May 2003 - mathfun@mailman.xmission.com

FW: Re: [math-fun] distinct and square-free factorizations
by James Tanton 13 Aug '03

13 Aug '03

Let D_k (n) = number of partitions of n with any number repeated at most k times. (Thus D_1(n) = D(n).) O_k (n) = number of partitions of n into numbers none of which is a multiple of k. ( Thus O_2(n) = O(n).) Then D_k (n) = O_(k+1) (n) for all n. The argument using generating functions to establish D(n) = O(n) easily generalises to establish this, but the kids (ages 11 - 17) at the Boston Math Circle recently came up with a nice fun proof of this in terms of representing numbers base k. - JT > [Original Message] > From: R. William Gosper <rwg(a)spnet.com> > To: <math-fun(a)mailman.xmission.com> > Date: 5/18/2003 4:11:57 PM > Subject: Re: [math-fun] distinct and square-free factorizations > > mlb>In fact, recalling this, I'd hoped to retrieve some collective memories! > > acw>> Gosper showed me an absurdly intricate _analytic_ proof of this, and > >>bewailed the lack of a _combinatorial_ proof; > > Here are my memories. Unfortunately, the TeX sources appear to to have been > abandoned at Xerox PARC. Shortly after the announcement of the factorization > of 2^2^2^3+1, I sent a(n unrelated) letter to the the usual gang of maniacs > (Dick Askey, George Andrews, George Gasper (I think), Rich&Hilarie, etc.) > that began: "Dear Thai food lovers and Rich," > answering Hilarie's challenge to find a bijection between the *complements* > of the odd partitions and distinct partitions. I.e., pair the not-all-differents > with the not-all-odds. A very young Peter Weyhrauch and I came up with one, > and then a PARC coworker, Ted Kaehler, came up with a different one! I > then found a generalization of the theorem to residue classes mod k, of which > #(some evens) = #(some duplicates) was the case k=2. This led to a nonobvious > generating function identity. > It would be nice to locate a hardcopy of that letter. > --rwg > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ========================= James Tanton mathcircle(a)earthlink.net P.O. Box 2344 Acton MA 01720 978 635 8036 FIND *MATH CIRCLE* ON THE WEB AT: www.themathcircle.org

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[math-fun] Signatures of irrational numbers
by lkmitch＠att.net 05 Jun '03

05 Jun '03

Hi all, Lately, I've been playing with the signature of irrational numbers. My understanding, from what I've read on the web, is this: For a positive irrational number x, form the numbers y = i + j*x, where i and j are both positive integers. Since x is irrational, no 2 y values will be the same for different i and j. Arrange the ys by size, then the sequence of i values is a fractal sequence, and the signature of x. For x = phi ~ 1.618034, the first few entries are: i j y 1 1 2.618033989 2 1 3.618033989 1 2 4.236067977 3 1 4.618033989 2 2 5.236067977 4 1 5.618033989 1 3 5.854101966 3 2 6.236067977 5 1 6.618033989 And the signature begins: 1, 2, 1, 3, 2, 4, 1, 3, 5. If you strike the first occurence of every integer in the sequence, you get the original sequence back, which makes this a fractal sequence. My questions are: - What about the j sequence? From what I've seen experimentally, it seems to be a fractal sequence, too. Why is the signature the i sequence as opposed to the j sequence? What's known about the relation of the j sequence to the i sequence? - For a limited set of integers (both i and j run from 1 to 50), I plotted i vs. j, and the result was very interesting (I thought). You can find the picture here: http://www.fractalus.com/kerry/sigofphi.html The plot is one continuous zig-zag line which seems to never cross itself. But, the angle of the line changes slightly, causing some areas to bunch up and appear darker, and others to spread out and appear lighter. The overall effect is of a series of rectangles drawn in different shades of gray. Can anyone point me to other work that has been done on this? Thanks, Kerry Mitchell -- lkmitch(a)att.net www.fractalus.com/kerry

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Re: [math-fun] problem with prime gap proof
by asimovd＠aol.com 31 May '03

31 May '03

Hi, Thane. I haven't forgotten your suggestion about getting together, and it happens that Marc Le Brun (mlb) who lives in Marin is also interested. He tends to get to Palo Alto where his mother is convalescing, temporarily. In the unlikely event he can make it tomorrow around noon, could you also make it? Otherwise, how does next Saturday, June 7 look? --Dan

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[math-fun] Interesting use of number square
by David Wilson 31 May '03

31 May '03

I've never seen this trick before. What properties of the subsquares make this trick work? Would it be possible to perform this trick with a square that has distinct entries? Entries 1 through 64? http://www.mdani.demon.co.uk/stunt/may98s1.htm

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[math-fun] George Bush speaks on Mathematics (fwd)
by Richard Guy 30 May '03

30 May '03

---------- Forwarded message ---------- Date: Wed, 28 May 2003 13:54:35 -0600 From: Ernest Enns <enns(a)math.ucalgary.ca> To: dept <dept(a)math.ucalgary.ca> Subject: George Bush speaks on Mathematics Apparently a teacher has been arrested in the UK in possession of compasses, protractor, and straight edge. It is claimed he is a member of the Al Gebra movement bearing weapons of math instruction.

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Re: [math-fun] [seqfan] Signatures of irrational numbers
by asimovd＠aol.com 30 May '03

30 May '03

Marc, I just disovered that AOL has been hiding the name accompanying the e-address under a little button called "Details". By clicking on that and reading through about 30 lines of tiny type, I found that I've been misspelling your first name. I hope you will accept my apologies. And I hope I will hear from you. --Dan

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Re: [math-fun] flame oops
by asimovd＠aol.com 30 May '03

30 May '03

Gene writes: << There are also wonderful resources hidden away in books; the problem is identifying the few good ones among all the trash. Something we could do to help the bright students stuck in lousy schools is to prepare reading lists of good books and post them on the internet. The lists should include politically incorrect material, not only because much that is scientifically correct is politically incorrect, but also because then the schools might then try to have the lists banned, and that would generate publicity that no amount of money could buy. >> I'm totally in favor of trying to improve the education of T.C. MITS (the common (hu)man in the street) and like the direction Gene's thought are running in. I am strongly against trying to maintain political correctness merely for the sake of not offending anyone. But let us suppose that, repeatedly, unbiased and well-performed studies of individuals of group A (e.g., an ethnicity or gender) showed them to be statistically less innately able to do task X than individuals not in group A. Perhaps that would be interesting information, but it strikes me as something that could far more likely do more harm than good if widely publicized. In particular, it would prejudice the easily-prejudiced (and that includes a lot of people) against anyone of group A, despite the fact that a very large number of group A may be vastly well-qualified at task X. * * * And I'm against "demonizing" plutonium or anything else, yet it is a very dangerous substance (I've heard convincing claims that a millionth of an ounce of Pu in one's lungs is virtually certain to cause lung cancer within 20 years). And as of about 1980, there were about 20 pounds of the stuff unaccounted for from our stockpiles of nuclear material. Now, with the breakup of the U.S.S.R. and Iraq behind us, who knows how much more is also unaccounted for. Without "demonizing" anything, I am realistically concerned that some rogue individuals or groups would gladly accumulate dangerous materials and use them to harm innocent people. And even non-rogue individuals are susceptible to bribes, threats, and simple mistakes. --Dan

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Re: [math-fun] [seqfan] Signatures of irrational numbers
by Henry Baker 29 May '03

29 May '03

Do these questions bear on the sequence generated by Broadway as it cuts across the rectangular grid of Manhattan? (Assuming that the tangent of the angle of Broadway is irrational...) At 01:59 PM 5/29/03 -0700, Marc LeBrun wrote: > >=Kerry Mitchell <lkmitch(a)att.net> > > [...] For a positive irrational number x, > > form the numbers y = i + j*x, where i and j are both positive integers. > > Since x is irrational, no 2 y values will be the same for different i and j. > > Arrange the ys by size... > > What about the j sequence? ... Why is the signature the i sequence as opposed to > > the j sequence? > >Funny you should mention it, since I just gave a talk that touched on this, >but with a twist... > >Instead of throwing away the js try mapping the pairs 1-1 into integers >(i,j)-->n. Many different pair mapping patterns, such as the common >"anti-diagonal scan", can be used. > >My current favorite mapping is n = A(i) + 2 A(j), where A is the Moser-de >Bruijn sequence, Sloane's A000695 > > 0 1 4 5 16 17 20 21 64 65... > >(which I notate 2[n]4, meaning "replace 2 with 4 in the binary expansion of >n"). > >This particular mapping just "interleaves the bits" of i and j. If you >visit the pairs in n-order it too traces a nice connected fractal path. > >Whatever mapping you choose, you can then sort the integers n via the >ordering of the y[n] values. The resulting sequence, a permutation of the >integers, is a "signature" that's characteristic of x. > >For example with the above mapping and x = sqrt 2 I get > > 0 1 2 4 3 8 5 6 9 16... > >(not yet in OEIS, sorry). > >I too would be very interested in an efficient algorithm to generate the >i+jx in order. (I currently just kludgily generate "enough" pairs to make >sure I don't miss any sequence elements, then sort--yuck). > >Anyway, besides giving signature sequences, this information-conserving >pair mapping approach also enables interesting abstract-arithmetic games >for quadratic x: just write the "numbrals" [n] for the objects y[n]. For >any given mapping scheme the addition table is independent of x while >multiplication is characteristic. This generalizes to algebraic x by >extending pairs to tuples, and so on...

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RE: [math-fun] Signatures of irrational numbers
by Kimberling, Clark 29 May '03

29 May '03

Hi mathfun - Lately, a student, JB, and I have been looking at the ranks of the numbers j*x when the numbers i+j*x, for any positive real x, are arranged in order. These rank sequences are closely related to signature sequences. If x is rational, then there are repeated terms, so we extend the notion of rank to "minrank" and "maxrank". For example, if you order the numbers i+j*2 for i>=0, j>=0, then minrank(2n) means the rank of the first appearance of 2n and maxrank(2n) means the rank of the last appearance of 2n. For x=2, maxrank(2n) gives squares; more generally, i*mj for integral m gives figurate numbers for maxrank and central polygonal numbers for minrank. Returning to i+j*x for rational x, limits provide insights in case x is irrational, and Farey fractions play a central role. Linked to rank sequences are notions of 1st partial complement of a sequence, 2nd partial complement, and so on, with applications to dispersions of sequences. Loosely speaking, if A is the dispersion of the complement of a rank sequence r of x>1, then (1) r is column 1 of A; (2) the 1st partial complement of r is, after the 1st term, row 1 of A; (3) row 1 of A is the rank sequence of 1/x. If the sequence r in column 1 is strictly increasing (but not necessarily a rank sequence), let t(r) denote the sequence in row 1; we expect to prove that t is involutory if and only if r is a rank sequence, and that the sequence r, t(t(r)), t(t(t(t(r)))), ... converges to a rank sequence. A preprint should be ready soon. Now, JB and I need enlightenment on origins of terminology. Was Lancelot Hogben first to give names to the sequences which he called central polygonal numbers? The first three of these sequences are Central polygonal numbers, A002061: 1,3,7,13,21,... Central triangular numbers, A005448: 1,4,10,19,31,... Central square numbers, A001844: 1,5,13,25,41,... Next, as far as I know, the terms "signature sequence" and "fractal sequence" (for an infinitive sequence that contains itself as a proper subsequence) first appeared in 1995. It would seems likely that someone considered signature sequences (or rank sequences) before 1995, perhaps under some other name. Any insights about these matters? As a final thought, consider the ranks of the numbers 3^j in the ordered sequence of the numbers (2^i)*(3^j). (What's x?) Best regards, Clark Kimberling -----Original Message----- From: lkmitch(a)att.net [mailto:lkmitch@att.net] Sent: Wednesday, May 28, 2003 1:57 PM To: math-fun Subject: [math-fun] Signatures of irrational numbers Hi all, Lately, I've been playing with the signature of irrational numbers. My understanding, from what I've read on the web, is this: For a positive irrational number x, form the numbers y = i + j*x, where i and j are both positive integers. Since x is irrational, no 2 y values will be the same for different i and j. Arrange the ys by size, then the sequence of i values is a fractal sequence, and the signature of x. For x = phi ~ 1.618034, the first few entries are: i j y 1 1 2.618033989 2 1 3.618033989 1 2 4.236067977 3 1 4.618033989 2 2 5.236067977 4 1 5.618033989 1 3 5.854101966 3 2 6.236067977 5 1 6.618033989 And the signature begins: 1, 2, 1, 3, 2, 4, 1, 3, 5. If you strike the first occurence of every integer in the sequence, you get the original sequence back, which makes this a fractal sequence. My questions are: - What about the j sequence? From what I've seen experimentally, it seems to be a fractal sequence, too. Why is the signature the i sequence as opposed to the j sequence? What's known about the relation of the j sequence to the i sequence? - For a limited set of integers (both i and j run from 1 to 50), I plotted i vs. j, and the result was very interesting (I thought). You can find the picture here: http://www.fractalus.com/kerry/sigofphi.html The plot is one continuous zig-zag line which seems to never cross itself. But, the angle of the line changes slightly, causing some areas to bunch up and appear darker, and others to spread out and appear lighter. The overall effect is of a series of rectangles drawn in different shades of gray. Can anyone point me to other work that has been done on this? Thanks, Kerry Mitchell -- lkmitch(a)att.net www.fractalus.com/kerry _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] knight problems from Elkies and Stanley
by Andy Latto 29 May '03

29 May '03

> -----Original Message----- > From: Michael B Greenwald [mailto:mbgreen@central.cis.upenn.edu] > Sent: Thursday, May 29, 2003 12:25 PM > To: math-fun > Cc: Andy Latto > Subject: Re: [math-fun] knight problems from Elkies and Stanley > > > Thu, 29 May 2003 10:39:44 -0400 > Andy Latto <Andy.Latto(a)gensym.com> > > Like many proofs by contradiction, I find that this proof that > there is no knight's tour on a 4xn chessboard hides a bit > of what's really going on. Here's a more direct version of the > same proof that I find more perspicuous. > > Call the squares in the top and bottom rows of a 4xn board > the "extreme" > squares. > > Since you can't get in one knight's move from one extreme > square to another, the maximum density > of extreme squares in a knight's tour is 50%, and this is > only attained if every other square in the tour is extreme. > > No; if the first and last moves of the tour are extreme, then you > can have two non-extreme squares in a row somewhere in the > middle. Those two moves allow you to "switch parity". If the > two consecutive non-extreme squares are in the center of the > tour, then you can (in theory) have a tour that first covers the > black extreme squares, then switches parity and does the mirror > image tour to cover the white extreme squares (and black > non-extreme). Right. Both the original proof and my reworking of it only prove the (true) statement that there is no closed knight's tour on a 4xn board, and not the (false) statement that there is no open knight's tour. An analysis of either proof shows that if there is an open knight's tour on a 4xn board, then exactly one link in the the tour connects two non-extreme squares, and that this must be the middle link of the tour. Andy Latto andy.latto(a)pobox.com

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