January 2003 - mathfun@mailman.xmission.com

[math-fun] Fibonacci power triangle
by Richard Schroeppel 06 Feb '03

06 Feb '03

(Everything that follows is unproven conjecture. It's probably old news to Fibonacci aficionados.) It occurred to me that squares of Fibonacci numbers should have a simple recurrence relation. After a bit of futzing, I found 2 2 2 2 F' = 2 F + 2 'F - ''F (I use ' to mean "next term" and prefix ' to mean "previous term".) So 169 = 2*64 + 2*25 - 9. Another way to say this is that the sequence of Fibonacci squares can be dotted with the vector (-1 2 2 -1) to produce 0. The vector for the plain Fibonacci sequence is (1 1 -1). Experimenting showed the vector for Fibonacci cubes to be (-1 -3 6 3 -1). I worked out fourth and fifth powers, and developed the Fibonacci-Pascal triangle, built from the vectors. 1 -1 1 1 -1 -1 2 2 -1 -1 -3 6 3 -1 1 -5 -15 15 5 -1 1 8 -40 -60 40 8 -1 -1 13 104 -260 -260 104 13 -1 The rule for the signs seems clear enough. The magnitudes turn out to be Fibonacci-Binomials. The outer diagonals are 1 1 1 1 ...; the next diagonals are just the plain Fibs; and the next diagonals are products of adjacent Fibs: 104 = 8*13, etc. The next diagonals turn out to be products of three adjacent Fibs divided by 2: 260 = 5*8*13/2. The right way to look at this seems to be 5*8*13/(1*1*2), a kind of binomial coefficent with the regular integers replaced by Fibonacci numbers. The next diagonals will be products of four adjacent Fibs divided by 1*1*2*3, and so on. This can probably be connected with q-binomial coefficients. Some other curiosities: (-1 2 2 -1) also works for F*F' (-1 -3 6 3 -1) works for F*F'*F'' (1 -5 -15 15 5 -1) works for F*F'*F''*F''' Fib(2N+1) = (1 1) dotted with F^2; Fib(2N) = (-1 0 1) dot F^2. Fib(3N) = (-1 1 1) dotted with F^3. Since (-1 2 2 -1) works with both F^2 and F*F', it will also work for squares of the Lucas sequence. Continuing this idea a bit further, the vectors work with powers of generalized Fibonacci sequences, where the starting pair of numbers is A,B, and the terms continue A+B, A+2B, 2A+3B, ... Rich rcs(a)cs.arizona.edu

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Re: [math-fun] Interesting unsolved problem in group theory
by Allan C. Wechsler 31 Jan '03

31 Jan '03

Apparently it is known that if B(2,5) is finite, its order is 5^34 or 582076609134674072265625. See http://dimacs.rutgers.edu/REU/1996/sims.html for a reference to this result. -A

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[math-fun] G/G^n
by Allan C. Wechsler 31 Jan '03

31 Jan '03

Whoops, I should read _all_ my mail before I jump in. Sorry, Michael. -A

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[math-fun] Interesting unsolved problem in group theory
by asimovd＠aol.com 31 Jan '03

31 Jan '03

I learned from some recent postings on sci.math.research that if F(r) is a free group (on r >= 2 generators), and if G^n denotes the subgroup generated by all nth powers of elements of a group G, then it's unknown whether the quotient group, B(r,n) = F(r) / F(r)^5 is finite or infinite. It is known that B(r,n) is always finite for n = 2,3,4, and 6 (cf. Marshall Hall's book, "Group Theory"), as well as for sufficiently large n (apparently independent of r). --Dan

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[math-fun] Help with a recursion
by Bernie Cosell 29 Jan '03

29 Jan '03

I know that I'm getting senile, and I'm growing to accept that, but... this little problem has been driving me *NUTS*. The problem whizzed by on another mailing list as a programming problem and I wasn't particularly interested in the programming aspect [which was actually simple enough given the right algorithm] but in the derivation of the algorithm... But I just *can't* do it, and so I'm looking for some help. I *must* be missing something because it just can't be as hard as I'm finding it. Here's the problem: You want to wager on the outcome of a multi-game [best n out of 2n-1] series, but your bookie will only allow bets on individual games. For a one-game series, this isn't an issue, obviously. But what I was surprised to see was that: For a best two-out-of-three, if you wager half what you want to risk [call that 'R'] on the first two games, then either the series is over and you've gotten the result of wagering R on the series, or the series is tied *and*you're*even*, so you just wager R on the last game and you're done. No matter what happens in the invidual games of the series you win or lose R. For a best three out of five series, it is actually easy enough to do a similar analysis brute force: Bet B on the first two games. If the teams split the series is now best two-out-of-three, and you're even, so use the above strategy. If either time wins both, then you're either ahead or behind by 2B. To make your overall position be 'R' if the same team wins the third game and sweeps, you need to bet R-2B on the third game. If the series goes on, it is now 2-1. But 2-1 in which a five game seeries is the *same* as 1-0 in a three game series and we know how the bets go in that: when a three-game series is at 1-0, you should be ahead [behind] by R/2. Since your current standing is [+/-](4B - R), you just set: R/2 = 4B - R and you solve for 'B' and find out that your first bet in the three out of five series should be 3R/8. It seems that this should just be a high-school simple recusion problem, but it is sure beating me -- I can't even find the brute-force solution for a best 4-out-of-7, much less the general formula for an arbitrary series. It is easy enough to show that all you need is to know the *first* bet -- given just the first bet for a particular series it is easy to figure out ALL the bets for that series, but I'm stuck on how to do that. Insight would be appreciated! :o) /Bernie\

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[math-fun] moving from academia to industry...
by wouter meeussen 28 Jan '03

28 Jan '03

more of you should consider it: it's a win-win proposal : the average intelligence of both populations increases! ;-)) Wouter.

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[math-fun] cell phone math
by Thane Plambeck 28 Jan '03

28 Jan '03

>From a Reuters news story today, "Cell Phones 'Blind' Drivers" Writing in the March issue of the Journal of Experimental Psychology: Applied, Strayer's group said use of a cell phone clearly distracted the drivers.... "It is like studies that show 90 percent of people think they are better-than-average drivers. Forty percent of them are wrong." Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm

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[math-fun] Test
by David Wilson 28 Jan '03

28 Jan '03

Test message

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[math-fun] I'm Back
by David Wilson 28 Jan '03

28 Jan '03

At the behest of my wife, I have resubscribed to math-fun, and am hoping to resubscribe to seqfan if I can get O. Gerard's address. I do not wish to be an OEIS editor, however. I do have comments on A055480. The energetic numbers are numbers that can be broken into substrings and expressed as a sum of powers of those substrings. For example, 8833 = 88^2 + 33^2 is energetic according to the example given in the sequence. If only one substring were allowed, then every number would be energetic, e.g, 44021 = 44021^1. Therefore we must assume that the number must be broken up into at least two substrings, in which case, the numbers 1 through 9, originally included in the sequence, should be omitted. Here is my updated/extended version of this sequence. %I A055480 %S A055480 24,43,63,89,100,101,102,103,104,105,106,107,108,109,132,135,142,153,175, %T A055480 209,224,226,262,264,267,283,284,332,333,334,357,370,371,372,373,374,375, %U A055480 376,377,378,379,407,445,463,518,568,598,629,739,794,809,849,935,994,1000 %N A055480 Energetic numbers. %O A055480 1,1

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RE: [math-fun] building an on-line math museum
by Kimberling, Clark 27 Jan '03

27 Jan '03

Ed (and others), Right on! Sketchpad (version 4) is a wonder - and so is the "automatic" JavaSketchpad. May I ask you to visit http://faculty.evansville.edu/ck6/GIAJSP/ . The interactive action-sketches show some things that can be done with the Version 4 JavaSketchpad much more readily than with Version 3. This is a new site, so there are bound to be things to be improved. Ed, if you or others visit the site and see errors or other opportunities for improvement, please let me know (ck6(a)evansville.edu) (The site cited above, called "G.I.A. Java Gallery", is associated with a book, Geometry in Action, released this month from Key College Publishing.) Thanks, Clark Kimberling -----Original Message----- From: ed pegg [mailto:ed@mathpuzzle.com] Sent: Sunday, January 26, 2003 6:32 PM To: math-fun(a)mailman.xmission.com Subject: RE: [math-fun] building an on-line math museum > >>This project would be part of an effort to create an on line web site > >>whose purpose would be to introduce members of the general public to > some > >>of the fascination connected with the subject of mathematics. There is > >>already a great deal of this sort of material scattered around on the > >>internet. After many hours of surfing I found a few examples which may > >>give an idea of what I have in mind. The first thing he should pick up is Geometer's Sketchpad. It's pretty effortless to go from a geometric construction, as in the triangle to square dissection, to a fully functional java applet. Really, it's just draw and save an applet. Geometry in Motion has examples: http://www15.addr.com/~dscher/ There are many good math sites out there. He should look at all the sites presented as columns at http://maa.org/news/columns.html . My site, mathworld.com, and Erich Friedman's site all have much more, especially in the way of links to much more. --Ed Pegg Jr. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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