June 2003 - mathfun@mailman.xmission.com

[math-fun] A sequence from Gardner's 6th book
by N. J. A. Sloane 17 Jun '03

17 Jun '03

Since there has been some discussion of this book, here is a sequence suggested by a problem in Chapter 4. Let S be a list of positive integers (with repeats allowed) such that every number from 1 to n can be obtained as a sum of 1 or 2 elements from S. (The minimal number of coins so that every amount from 1 through n can made using 1 or 2 coins) What are the values of a(n) = min size of any such S , for n = 1, 2, ...? (I don't know the answer.) M. G. says that a(100) <= 16, using S = { 1 3 4 5 8 14 20 26 32 38 44 47 48 49 51 52} found by Peter Wegner. NJAS

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Re: [math-fun] Gardner's 6th book of games
by asimovd＠aol.com 16 Jun '03

16 Jun '03

<< Is Martin Gardner's "Sixth book of mathematical games/diversions" also known by a different name? >> Quite likely, assuming it's still in print. (That's one of the two where my name appears.) Regards, Dan Asimov

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[math-fun] Gardner's 6th book of games
by Jud McCranie 16 Jun '03

16 Jun '03

Is Martin Gardner's "Sixth book of mathematical games/diversions" also known by a different name?

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[math-fun] Re: some puzzle sequences not in OEIS!
by David Wilson 16 Jun '03

16 Jun '03

17: 36, 864, 21060, 544320, ? If we index starting at 1, we have a(n) = a(n-1) ((n+10)(9n-12)/(5n-7)). This gives ? = 14968800. However, the subsequent term is noninteger. Yeah, it's ugly, what can I say.

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[math-fun] mystery constant .32810656687498
by R. William Gosper 15 Jun '03

15 Jun '03

is the largest nome, q, for which the closed curve x=theta_2(t,q), y=theta_1(t,q) is convex. Plouffe's inverter gives me 3281065668749782 = (m405) 1/NineConst^(1/2) Trying to parse "NineConst", I tried both squaring and reciprocating, and got "Not found"! And Simon seems to be away fom his email. ? --rwg

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[math-fun] Jaynes: "Probability Theory"
by Eugene Salamin 13 Jun '03

13 Jun '03

The book is now available. ISBN = 0521592712. Probability theory is developed as an extension of deductive logic to statements which are not true or false, but rather have degrees of plausibility. If you use statistical methods, and would like to understand how and why (and maybe even if) they work, you will want to read this book. Gene __________________________________ Do you Yahoo!? Yahoo! Calendar - Free online calendar with sync to Outlook(TM). http://calendar.yahoo.com

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[math-fun] Monckton & creation of hard tessellation puzzles
by Thane Plambeck 13 Jun '03

13 Jun '03

Monckton's Eternity Puzzle went bad on him, but is it possible to create similar puzzles with a reasonable expectation that no one is going to solve them over, say, a three year time frame, even with computers and extreme cleverness? It does seem to be a good idea to sell puzzles with big prizes for solutions if only one could know that they're attractive enough to sell and unlikely to be solved before the time limit... * * ** * * Old news from about 2000: Christopher Monckton thought he was on to a winner when he invented a new puzzle and offered 1 million pounds to the first person who solved it. Alas, his brainchild looks set to be a ruinous loser... http://showcase.posiweb.net/media/property01.pdf ********* Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/ehome.htm

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[math-fun] no. of tilings by rhombs of the regular 2n-gon
by Alan Schoen 13 Jun '03

13 Jun '03

Coxeter mentioned (I believe in the Rouse Ball book) that one would like to know the number of different tilings by rhombs of the regular 2n-gon. I never asked him whether he believed that it might be possible to find an expression valid for any n. (I imagine he had no such belief.) It doesn't take more than a moment to prove that (ignoring rotations and reflections) there are six tilings of the 10-gon, but doing the counting for the 12-gon is tedious. In 1995 I wrote a program called Ovalturn that scans the n connected strips ('ladders' is the informal name Branko Grunbaum suggested for them in 1979 or 1980) of n-1 rhombs between opposite boundary edges, and performs a halfturn rotation on every 'oval' (hexagon tiled by three rhombs) it encounters. After it has once traversed all n ladders (in cyclic order), it goes around again (etc.). I let this program run for a while on the 12-gon and printed out the tiling after every ovalturn. When I cut out the printouts of all these tilings and sorted them into piles, I found I had 49 different patterns. I believe I caught them all, but it would be nice to have a better way to do the counting! Alan Schoen

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[math-fun] Problem about tiling regular 2n-gon with rhombi
by asimovd＠aol.com 12 Jun '03

12 Jun '03

A recent local math olympiad problem showed how to tile a regular hexagon and octagon using rhombi (with each polygon edge coinciding with a rhombus edge), and asked "What is the number of rhombi needed for such a tiling of a 2n-gon ?" This wasn't hard to answer when the given tiling method was generalized to the 2n-gon. But implicit in the problem was the fact that *any* such rhombus tiling of a 2n-gon uses the same number of rhombi. I've since found the answer proved in passing in a book, but y'all, or y'some, may find this an interesting puzzle. (Also, I suspect there must be a more straightforward proof than the one I happened upon.) PUZZLE: Given a rhombus tiling of a 2n-gon (so that each polygon edge is a rhombus edge), show that the number of rhombi is determined by n. --Dan

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Re: [math-fun] Borromean Rings
by asimovd＠aol.com 12 Jun '03

12 Jun '03

David Wilson writes: << I thought it was kind of neat that Borromean rings cannot be formed using circular rings (of equal or differing sizes). A related question: Given three rigid rings which are not all circular and are reasonably smooth locally, is it always possible to form Borromean rings? I was inclined to think this was true, but then I thought it might not be possible if the rings were sufficiently tangled up that any attempt to form Borromean rings would result in a more complicated linkage. Is this question easier if we require the rings to be planar? >> Matthew Cook has a Borromean rings page: <http://www.paradise.caltech.edu/~cook/Workshop/Math/Borromean/Borrring.html> in which he presents several proofs of the impossibility of making the rings from perfect planar circles of any radii, and also makes the same conjecture (without any smoothness condition). --Dan

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