[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@att.net www.fractalus.com/kerry
--- lkmitch@att.net wrote:
... 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. ...
What is a fractal sequence? The sequence 1,1,1,1,... also possesses the above mentioned property, but would not reasonably be called fractal. __________________________________ Do you Yahoo!? Yahoo! Calendar - Free online calendar with sync to Outlook(TM). http://calendar.yahoo.com
=Kerry Mitchell <lkmitch@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...
Here's an excerpt from another mailing list http://groups.yahoo.com/group/mathpuzzle/
From "Jean-Charles Meyrignac" <euler@free.fr> 20 knights, each attacks 3 others, on a 7x7 board. 32 knights, each attacks 0 others, on a 8x8 board. 32 knights, each attacks 1 others, on a 8x8 board. 32 knights, each attacks 2 others, on a 8x8 board. (single path possible) 32 knights, each attacks 3 others, on a 8x8 board. (the solution is unique). 36 knights, each attacks 3 others, on a 9x9 board.
Jean runs the site http://euler.free.fr/ I haven't noticed the 0, 1, 2, 3 together. I think I've usually just seen the 0 solution (I remember it in Gardner's works.) --Ed Pegg Jr, www.mathpuzzle.com
participants (4)
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Ed Pegg Jr -
Eugene Salamin -
lkmitch@att.net -
Marc LeBrun