March 2004 - mathfun@mailman.xmission.com

[math-fun] odometer word game
by Richard Schroeppel 03 Jun '04

03 Jun '04

zimb zinc chromatic chromatid Rich

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[math-fun] Torus
by Richard Guy 09 Apr '04

09 Apr '04

Yesterday we watched the video Outside In. Is it possible to turn a torus inside-out without breaking or creasing it? R.

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[math-fun] 4 pile nim
by Thane Plambeck 01 Apr '04

01 Apr '04

David Gale writes: > Before I start here's something more in the math-fun spirit. I'm offering two hundred bucks ($200) for a solution of the following Nim variant. > There are 4 piles, ordinary Nim rules, the only difference being that the game ends when three of the four piles are empty. OK---I'll bite, if no one else will. Case I: (Single non-empty heap) Answer: *0 (the game is already over) everything is a P position, ie 2nd player to move wins. Case II: (Two nonempty heaps, sizes m and n). Everything is an N-position. A wiinning move to take everything in one of the heaps. For extra credit (but not for the $200), after a few minutes of noodling I got the nim heap equivalent is BitXor[m-1,n-1] +1. Case III: (Three nonempty heaps, sizes r,s,t). Assume r <= s <= t. Every move involving taking a whole heap leads to a (Case II) N position. What are the three heap P positions? 1,1,1 is a P position for example. Then 1,1,t for other t>1 is always an N position, since there's a move to 1,1,1. It looks like the P positions are in one-to-one correspondence with every pair (a,b) = (s-r, t-s) with a, b nonnegative. Ie, the P position (1,1,1) corresponds to the choice (a,b) = (0,0), and (for example), the P position (2,3,4) corresponds to the choice (1,1), (and for another example), (4,6,7) corresponds to (2,1), ET CETERA, where I don't really know what ET CETERA means. Assuming this correspondence is valid, how to recover r from a and b? This should be achievable---but not after two glasses of chianti. Is it hopeless to give a formula for the Case III nim equivalent? Case IV: (Four nonempty heaps). Hmmm.... Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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[math-fun] april 1 letter in Nature
by Thane Plambeck 31 Mar '04

31 Mar '04

Irrelevance, buzzwords, coffee: it's the virtual PI Sir - Several graduate students would like to suggest a modest proposal to the authors of the recent article "Functional genomic hypothesis generation and experimentation by a robot scientist" (Nature 427, 247-252; 2004). We've been doing a little research of our own and have generated an exciting hypothesis: we can build a Virtual Principal Investigator. The software platform will consist of a tiny amount of background knowledge about a completely different system from the one being actively investigated, a buzzword-invention engine, coffee-consumption apparatus, a grant-writing node and a cat-o'-nine-tails that will tie everything together. For much less money than an actual investigator, we created a machine that resides in an office, drinking coffee at a steady rate. At bi-monthly intervals, the Virtual Principal Investigator spits out grants that are derivatives of highly successful grants from the past. Between these time points, it will play an endless loop of inspirational messages and test the effectiveness of the whip as a method of controlling subservient organisms. Initial tests of our Virtual Principal Investigator have revealed no differences from our existing model of an investigative scientist, but we have yet to perform an experiment that was not naive or random. We look forward to discussing our results with you. Several non-robot scientists (names supplied) Department of Biological Chemistry and Molecular Pharmacology, Harvard Medical School, 240 Longwood Avenue, Boston, Massachusetts 02115, USA Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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[math-fun] Buying a PhD from a University that doesn't exist
by Thane Plambeck 30 Mar '04

30 Mar '04

Amusing research (123 slides!) by George Gollin http://www.hep.uiuc.edu/home/g-gollin/diploma_mills.pdf Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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[math-fun] Are you sharp
by Thane Plambeck 25 Mar '04

25 Mar '04

Once Bob Floyd saw a stanford newsgroup message sent by someone named Tuna Ertemalp. He immediately typed in a response that said "This is an obvious pseudonym. Aunt Palmtree, I have found you out!" Yet it was no pseudonym (I think Tuna works at Microsoft). Now I read in the introduction to Atiyah and MacDonald's Introduction to Commutative Algebra "We have provided a substantial number of exercises at the end of each chapter. Some of them are easy and some are hard. Usually we have provided hints, and sometimes complete solutions, to the hard ones. We are indebted to Mr. R. Y. Sharp, who worked through them all..." Aha! Caught them, I thought. But no, he's the current head of the math dept at Sheffield. Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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[math-fun] some exact formulas for integer sequences,
by Simon Plouffe 23 Mar '04

23 Mar '04

hello, (this is old material from 1993), I collected some of my notes about exact formulas for integer sequences. They are somewhat usefull since it can be used to compute the n'th term of a sequence. it uses [ ] , { } and some constants like exp(1). http://www.lacim.uqam.ca/%7Eplouffe/exact.htm note : most of those formulas do not appear in the current On-line Encyclopedia of integer sequences. By the way, can anybody do this : Can we find an EXACT formula for the n'th term of A001006 (Motzkin numbers), the only useful formula we have is a recurrence but with the index that grows with n which makes it difficult to compute for large n. http://www.research.att.com/projects/OEIS?Anum=A001006 is 1,1,2,4,9,21,51,127,323,835,2188,5798, ... and it grows roughly like 3^n. Simon Plouffe

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[math-fun] exact formulas for Motzkin numbers
by N. J. A. Sloane 23 Mar '04

23 Mar '04

Maybe Simon was looking at a very old copy of the OEIS. There are now several exact formulas for A001006. See the entry for more details. Here are two: %F A001006 a(n) = (-1/2) SUM (-3)^i C(1/2,i) C(1/2, j); i+j=n+2, i >= 0, j >= 0. %F A001006 a(n) = (1/(n+1)) * Sum_{i} (n+1)!/(i!*(i+1)!*(n-2*i)!) - Bernhart. NJAS

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[math-fun] weird continued fraction
by R. William Gosper 19 Mar '04

19 Mar '04

Benoit Cloitre sent me a draft containing a recurrence identity implying a convergent matrix product: [ 2 2 ] [ --------- --------- ] [ sqrt(%pi) sqrt(%pi) ] [ ] = [ %pi 2 %pi 2 ] [ sqrt(---) - sqrt(---) sqrt(---) - sqrt(---) ] [ 2 %pi 2 %pi ] inf [ sqrt(n + 1) ] /===\ [ ------------- 1 ] | | [ n sqrt(n + 2) ] | | [ ] , | | [ sqrt(n) ] n = 1 [ ----------- 0 ] [ sqrt(n + 2) ] giving the c.f. (if you'll pardon my ascii), _ /2 /3 2 /2 / - / - v V 1 V 1 ------ = ---- + ------------------ . pi - 2 1 /3 /4 / - / - V 2 V 2 ---- + ----------- 2 /4 /5 / - / - V 3 V 3 ---- + ---- 3 . . . Nontriangular 2x2 products usually diverge, and you get the c.f. as the limit of the top row over the bottom. If we can find decent asymptotics to c.f. numerators (and denominators), we might scale them out of the matrix to produce more interesting identities of which the c.f. is a projection (which, in the case above, conceals a sqrt(pi)). --Bill Gosper P.S., note the near "recurrence" of our old friend, pi/2+2/pi, the expected magnitude of a 3D rotation, and once-believed optimum "sofa problem" area.

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[math-fun] Re: Number derivative
by Mike Stay 18 Mar '04

18 Mar '04

You can take q-deformed number derivatives of field elements. Work in GF(p^n). Define [a] as (q^a - 1)/(q - 1). Let x be a generator of the field. D( x^a ) = ((qx)^a - x^a) / (qx - x) = (q^a - 1)/(q-1) x^(a-1) = [a] x^(a-1) If x^a + x^b = x^c, then [a] x^(a-1) + [b] x^(b-1) (qx)^a + (qx)^b - (x^a + x^b) = ----------------------------- qx - x So if q is such that (qx)^a + (qx)^b = (qx)^c, then we have (qx)^c - x^c = ------------ qx - x = [c] x^(c-1) and the derivative is linear. Setting q = x^(p^k - 1) will satisfy the equation, where 1 < k < n. Of course, you can do "differential equations" this way, too. For example, take GF(2^3) mod x^3+x+1; then x is a generator and D(x^2+x+1) = x^2+x+1. -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

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