December 2005 - mathfun@mailman.xmission.com

[math-fun] G4G7
by Michael Kleber 08 Dec '05

08 Dec '05

This being my first time, I'm still a bit giddy from the fact that I received an invitation to the Gathering for Gardner next spring. Wow. I mean, more cool people than you can shake a stick with fractal branching at! Picture me lying awake nights trying to think of sufficiently interesting things to bring... (Er, I'm not entirely clear on the nondisclosure situation: it seems one is not supposed to talk about the G4G publicly beforehand, at least to some extent. I suppose it's in the nature of these you-don't-apply-we-just-invite-you events. So to be clear, I'm explicitly invoking the math-fun privacy agreement: this is private communication, do not forward or quote without permission, stuff like that.) Anyway, certainly some funsters are past participants, and the darn thing is expensive. Anyone out there planning to go and interesting in sharing a hotel room? --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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Re: [math-fun] Worst packing among all planar convex sets
by dasimov＠earthlink.net 08 Dec '05

08 Dec '05

Gareth quotes from MathWorld: << Surprisingly, the circular disk is not the least economical region for packing the plane. The "worst" packing shape is not known, but among centrally symmetric plane regions, the conjectured candidate is the so-called smoothed octagon. >> (I'll assume this is meant to be among all *convex* centrally symmetric plane regions. Otherwise, we could use a huge figure-eight painted with a thin brush, with symmetrical, thinner, channels removed at the two most-distant points. Such can have arbitrarily low best packing density.) OK, but what about the problem where the only restrictions on the planar set is that it's connected and convex? Does anyone know the sup packing density for each of the regular polygons? How about just the pentagon and heptagon? Is this number monotone decreasing from the hexagon onward? --Dan

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Re: [math-fun] Sum of last ten digits
by mbeeler＠csc.com 08 Dec '05

08 Dec '05

A better program filled in missing values of loop length, caught 2 errors in my manual work (for s,b = 5,4 and 7,3), and extended the results to s (number of digits) and b (radix) = 16. (All of these start with s-1 zeros followed by a one.) s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9 s=10 b=2 3 3 1 3 2 1 1 1 5 b=3 4 4 3 7 5 4 14 16 20 b=4 3 3 4 42 13 36 1 5 58 b=5 5 5 1 43 4 46 5 10 34 b=6 3 12 15 110 31 154 406 5 197 b=7 4 9 6 34 13 33 26 1440 104 b=8 6 6 45 84 41 249 171 6458 2801 b=9 9 30 14 52 16 74 20 14654 24 b=10 8 4 50 171 14 461 78 12203 312 b=11 4 14 30 10 36 332 666 16294 4686 b=12 5 10 26 116 39 603 120 6750 16105 b=13 7 9 61 57 9 263 130 13536 312 b=14 8 30 41 70 83 466 84 20008 16578 b=15 11 15 8 209 31 249 1010 31320 16806 b=16 9 12 54 224 25 666 312 19107 26294 s=11 s=12 s=13 s=14 s=15 s=16 b=2 6 5 7 7 5 6 b=3 1 1 11 6 6 8 b=4 5 6 7 161 8 70 b=5 6 12 374 12 409 7 b=6 101 396 7 937 311 968 b=7 36 249 753 235 1478 794 b=8 687 88 676 35 3129 2533 b=9 2211 28 28 32 32 1093 b=10 318880 2184 57725 5804 1401 9722 b=11 220110 1456 3666650 99291 8 69188 b=12 9076 354312 576360 161050 1044670 26256 b=13 188448 728 23487352 1456 21972045 720 b=14 1541063 9520 226440 10248 big big b=15 1396913 2394 3281355 1098056 10309491 13680 b=16 1013143 1456 28640 852852 24039416 976560 "big" means no loop had been detected when I stopped the program after 44 million terms, so tail length plus twice the loop length is > 44000000. (Thanks to Bill Gosper for the loop detector in HAKMEM item 132.) We don't know the tail length, so loop length can still be anything. A new large tail-to-loop ratio, for s=15, b=11, is 33668/8 = 4208.5. The loop is 99, 99, 99, 99, 104, 99, 99, a3 (base 11; a=ten). The loop lengths that occur above look rather random, getting sparser as they get larger, except for 5 values: loop length = 171 for s,b = 5,10 and 8,8 loop length = 249 for s,b = 7,8 and 7,15 and 12,7 loop length = 312 for s,b = 8,16 and 10,10 and 10,13 loop length = 666 for s,b = 7,16 and 8,11 loop length = 1456 for s,b = 12,11 and 12,16 and 14,13 Are these replications mere coincidence? Or is there a simple mapping of the structure between loops of the same length? -- Mike

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Re : [math-fun] Sum of last ten digits
by Alexandre Wajnberg 08 Dec '05

08 Dec '05

OK, I did'nt understood you made same seqs with other beginnings than ours... So I agree, there are a lot of loops around there... :-) Thanx Alexandre De : Graeme McRae <g_m(a)mcraefamily.com> Date : Thu, 8 Dec 2005 10:28:11 -0800 À : Alexandre Wajnberg <alexandre.wajnberg(a)skynet.be> Objet : Re: [math-fun] Sum of last ten digits Just taking the first one, if the sequence begins 0, 6, 1, 8, 7, 8, 6, 6, 3, 0, then the sequence through the repetition (loop length 8) is 0, 6, 1, 8, 7, 8, 6, 6, 3, 0, 45, 48, 51, 42, 36, 42, 39, 39, 45, 48, 51, 51, 45, 42, 39, 39, 45, 48, 51, 51, 45. --Graeme > > ----- Original Message ----- > > > De : Graeme McRae <g_m(a)mcraefamily.com> > Date : Thu, 8 Dec 2005 07:47:23 -0800 > > Searching randomly, I found that: > A loop of length 8 is possible, starting at 0, 6, 1, 8, 7, 8, 6, 6, 3, 0. > A loop of length 24 is possible, starting at 0, 4, 2, 1, 9, 7, 1, 7, 7, 4. > A loop of length 26 is possible, starting at 2, 3, 2, 7, 0, 9, 8, 7, 8, 4. > A loop of length 78 is possible, starting at 2, 2, 5, 2, 6, 0, 3, 8, 5, 7. > --Graeme > >> >> The nine sequences (beginning with ...0 1; ...2; 9) involve digits in their >> construction and our loops are made of terms (numbers). >> >> I tried your digit-propositions without success, no "digit loops" hidden >> inside. >> I did this way: >> -pasting the nine complete sequences (the beginning + the first loop + the >> ten first terms of the second loop) on a new page of my "word" program; >> -compacting them (removing all the "blanks" and the ",") >> -compacting the same way your ten digits propositions; >> -searching, using the "find" function of my "word" program. >> I found only my control. >> So I don't understand. May be I miss a point? >> Are you shure of your finding? Or of your sequences? >> I send you separately our complete sequences. >> Thank you for your interest. >> >> Alexandre >>

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RE: [math-fun] Sum of last ten digits
by Alexandre Wajnberg 08 Dec '05

08 Dec '05

De : Graeme McRae <g_m(a)mcraefamily.com> Date : Thu, 8 Dec 2005 07:47:23 -0800 Searching randomly, I found that: A loop of length 8 is possible, starting at 0, 6, 1, 8, 7, 8, 6, 6, 3, 0. A loop of length 24 is possible, starting at 0, 4, 2, 1, 9, 7, 1, 7, 7, 4. A loop of length 26 is possible, starting at 2, 3, 2, 7, 0, 9, 8, 7, 8, 4. A loop of length 78 is possible, starting at 2, 2, 5, 2, 6, 0, 3, 8, 5, 7. --Graeme > > > > The nine sequences (beginning with ...0 1; ...2; 9) involve digits in their > construction and our loops are made of terms (numbers). > > I tried your digit-propositions without success, no "digit loops" hidden > inside. > I did this way: > -pasting the nine complete sequences (the beginning + the first loop + the ten > first terms of the second loop) on a new page of my "word" program; > -compacting them (removing all the "blanks" and the ",") > -compacting the same way your ten digits propositions; > -searching, using the "find" function of my "word" program. > I found only my control. > So I don't understand. May be I miss a point? > Are you shure of your finding? Or of your sequences? > I send you separately our complete sequences. > Thank you for your interest. > > Alexandre >

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Re : [math-fun] Sum of last ten digits
by Alexandre Wajnberg 08 Dec '05

08 Dec '05

Hi, Neil said: %S A112402 0,0,0,0,0,0,0,0,0,1,1,2,4 > %N A112402 Next term is the sum of the last 10 digits in the sequence. > %C A112402 Digits, not terms! > %C A112402 There are only 10^10 possibilities for the last 10 digits, so the > sequence must eventually cycle. Hans: In fact, terms 19-23 (44, 40, 37, 42, 38) are repeated by terms 331-335 already. Me: It seems the shortest loop of Éric Angelini's < sum of last ten digits > can be found beginning with 9, and has a lenght of 1 term: 0 0 0 0 0 0 0 0 0 9 9 18 27 36 45 54 45 45 45 45 45 45 45 45 45 It should be interesting to taste the eight other (beginning with 2, 3 ...8) Thanks to Gilles Sadowski's computing, the seven other loops are, beginning with: 1: loop at n(19)=44, 312 terms (Hans) 2: loop at n(17)=38, 312 terms 3: loop at n(21)=36, 104 terms 4: loop at n(16)=38, 312 terms 5: loop at n(28)=42, 312 terms 6: loop at n(28)=36, 104 terms 7: loop at n(18)=44, 312 terms 8: loop at n(32)=37, 312 terms 9: loop at n(17)=45, 1 term (with offset 1,) I'm submitting them. I don't know if this is of interest: -only loops by lenght of 104 and 312 (and 1); -a pattern: [312 312 104]; -loop 4 == loop 2; loop 6 = loop 3 but beginning at position 47 of loop 3; What's happening beginning with 10, 11, 12... ? Alexandre Exemple: 2: loop at n(17)=38, 312 terms 0 0 0 0 0 0 0 0 0 2 2 4 8 16 23 28 38 45 42 41 41 36 34 32 31 30 28 29 33 34 37 44 42 37 41 39 41 38 43 40 39 39 46 45 47 54 51 45 44 43 39 42 42 39 43 43 38 43 44 40 37 40 33 32 29 36 35 39 45 49 51 48 52 47 49 49 56 55 58 60 53 48 49 52 46 50 47 46 43 43 40 39 40 34 34 34 37 35 39 44 45 47 48 52 47 50 46 45 42 41 35 38 39 42 42 43 42 37 35 37 41 39 45 44 44 42 43 38 40 36 37 41 39 40 40 35 33 34 29 36 41 38 43 43 39 42 43 39 44 45 42 42 41 34 33 30 27 30 28 31 29 37 38 46 46 52 48 50 44 42 38 42 36 40 36 39 40 38 40 40 35 31 31 24 26 30 25 28 34 35 35 40 37 37 40 36 37 43 40 34 37 38 39 44 48 53 51 46 44 44 40 36 39 41 38 41 42 39 39 46 45 49 56 55 53 51 48 47 47 48 52 53 49 51 46 44 45 46 43 44 42 40 35 33 32 29 34 37 39 45 49 51 50 45 42 39 38 43 45 45 48 48 49 55 56 58 59 61 55 55 54 50 41 39 41 36 36 40 39 39 46 47 49 58 59 61 58 60 53 48 46 49 49 56 59 61 58 58 58 60 52 52 46 43 37 41 39 44 42 41 36 40 32 29 34 36 36 41 41 35 36 36 36 40 39 43 41 37 38 45 42 41 41 36

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[math-fun] Worst packing among all planar convex sets
by dasimov＠earthlink.net 08 Dec '05

08 Dec '05

The recent discussion re packing of various equiareal ellipses reminds me of a question discussed in this venue some years ago -- but I don't recall the upshot: Which bounded planar convex shape W, packed as well as possible, has the lowest density? (I'm not sure that there's always a packing that realizes the sup over all packings of a given planar shape, so just in case there isn't, let's say we're looking for the shape W whose sup packing density is least over all shapes. And it seems quite possible that there could be many W's tied for this dubious honor. So the real question is, What are all of these? (We want a shape whose sup-best packing density is least among all bdd convex planar shapes is least. Since the space of bdd convex shapes is closed, it seems there will always be an actual such shape that realizes the inf of such sup-best densities.) ------------------------------------------------------------------ As a start, since the sups of a shape K and any affine transformation T applied to it are the same, let's only consider shapes which cannot be made "rounder" by any affine tsfmn. Among many ways to define this, perhaps an affine tsfmn T that makes T(K)'s 2D correlation matrix equal to the identity (when T(K) is viewed as a uniform probability density in R^2) would be the way to go. (What is the word for this matrix in physics?) Now, there may be a natural 2nd-order measure of roundness, and I'd suspect that the worst packer among these "uncorrelated" shapes would be among those whose "2nd-order roundness" is minimum. One obvious 2nd-order measure of roundness is the standard deviation of the radius, as measured from the center of gravity as a function of angle. (To normalize such measures, we should probably assume the average radius -- as a function of angle from the c.o.g. -- is 1.) (On the other hand, such an uncorrelated shape with highest s.d. may well be an equilateral triangle -- which doesn't pack that badly at all.) --Dan

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[math-fun] fractal continued fractions
by James Propp 08 Dec '05

08 Dec '05

I passed Ed Pegg's email about fractal continued fractions to Henry Cohn, who knows something about the topic. Henry replied as follows to Ed (cc-ing me), and he has given me permission to post his email to Ed: ************************************************************************ I don't currently subscribe to math-fun, but Jim Propp forwarded me your e-mail message about fractal continued fractions. >Does anyone know of other numbers with a fractal sequence for their >continued fraction? For any sequence of positive integers a_n such that a_n^2 divides a_(n+1), the sum of 1/a_n will have such a continued fraction (the Liouville number is a special case). This was discovered independently by Kmosek and Shallit in the 1970's, and as far as I know had never been observed before even for the Liouville number (but there's so much related literature that nobody could search exhaustively). Nine years ago I wrote a paper (available at http://front.math.ucdavis.edu/math.NT/0008221) which generalizes these constructions. My favorite examples are the sum of 1/T_(4^n)(2), where T_k(x) = cos(k*arccos(x)) is the k-th Chebyshev polynomial, and the sum of 1/2^(2^n) + 1/8^(8^n). In each case there is self-similarity but the pattern is a little longer than for Liouville's constant. Henry ************************************************************************ Jim Propp

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Re: [math-fun] A few self-descriptive fractal sequences
by dasimov＠earthlink.net 08 Dec '05

08 Dec '05

Kerry writes: << . . . [several interesting examples of "fractal" sequences, and] . . . Since the exact manner in which a sequence can contain copies of itself varies, I don't know that there is an exact definition of a fractal sequence. >> Hmm. Suppose we take a random sequence of 0's and 1's. With probability 1, every initial string of this sequence is later repeated, nay, repeated infinitely often. Does that mean that except for a set of measure zero (in the product space {0,1}^N endowed with the product measure {1/2,1/2}^N), every sequence is fractal? (N is your favorite definition of the natural numbers.) --Dan

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[math-fun] Re: ellipse packing
by Dean Hickerson 08 Dec '05

08 Dec '05

Richard Schroeppel asked: > Puzzle: Choose some specific ellipse, say a=2 b=1. > Find any packing that has a higher packing density > than circles in the plane (the simple hex packing). And I generalized the question to: > Suppose we're allowed to use more than one shape of ellipse, as long as > all of them have the same area. Can we get a higher density than the > hex packing of circles? I asked Sherman Stein, who asked Don Chakerian, who found a reference that gives negative answers to both questions: > I found in Fejes-Toth's Lagerungen in der Ebene, auf der Kugel, und im > Raum, a proof that the density of the closest packing of plane convex > sets all congruent to a given set K is at most area(K)/area(H), where H > is a hexagon of least area circumscribed about K. Then on p.88 he > remarks that his proof extends to equiareal affine images of a given > convex set, from which one deduces that ellipses of equal area cannot be > packed denser than congruent circular disks. I don't have access to a good math library, so I won't be able to look for the book. Dean Hickerson dean(a)math.ucdavis.edu

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