August 2005 - mathfun@mailman.xmission.com

[math-fun] Continued fraction expansion
by andrew＠DMS.UMontreal.CA 09 Aug '05

09 Aug '05

Dear Mr Gosper (I hope that it is indeed you who receives this) In an email correspondence with Hendrik Lenstra we proved, by accident, something amusing about continued fractions that I do not remember seeing anywhere, but maybe this is well-known to you? (I think you are one of the few remaining experts on continued fractions) If so can you please point us in the right direction. The result is as follows: Let a and b be pairwise coprime positive integers. The sign of the least residue, in absolute value, of a^(-1) mod b EQUALS (-1)^(length of the continued fraction of a/b) Proving these two are equal is trivial by induction, so there is nothing deep here, just amusing. Thanks for any references you can give Andrew Granville

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Hartmut Holzwart/IS-F1 A1/Allianz-Leben ist außer Haus. : [math-fun] A000701 generating function
by hartmut.holzwart＠allianz.de 08 Aug '05

08 Aug '05

Ich werde ab 06.08.2005 nicht im Büro sein. Ich kehre zurück am 01.09.2005. Bitte wenden Sie sich an Herrn Dr. Stefan Krause (allgemein, IS-C4 PM), Herrn Frank Tiemann (SOX, SCD) , Frau Bettina Koch (SOX) Ursprüngliches Thema: [math-fun] A000701 generating function

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[math-fun] clarification, trapezohedra
by R. William Gosper 08 Aug '05

08 Aug '05

I said, Recall the odd (mod 3) squarefree sequence d(k) = 0 1 2 1 0 1 2 0 2 1 0 2 0 1 2 1 0 1 2 0 1 0 2 1 2 0 2 1 0 ... defined as the left-to-right alternating sum of the nonzero trits of k. In *balanced* ternary. (Thanks, Neil!) Theorem: if any trapezohedron be scaled to inscribe in a sphere, its faces are "right kites", i.e., they have two square vertices. Mathworld's trapezohedron article just lists areas and volumes of a few special cases (scaled inconsistently). For an n-trapezohedron inscribed in a spheroid with polar radius a, equatorial radius b, 2 2 2 2 2 2 8 c (b c - b + 2 a ) 2 sqrt(c (a c + b )) s = sqrt(4 a - ----------------------), S = ---------------------, 2 c + 1 (c + 1) 2 2 2 %pi 8 b sqrt(b c + a c) sin(---) n a b c n inradius = ------------------, area = ---------------------------------, 2 2 2 2 sqrt(b c + a c) (c + 1) 2 %pi 8 a b c sin(---) n n 2 b sqrt(c) volume = -------------------, tradius = -----------, 2 c + 1 3 (c + 1) 2 2 2 2 2 2 (c + 1) ((2 - 2 c) S - s ) (c + 1) ((c - 1) S - c s ) sqrt(---------------------------) sqrt(-----------------------------) (1 - c) (2 c - 1) (c - 1) c (2 c - 1) a = ---------------------------------, b = -----------------------------------, 2 2 2 2 %pi 2 b cos(-----) + a c n angles = [acos(--------------------), 2 2 a c + b 2 2 (b - a ) sqrt(1 - c) sqrt(c) acos(-----------------------------------------), 2 2 2 2 sqrt(a c + b ) sqrt(2 b c + a (1 - c)) 2 2 2 2 b c - a (1 - c) %pi - acos(--------------------)], 2 2 2 b c + a (1 - c) 2 3/2 b c --------------------------- + a 2 2 sqrt(a c + b ) + a sqrt(c) apex_solid_angle = 2 acos(-------------------------------) n, 2 2 sqrt(b c + a ) 2 2 %pi 2 girdpt_solid_angle = acos(((a cos(-----) + b c) n 4 2 %pi 4 2 2 2 (a cos(-----) + b c - 2 a b c (2 c + 1)) n 2 2 2 3/2 2 2 + 4 a (b - a ) sqrt(1 - c) c (c + 1) sqrt(a c + b ) 2 2 2 2 3 sqrt(2 b c + a (1 - c)))/(b c + a ) ), c + 1 natural_a = --------------------------------------------, %pi 2 sqrt(2) sqrt(1 - c) sqrt(2 c + 1) sin(---) n c + 1 natural_b = ------------------, %pi 4 sqrt(c) sin(---) n where s:=short side, S:=long side, c:=cos(%pi/n), tradius := the "tubular" or girth radius, and natural_a and natural_b are the spheroid radii of the trapezohedron which is the dual of the antiprism with unit edges. The face angle order is [apex, upper girdle, lower girdle]. Note vanishing cosine of upper girdle when a=b. Maybe somebody can simplify the girdle vertex solid angle. When a=b, it boils all the way down to acos(1 - 2 c). (This all started when I added prisms, pyramids, etc. to Macsyma's rather meager PLOT_3D_FIGURE for a geomety lesson.) --rwg PS, while the cube is the only regularhedron whose vertex solid angle is a rational fraction (1/8) of the sphere, the tetrahedral and octahedral vertex solid angles are 2 atan(sqrt(2)/5) and 4 atan(sqrt(2)/4). Since 3 atan(sqrt(2)/4) + 2 atan(sqrt(2)/5) = pi/2, six octahedra and eight tetrahdra come together at each vertex in the octahedral-tetrahedral tessellation of 3-space. This is a (rather lame) example of a pi arctan formula a la Machin's, but with sqrt(rational) series multipliers, which are considered kosher in the game of maximizing digits/term of hypergeometric pi series. So, with sqrts, (but not surds), can we beat Machin?

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RE: [math-fun] octonians - cubic extension?
by Richard Schroeppel 04 Aug '05

04 Aug '05

me> Is there an interesting cubic extension of the octonians? various> Huh? dan> Please, folks: it's spelled "octonions". Obviously, by analogy with the Realions, Complexions, and Quaternions. We discussed the doubling process(es) before, giving systems with dimensiyn 8, 16, 32, etc. Is it possible to take a 3-step, going from octoniens to a system of dimensiun 24? There's be 24 unit elements, one of them the Real(ion)s, and various subsets isomorphic to the Cs, Qs, Octs. I'm imagining some relationship to the Monster group. Evidence against, is the absense of a degree 3 extensiin of R of this kind. Is this concluswve? Rich rcs(a)cs.arizona.edu

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[math-fun] Busy Beaver -- news
by Richard Schroeppel 04 Aug '05

04 Aug '05

There's been some recent news on the Busy Beaver problem. [Consider all Turing Machines with C symbols and S states WHICH HALT when started on a blank tape. The problem is to find the longest running machine, and the one which leaves the most non-blank stuff on the tape. These might be different. The general problem is of course unsolvable. C=2 S=4 is known. Proposed by Rado in the 1960s.] The current record for 2 symbols & 6 states is 10^1730 steps and 10^865 non-blank symbols. The "magic" is that the tape has a state variable (in unary?), and a quadratic function like 2x^2+3 is applied to it, halting when some modular condition (like x=3 mod 5) is reached. The record 2&5 machine is similar, but the iterated function is only linear, such as 8x+3. The step count is 47 million, and it prints 4098 non-blanks. Marxen has posted his 1990 paper with these results. The record for 2 symbols & 5 states has stood for 15 years, but there must be a small residuum of unknown-outcome machines that run for a long time and seem to be chaotic. It's not clear if there's been any progress here, even toward reducing the number of unknowns. The record 3&3 machines produce 36K non-blanks, and (a different machine) run 544M steps (Myron Souris). See Heiner Marxen's page for details. http://www.drb.insel.de/~heiner/BB/ Perhaps the 2&7 winner will iterate 2^x+3. But if the 2&5s can't be polished off, then the larger ones are beyond hope. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] octonians - cubic extension?
by dasimov＠earthlink.net 04 Aug '05

04 Aug '05

Jim wrote: >> Rich wrote: >Is there an interesting cubic extension of the octonians? I'm not sure what Rich means by a "cubic" extension, but there's an extension of the octonians called the sedenions, and some folks go even farther. >> Please, folks: it's spelled "octonions". (I care only because they're sacred!) Nor do I know what Rich means by "a cubic extension". (The abstract Jim posted was, um, rather abstract.) But indeed there is a doubling procedure ("Cayley-Dickson") that, repeatedly applied, starting with the reals, gives the complexes, then the quaternions, then the octonions, and can be continued indefinitely to give a real algebra of dimension 2^n, for each n. (A *real algebra* of dimension k is just a bilinear function R^k x R^k -> R.) (Cf. http://planetmath.org/encyclopedia/DoublingProcess.html. Note they use "non-associative algebra" to mean a "real algebra" as above; it isn't *required* to be associative, but is allowed to be.) But for n >= 4, all these algebras have zero-divisors. And a theorem of J. Frank Adams, c. 1956 -- that AFAIK has only a topological proof and no known purely algebraic one -- shows that real algebras without zero-divisors (aka real division algebras) exist *only* in dimensions 1,2,4,8. (Surprisingly, there are real division algebras in dimensions 2,4,8 other than the complexes, quaternions, or octonions.)

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[math-fun] octonians - cubic extension?
by James Propp 04 Aug '05

04 Aug '05

>Is there an interesting cubic extension of the octonians? I'm not sure what Rich means by a "cubic" extension, but there's an extension of the octonians called the sedenions, and some folks go even farther. See for instance the article math.RA/0207003 by Robert P. C. de Marrais, entitled "Flying Higher Than a Box-Kite: Kite-Chain Middens, Sand Mandalas, and Zero-Divisor Patterns in the 2n-ions Beyond the Sedenions". Abstract: Methods for studying zero-divisors (ZD's) in 2n-ions generated by Cayley-Dickson process beyond the Sedenions are explored. Prior work showed a ZD system in the Sedenions, based on 7 octahedral lattices ("Box-Kites"), whose 6 vertices collect and partition the "42 Assessors" (pairs of diagonals in planes spanned by pure imaginaries, one a pure Octonion, hence of subscript < 8, the other a Sedenion of subscript > 8 and not the XOR with 8 of the chosen Octonion). Potential connections to fundamental objects in physics (e.g., the curvature tensor and pair creation) are suggested. Structures found in the 32-ions ("Pathions") are elicited next. Harmonics of Box-Kites, called here "Kite-Chain Middens," are shown to extend indefinitely into higher forms of 2n-ions. All non-Midden-collected ZD diagonals in the Pathions, meanwhile, are seen belonging to a set of 15 "emanation tables," dubbed "sand mandalas." Showcasing the workings of the DMZ's (dyads making zero) among the products of each of their 14 Assessors with each other, they house 168 fillable cells each (the number of elements in the simple group PSL(2,7) governing Octonion multiplication). 7 of these emanation tables, whose "inner XOR" of their axis-pairs' indices exceed 24, indicate modes of collapsing from higher to lower 2n-ion forms, as they can be "folded up" in a 1-to-1 manner onto the 7 Sedenion Box-Kites. These same 7 also display surprising patterns of DMZ sparsity (with but 72 of 168 available cells filled), with the animation-like sequencing obtaining between these 7 "still-shots" indicating an entry-point for cellular-automata-like thinking into the foundations of number theory. I haven't done more than skim de Marrais' article; I'd be curious to know what others think of it. Jim Propp "Then he almost fell flat on his face on the floor When I picked up the chalk and drew one letter more A letter he had never dreamed of before And I said 'You can stop if you want with the Z Most people stop with the Z, but not me!" --- "On Beyond Zebra", by Dr. Seuss

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[math-fun] What is 'positive' ?
by MCKAY＠vax2.concordia.ca 04 Aug '05

04 Aug '05

Today, what the "positive" numbers are depends on geography and discipline. I think the sign bit of the 2-s complement arithmetic has led to zero as being considered positive. Certainly in an international context one needs to be careful! "Non-negative" is well-defined. John McKay

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[math-fun] octonians - cubic extension?
by Richard Schroeppel 04 Aug '05

04 Aug '05

Is there an interesting cubic extension of the octonians? Rich rcs(a)cs.arizona.edu PS to Dean: The Q(n) mod 64 result is amazing.

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[math-fun] Re: even Q(n)
by Dean Hickerson 04 Aug '05

04 Aug '05

Bill Gosper wrote: > George Andrews kindly informs me that any fixed power of 2 divides almost > all Q(n) (usually written q(n), to confuse you with the nome). I looked at Q(n) mod small powers of 2 a few years ago. I found that the value of Q(n) mod 64 is determined by representations of 24n+1 by the binary quadratic form x^2 + 24 y^2. Specifically: Let f(x,y) be determined by the values of x mod 192 and y mod 2: x mod 192 1 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 f(x,even) 1 51 55 29 5 15 27 25 41 11 31 53 45 39 3 49 f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23 x mod 192 49 53 55 59 61 65 67 71 73 77 79 83 85 89 91 95 f(x,even) 17 35 7 13 21 63 43 9 57 59 47 37 61 23 19 33 f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7 x mod 192 97 101 103 107 109 113 115 119 121 125 127 131 133 137 139 143 f(x,even) 33 19 23 61 37 47 59 57 9 43 63 21 13 7 35 17 f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23 x mod 192 145 149 151 155 157 161 163 167 169 173 175 179 181 185 187 191 f(x,even) 49 3 39 45 53 31 11 41 25 27 15 5 29 55 51 1 f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7 Let b(n) be the sum of f(x,y) over all solutions of 24n+1 = x^2 + 24 y^2, x>0. (0) Then Q(n) == b(n) (mod 64). (1) For example, for n=5 there are 3 solutions of (0): (x,y) = (11,0) or (5,2) or (5,-2). So Q(5) == f(11,0) + f(5,2) + f(5,-2) = 29 + 51 + 51 = 131 (mod 64); in fact Q(n) = 3. For most values of n, there are no solutions of (0), so Q(n) is divisible by 64. The values of f(x,odd) only range from 1 to 31, while those of f(x,even) range from 1 to 63. That's because, if (x,y) is a solution of (0) in which y is odd, then (x,-y) is another solution. Together these contribute 2 f(x,odd) to the value of b(n), so only the value of f(x,odd) mod 32 matters. This doubling also happens if y is even and nonzero. The only time that we need the value of f(x,y) mod 64 rather than just mod 32 is when y=0; i.e. when 24n+1 is a square; i.e. when n is pentagonal. If we're willing to ignore such values of n, then we can simplify the table above by reducing f(x,even) mod 32, which implies that we only need the value of x mod 96 rather than mod 192, since f(x+96,0) == f(x,0) (mod 32) for all x. There's no hope of modifying f(x,y) to give Q(n) mod 128: For n=20, there are no solutions of (0), so b(n) would be 0 no matter how we define f(x,y). Yet Q(20) = 64 is not divisible by 128. (I think I showed that Q(n) mod 128 was determined by representations of 24n+1 by (0) and by the quadratic form x^2 + 48 y^2, but I don't see the details in my notes.) I don't know if the value of Q(n) mod 2^k can be expressed in terms of binary quadratic forms for larger values of k, but it's not hard to show that it can be expressed in terms of representations of 24n+1 by quadratic forms in at most k variables. Outline of proof: The generating function for Q(n) can be written as J[1]/J[1,2] (2) where n n n(3n+1)/2 J[1] = PRODUCT (1 - q ) = SUM (-1) q (3) n>=1 n and 2n-1 2 2n J[1,2] = PRODUCT (1 - q ) (1 - q ) = 1 + 2A (4) n>=1 where 2 n n A = SUM (-1) q . (5) n>=1 Thus J[1]/J[1,2] = J[1]/(1 + 2A) 2 3 4 = J[1] (1 + 2 A + 4 A + 8 A + 16 A + ...). (6) Working mod 2^k, we only need the terms on the right up to A^(k-1). And the coefficient of n in J[1] A^i is determined by representations of 24n+1 by the quadratic form 2 2 2 r + 24 (s + ... + s ). (7) 1 i Dean Hickerson dean(a)math.ucdavis.edu

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