November 2002 - mathfun@mailman.xmission.com

[math-fun] funny little limit on central binomial coefficients
by Meeussen Wouter (bkarnd) 19 Nov '02

19 Nov '02

it seems to be 3, but why?: (2n choose n) /( Sum(k=0..n-1, (2k choose k) ) or (2n choose n) /(1+Sum(k=0..n-1, (2k choose k) ) hmmm, Mathematica 4.0.1.0 doesn't seem to handle it. Wouter. email: wouter.meeussen(a)vandemoortele.com =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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RE: [math-fun] Dominos and non-attacking rooks.
by Andy Latto 19 Nov '02

19 Nov '02

> -----Original Message----- > From: ed pegg [mailto:ed@mathpuzzle.com] > Sent: Tuesday, November 19, 2002 12:07 PM > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Dominos and non-attacking rooks. > > > > Can a 6x6 board be covered with 15 dominoes and 6 > non-attacking rooks? > > I have a nice little proof that it can't be, but I'm > wondering if it is a new > problem. Here's my proof: I'm curious if it is the same as yours. Define a function f by f(n) = rank of the rook in file n. None of the rooks attack each other, so f is a permutation on S = {1,2,3,4,5,6}. So the number of elements of f that map an even element of S to an odd element is equal to the number that map an odd element to an even one. So the total number of elements of S that are mapped to an element of opposite parity by f is even. But this is just the number of rooks on white squares, which must be odd, since the number of rooks on white and black squares must be equal (since each domino covers a white and a black square). This proof works for any square chessboard of size 4k+2. Is this the same as your proof? No idea whether it's a new problem. But it's a nice one.

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[math-fun] Domino problem
by asimovd＠aol.com 19 Nov '02

19 Nov '02

I thank Richard Guy for his answer (negative) to the converse domino question I posed. Can someone please tell me where to find the "Inquisiive Problem Solver" ? I don't even know whether it's a magazine or website or what. Somehow I don't think I associate Sol LeWitt's name with the problem I'm trying to recall. It was posted by someone whose name I don't think I recognized then and don't recall now. (If you are that person, please forgive me.) But it had to do with 4x4 arrangements of black & white squares such that each row and each column contains 2 W and 2 B squares. The question was whether all such arrangements could be obtained in a certain way, which I don't remember, either. --Dan

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Re: [math-fun] Origins of a problem.
by asimovd＠aol.com 19 Nov '02

19 Nov '02

<< >From my site, a few years ago ... At the dinner, I met Bernardo Recaman Santos of Columbia. >> Just wondering -- is the Columbia University, or Colombia, the country? --Dan

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[math-fun] Lego Escher
by Helger Lipmaa 18 Nov '02

18 Nov '02

http://www.lipsons.pwp.blueyonder.co.uk/lego.htm Nice. Especially the "Ascending and Descending" is brilliant. Helger

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Re: [math-fun] Origins of a problem.
by asimovd＠aol.com 18 Nov '02

18 Nov '02

Ed Pegg Jr. writes: << Joe Kisenwether has upped the ante. Arrange the numbers 1-32 in a circle so that any two adjacent numbers sum to a square number. The answer is unique. >> That's an interesting circle. I'd like to see a picture of the circle of numbers 1-32 in numerical order, with 32 chords indicating adjacency in the Kisenwether ordering -- especially with that ordering being unique. --Dan

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[math-fun] Origins of a problem.
by Richard Guy 18 Nov '02

18 Nov '02

Can anyone provide any reference to the following problem? For which n is is possible to arrange the numbers from 1 to n in a chain so that the sums of adjacent links are always squares? Partial answer: n = 15, 16, 17: (16-)9-7-2-14-11-5-4-12-13-3-6-10-15-1-8(-17) Elwyn Berlekamp & I got it from Yoshiyuki Kotani in Edmonton last July, who was examining the corresponding problem with cubes in place of squares, and said that he had seen the squares problem somewhere. I think the answer for any power is `for all sufficiently large n'. We proposed the corresponding problem with Fibonacci numbers in place of squares and have the theorem: There is a chain with adjacent links adding to Fibonacci numbers just if n = 9, 11 or F_k or F_k - 1 where F_k is a Fibonacci number with k > 3. Before publishing formally, we'd like to trace the origins, if we can. R.

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[math-fun] limits, batting averages
by Richard Schroeppel 18 Nov '02

18 Nov '02

>(2n choose n) /( Sum(k=0..n-1, (2k choose k) ) This is 3, beacuse lim (2n choose n)/(2n-2 choose n-1) is 4. My favorite form of Simpson's paradox, using baseball batting averages, comes to mind. In the first half of the baseball season, A has a better batting average than B (.275 vs. .250); and the same for the second half of the season (.325 vs. .300). But overall, B can have a better average, due to hidden bias: (25+100)/(100+300) > (55+65)/(200+200). Err, this example is wrong. Bs average for the second part of the season is 100/300 = .333, bigger than As .325. Rich rcs(a)cs.arizona.edu

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Antwort: [math-fun] The two Germanys
by Hartmut.Holzwart 18 Nov '02

18 Nov '02

The correct signature would be 'Mit besten Gr"ussen', so neither version is completely correct. 'Mit freundlichen Gr"ussen' is also very common. Hope this helps, kind regards, Hartmut thane@best.com@mailman.xmission.com on 16.11.2002 06:33:20 Bitte antworten an math-fun(a)mailman.xmission.com Gesendet von: math-fun-admin(a)mailman.xmission.com An: math-fun(a)mailman.xmission.com Kopie: Thema: [math-fun] The two Germanys This is a multi-part message in MIME format. --------------------------------------------------------------------------- ---- These two signatures from recent math-fun postings challenge my passing familiarity with German plurals and/or the treatment of the dative case. Which is correct? Or perhaps both? Mit besten Gr"usse, Aviezri Fraenkel. Mit bestem Gr"usse, Richard K. Guy. Alles Gute! Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm --------------------------------------------------------------------------- ---- _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Nice knowing you all
by David Wilson 16 Nov '02

16 Nov '02

A goodbye note. Given my job search success over the past 7 months, it would be irresponsible for me to continue to market myself as a techie. A job search these days is tantamount to a sales job without commission, and apparently I have a pretty poor product to sell. If I am going to be a salesman, then dammit, I'm going to choose my own market and reap my own benefits. Therefore, I have decided to start my own business (my first foray into the business world will be selling gift baskets. No opinions on my probability of success will be tolerated. I will allow you to wish me good luck if you feel so inclined). I am also dropping mathematics as a hobby. My contribution is small, my time limited, my tools inadequate, and my interest waning. I have a moderate collection of mathematics and computer science titles gathering dust on my shelves, I plan to trash these. If there is interest I will post titles and make them available at shipping cost. I respectfully request to be dropped from the math-fun and seqfan mailing lists. As of late, I have not been able to follow the discussions with much interest, being preoccupied as I am with my family's future well-being. I have also not been fulfilling my role as OEIS editor in any substantive way, and request to be dropped from the editor list. It was an honor to be chosen, Neil. (I might still be induced to fix or extend a sequence now and then). Thank you all for your kindness and consideration over the years. I have been feeling quite bitter as of late, and you good friends do not deserve to hear my vitriol on a recurring basis. I feel I must turn my whole attention to other endeavors. I remain available at davidwwilson(a)attbi.com. ;-) - Dave Wilson

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