >> http://www.science.org.au/nova/073/073key.htm

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Re: [math-fun] Ethnomathematics
by lkmitch＠att.net 20 Jun '05

20 Jun '05

I don't know how similar this is, but it's by the same author and it's on the same subject, and it's freely accessible: http://college.hmco.com/currentconflict/students/multicultural/ravitch_arti… Kerry -------------- Original message from Henry Baker <hbaker1(a)pipeline.com>: -------------- > In this morning's Wall Street Journal, Diane Ravitch has an editorial on > "Ethnomathematics". I don't have access to the WSJ Online, so I can't > include this short editorial, but you can get the gist of "ethnomathematics" > from this website: > > http://www.science.org.au/nova/073/073key.htm

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[math-fun] viruses etc.
by Richard Schroeppel 20 Jun '05

20 Jun '05

I'm not inclined to impose extra requirements on Math-Fun subscribers without a very good reason. My opinion is that requiring virus-protection software won't do much to prevent the problem of subscriber email addresses showing up as bogus senders of viruses. The people who run Windows already have their hands full with security troubles, and are doing what they can to cope. Many of them have no effective control over their configuration, being stuck with whatever work or school provides. Xmission provides fairly good spam & virus control: As moderator, I have to discard a modest number of junk messages, but it's minor. So you probably won't get spam/viruses directly from the list. (No promises.) But the problem David Wilson describes, a presumed triangle of victims of various kinds, poses a big problem for Whitelisting, an otherwise very effective defense method. Many of us are already in a lot of address books, and thinning out the math-fun subscribers won't make much difference. I've had this problem in an odd kind of ghosting mode: Spam from one of my email accounts to another. It took a while for me to figure out that the From address was simply a forgery. I'll let people discuss this *if they must* for 3 days. If you just can't stop yourself from posting, please try to say something new, that we haven't heard before. Remember that very few people signed up for this list to discuss mailing list security problems. I'll be away for 10 days starting Saturday, so you're on your honor to behave. If the security discussion gets out of hand, you can try using the web interface to unsubscribe or temporarily turn things off, or switch to digest mode. Rich rcs(a)cs.arizona.edu

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[math-fun] Virus mail and such
by David Wilson 15 Jun '05

15 Jun '05

Creighton Dement recently received a virus mail that appeared to be from me. I know that my address is on somebody's infected computer, since I regularly receive virus mail bounces from random servers. I have taken all reasonable steps within my knowledge and power to ensure that that infected computer is not mine. I run ZoneAlarm firewall and AVG antivirus, and do occasional manual checks of my registry for unwanted DLLs, BHOs and startup programs. I rarely send attachments, if I do, they will be innocuous formats such as .jpg or .pdf files. If you get a message from me with an attached executable, such as a .exe, .pif, or .scr file, assume it's not from me. Almost all viruses read address books, but I'ven never of one that adds addresses to it. Therefore, I assume that Creighton's and my own address had been added to the address book of the offending machine in the normal fashion. This argues fairly strongly for a seqfan or math-fun subscriber with an infected computer. Maybe math-fun and seqfan should require a certain level of virus protection from their subscribers. Just a thought. -------------------------------- - David Wilson

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P.S. RE: [math-fun] Nonexistence of cubed cube
by dasimov＠earthlink.net 14 Jun '05

14 Jun '05

<< But this reminded me of a question I've had for a long time: QUESTION: Can there be a cubed 3-torus? Start with a cubical 3-torus: an NxNxN cube with opposite faces identified. Can this 3-torus be tiled by unequal integer cubes (each of which is a union of some of the N^3 unit cubes that constitute the original cube) ? >> Here's a couple of related questions: 1. Can 3-space be tiled by unequal cubes? Assume that a) the tiling is locally finite, i.e., that any sphere contains at most finitely many cubes, and b) each face of each cube is parallel to a coordinate plane. 2. What if we drop condition a) and/or b) ? ----------------------------------------------------------------- (Just to be precise: We want a collection of closed cubes {C_j} in 3-space whose union is all of 3-space and whose interiors are disjoint.) --Dan

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[math-fun] Nonexistence of cubed cube
by dasimov＠earthlink.net 14 Jun '05

14 Jun '05

I was scanning the cool Geometry Junkyard website (http://www.ics.uci.edu/~eppstein/junkyard/3d.html) and came across this link http://www.ics.uci.edu/~eppstein/junkyard/no-cubed-cube.html to a post by David Moews that explains why there can be no cubed cube (the same argument that appeared in Martin Gardner's Second Scientific American Book of Mathematical Puzzles and Diversions). But this reminded me of a question I've had for a long time: QUESTION: Can there be a cubed 3-torus? Start with a cubical 3-torus: an NxNxN cube with opposite faces identified. Can this 3-torus be tiled by unequal integer cubes (each of which is a union of some of the N^3 unit cubes that constitute the original cube) ? --Dan

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[math-fun] Self-Belgian English Number-Words
by Eric Angelini 14 Jun '05

14 Jun '05

... wait to be discovered here: http://angelink.be/?BelgianWords Best, É.

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[math-fun] naive observation
by wouter meeussen 13 Jun '05

13 Jun '05

given that Sum[ EulerPhi[k]/k^s , {k,Infinity}] equals Zeta[s-1]/Zeta[s], and that Sum[ DivisorSigma[p,k]/k^s , {k,Infinity}] equals Zeta[s-p]*Zeta[s], it is evident that Sum[(Zeta[3] EulerPhi[k] - DivisorSigma[1, k]/Zeta[3])/k^3, {k,Infinity}] =0 since it boils down to 'Zeta[2]-Zeta[2]'. Being well aware of the 'indefiniteness \infinitness' of Zeta[1], I was amused by the behaviour of 'Zeta[1]-Zeta[1]' as in: Sum[(Zeta[2] EulerPhi[k] - DivisorSigma[1, k]/Zeta[2])/k^2, {k,Infinity}]. Its numerical values dance 'round zero. I naively expected it to sum to zero anyhow, but it refuses to oblige. It gets stuck at about 1.1399.. , not a familiar value over at my place. W.

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