December 2009 - mathfun@mailman.xmission.com

[math-fun] Fractional powers considered harmful to the cold blooded
by rwg＠sdf.lonestar.org 19 Dec '09

19 Dec '09

MW (not a typo): Main Entry:rotenone Pronunciation:*r*t*n**n Function:noun Inflected Form:-s Etymology:International Scientific Vocabulary roten- (from Japanese roten derris plant) + -one [etymologies are approximate!--rwg] : a crystalline pentacyclic compound C23H22O6 that is related to isoflavone, found especially in derris and cube roots, used in insecticides and in primitive fish and arrow poisons, and of low toxicity to warm-blooded animals --rwg [So the arrows are only good against big reptiles? And obviously first degree murderers.]

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[math-fun] help needed (for OEIS) with Mediawiki's Lucene-search extension
by N. J. A. Sloane 17 Dec '09

17 Dec '09

This is most definitely not fun, but we could use some help. As you may have heard, we are moving the OEIS to a wiki format, to be launched at the AMS meeting Jan 13 in San Francisco (visit the OEIS Foundation at Booth 813!). The trouble is, we are essentially forced to use Mediawiki's Lucene-search extension to do the searches. Is anyone familiar with the internal workings of this program? It turns out that it cannot handle phrase queries with commas, like "1, 2, 1, 2, 2, 3, 1". Of course this is terrible for the OEIS! Neil Sloane (njas(a)research.att.com)

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[math-fun] Alice's Adventures in Algebra
by Hans Havermann 16 Dec '09

16 Dec '09

"Alice's adventures in algebra: Wonderland solved" in New Scientist, 16 December 2009, by Melanie Bayley: http://www.newscientist.com/article/mg20427391.600-alices-adventures-in-alg…

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Re: [math-fun] Lie-sphere lurker
by Dan Asimov 16 Dec '09

16 Dec '09

Fred wrote: << . . . Unfortunately, quaternions could be modelled by the Clifford algebra Cl(0,2); but it's a very bad idea because their symmetry is then lost. . . . >> What does "could be modelled by" mean, or for that matter, "is modelled by" ? (And does "their" refer to the quaternions or to Cl(0,2) ?) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Lie-sphere lurker
by Fred lunnon 16 Dec '09

16 Dec '09

Feedback invited: errors, obscurities, typos, "stupid questions" and all. Text key: "^" for superscript, "_" for subscript, "\" for boldface. First instalment is a skeletal summary introducing 1. Clifford Algebras ____________________ A (nondegenerate) Clifford algebra Cl(p,q) over some "scalar" field --- usually the real numbers \R, occasionally the complex numbers \C --- extends it to polynomials in a set of transcendental "generators" subject to the usual arithmetic laws of polynomial rings, except that: multiplication of generators is anti-commutative; and generators square to +1 (p of them) or to -1 (q of them). Scalars are embedded in the algebra in the natural fashion, as elements of grade 0 (constant polynomials) or -1 (zero); vectors of dimension p+q as elements of grade 1 (linear polynomials). A "versor" is any element which conjugates vectors to vectors: if F is linear, then so is G = A^+ F A. Here A^+ denotes "reversion" of A, which acts as an inverse to the extent that the "magnitude" ||A|| == A^+ A is scalar. For most purposes, nonzero scalar factors are irrelevant; zero magnitude on the other hand indicates a non-invertible projection. It's fortunate that the reversion is easily computed: if the terms of A are broken down into subsums <A>_k by grade (number of generators) A = <A>_0 + <A>_1 + <A>_2 + <A>_3 + <A>_4 + ... then A^+ == <A>_0 + <A>_1 - <A>_2 - <A>_3 + <A>_4 + ... Finally, A is invertible just when ||A|| is nonzero. WFL

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Re: [math-fun] [seqfan] Divisibility sequences in OEIS. (fwd)
by Richard Guy 15 Dec '09

15 Dec '09

It may be that someone else is interested in (parts of) this reply that I just made. Can anyone prove that A047817 in OEIS is in fact a divisibility sequence? R. ---------- Forwarded message ---------- Date: Tue, 15 Dec 2009 13:57:08 -0700 (MST) From: Richard Guy <rkg(a)cpsc.ucalgary.ca> To: Georgi Guninski <guninski(a)guninski.com> Cc: Richard Guy <rkg(a)cpsc.ucalgary.ca> Subject: Re: [seqfan] Divisibility sequences in OEIS. The Hurwitz numbers evidently form a div seq, but I don't know what they are, so don't know how to prove it. [snip] More exciting is that the numbers of points (including the pt at infty) on any elliptic curve over the field F(q^n) form a fourth order divisibility sequence, with nth term q^n + 1 - t_n where {t_n} is a 2nd order recurring sequence generated by x^2 - tx + q, with t_0 = 2 (giving a(0) = q^0 + 1 - 2 = 0) and t_1 = t where t is the `trace of Frobenius' (Hecke eigenvalue) for the curve mod q. a(1) = q+1-t is a common factor of all the terms. It's an interesting (at least to me) paradox that 2 + 2 = 2 x 2. I.e., the sum of two second order recurring sequences [ q^n + 1 satisfies the recurrence x^2 - (q+1)x + q ] form a 4th order one, which is the product of 2 second order div seqs (of integers in a quadratic extension field of Q), the 4 roots being the products of the 2 roots of each of the two quadratics. There's a 2-parameter (q \& t) infinite family of such sequences, but A002248 is the only one I've spotted in OEIS. Your search didn't pick this one up. Several hundred should be in OEIS, e.g., for q = the first 8 or 10 primes and t = 0, +-1, +-2, ... Note that although Hasse tells us that |t| <= 2 * sqrt(q), you still get divisibility seqs with t outside that range. The fact that it's a divisibility sequence follows from the group structure of points on an elliptic curve. Also my colleague Hugh Williams has recently proved that a large (3-parameter) family of 4th order recurring sequences, which includes these as a particular case, are div seqs. [snip]

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[math-fun] Divisibility sequences in OEIS.
by Richard Guy 15 Dec '09

15 Dec '09

Apologies for multiple copies of this rather long message, but I'm keen to find a few potential editors for the following scheme. If you are able to help, send to me personally, not to bother the rest of the lists. I can then send you attachments of even longer files, and we can divvy up the work so that there is only enough duplication to ensure checking. Here is the preface to a file: This is a list, in OEIS numerical order, of divisibility sequences. The suggestion is that these should be given the codeword ``div'' (or ``divi'' if codewords all have four letters). Where this is not already done, it should be remarked that they are divisibility sequences. Many are values of Chebyshev polynomials: for these the polynomial should be quoted. Those which satisfy recurrence relations should include the recurrence and the generating function. A `closed' formula is also always possible. Divisibility sequences satisfying recurrence relations of order 4 or higher may contain prime divisors with more than one rank of apparition. At least the first few examples of this should be noted. Many higher order sequences factor into ones of lower order: this should be noted. Check list for div seqs in (and not yet in) OEIS: 1. Is it stated that it's a divisibility seq ? Should there be a keyword, say `divi' ? 2. Check the offset. The ``right'' offset is $a(0)=0$, but there may be a good reason for having two entries with different offsets. For example, the Fibonacci sequence would normally be given as $a(0)=0$, 1, 1, 2, 3, 5, \ldots, but it often manifests itself as $a(0)=1$, 1, 2, 3, 5, \ldots, e.g., the number of ways of packing $n$ dominoes in a $2\times n$ box. 3. If a divisibility sequence is multiplied or divided through by an integer, it remains a divisibility sequence, so, in order to be sure of locating all manifestations, if a (the) ``natural'' occurrence of the sequence has $a(1)=k$, then the division by $k$, starting $a(0)=0$, $a(1)=1$ should also be given, and often the versions with $a(1)=2$, 3, etc., up to a point, should be given as well. 4. Is there a recurrence relation ? Note that, paradoxically, there may be more than one, and of different orders, because of factorization. 5. Is the generating function given ? 6. Is there a ``closed'' formula of type $A\alpha^n+B\beta^n+C\gamma^n+\cdots$ ? 7. Does the sequence factor into recurring sequences of lower order (which are not necessarily divisibility sequences) over the integers ? over other number fields ? 8. Are the terms values of one or more Chebyshev polynomials ? 9. Note that every second term forms a divisibility sequence which may be worthy of mention in its own right. Similarly for every third term, every fourth term, \ldots . 10. Numbertheoretic properties, e.g., the rank(s) of apparition of primes, may be worthy of note. In sequences of order $2^k$ a prime may have as many as $k$ ranks of apparition. Second order linear recurrences are often numerators or denominators of convergents to continued fractions of quadratic irrationals. Many 2nd order sequences are related to solutions of Bramahgupta-Bhaskara-Pell equations 11. Cross-references to other sequences, especially those arising from 2., 3., 6., 8., \& 9.\ above. 12. Does the sequence arise ``in real life'' as the number of points on an elliptic curve; or the number of perfect matchings in a family of graphs; or the numer of spanning trees (sometimes called the `complexity') thereof, or in other ways? Note that perfect matchings are also called dimer problems or domino tilings or packings. 13. If it's a second order sequence, then I believe that $a_n=F_{rn}/s$ for some $(r,s)$, where $F_n$ is the $n$\,th Fibonacci number. If so, then $r$ and $s$ should be given. 14. Other things that I've missed. For a start, compare and contrast the descriptions of A010892, A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362, A007655, A078364, A077412, A078366, A049660, A078368, A075843, A092449, A077421, A097778, A077423, A097780, A097309, A097781, A097311, A097782, A097313, and note that the varied properties that are mentioned are in fact shared by ALL of them, though rarely are more than one or two mentioned. When, as with the above examples, a sequence belongs to a family, it would be good to give the family a name, and say which member it is. The companion sequence of sequences is A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A041041, A049666, A041061, A140455, A041085, A154597, A041113, missing, A041145, missing, A041181, missing, A041221, missing, A041265, missing, A041313, missing, A041365, A049667, A041421, missing, A041481, missing, A041545, missing, A041613, missing, A041685, missing, A041761, missing, A041841, missing, A041925, missing, A042013, missing, A042105, missing, A042201, missing, A042301, missing, A042405, missing, A042513, missing, A042625, missing, A042741, [perhaps itself a more interesting sequence than many that are submitted?] The major interest of divisibility sequences is the factorization of the terms. It would be good to have this, perhaps on a separate page. In this sense, all of the 'Cunningham project' should be accessible via OEIS. Thankyou for your patience, if you've got as far as this. R.

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Re: [math-fun] random rational numbers
by James Propp 14 Dec '09

14 Dec '09

Consider the Stern-Brocot scheme for generating the positive rationals (with 0 and infinity thrown in): 0/1 < 1/0 0/1 < 1/1 < 1/0 0/1 < 1/2 < 1/1 < 2/1 < 1/0 0/1 < 1/3 < 1/2 < 2/3 < 1/1 < 3/2 < 2/1 < 3/1 < 1/0 etc. (To obtain the n+1st row, we take the nth row and insert the mediant (a+c)/(b+d) of any two adjacent entries a/b and c/d.) Based on this scheme, let's say that 1/2 of the positive rationals are beween 0 and 1, 1/2 of the positive rationals are beween 1 and infinity, 1/4 of the positive rationals are beween 0 and 1/2, 1/4 of the positive rationals are beween 1/2 and 1, 1/4 of the positive rationals are beween 1 and 2, 1/4 of the positive rationals are beween 2 and infinity, 1/8 of the positive rationals are beween 0 and 1/3, etc. If you're suspicious of non-countably-additive measures, you can put a probability measure mu on the positive REALS that assigns measure 1/2 to the intervals [0,1] and [1,infinity], measure 1/4 to the intervals [0,1/2], [1/2,1], [1,2], and [2,infinity], etc. (In fact, the restriction of mu to [0,1] is nothing other than the pullback of Lebesgue measure on [0,1] via Minkowski's question mark function, if you want to be a sophisticate. Alternatively, if you're a fan of the Stern enumeration of the positive rationals (rediscovered by Calkin and Wilf), you can just say, as Reznick does, that the mu-measure of an interval [a,b] is the limit, as n goes to infinity, of the proportion of the first n terms of the Stern sequence that fall in [a,b].) This measure mu has two really nice properties: mu(1/S) = mu(S) and mu(S+1) = mu(S)/2. The first of these you can read off from the 180-degree rotational symmetry of each row of the Stern-Brocot scheme (exchange left with right and numerators with denominators). To see why the second is true, it suffices to note that the last half of each row (leaving aside the final term) is gotten from the previous row (again leaving aside the final term) by adding 1 to each fraction. >From this, it's easy to show that the average positive rational is 3/2! One way is to note that the center of mu-mass of each interval [n,n+1] is at n+1/2. So the center of mu-mass of [0,infinity) is (1/2)(1/2) + (1/4)(3/2) + (1/8)(5/2) + (1/16)(7/2) + ... and you'll see instantly that this is 3/2 if you're von Neuman. Alternatively, note that the center of mass of [1,infinity) is 1 to the right of the center of mass of [0,infinity) (this follows from the formula mu(S+1) = mu(S)/2). So if we split [0,infinity) into two equal-mass sub-pieces [0,1] and [1,infinity), we get (*) x = (1/2)(1/2) + (1/2)(x+1), where x is the center of mass of [0,infinity), 1/2 is the center of mass of [0,1], and x+1 is the center of mass of[1,infinity). >From (*) we get the solution x = 3/2. Can anyone see how to compute the center of mass of [0,1/2]? For that matter, what is mean-squared positive rational? Jim

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[math-fun] analyzing a strategy game
by Dave Dyer 14 Dec '09

14 Dec '09

Today's brain twister is how to think about strategy for this game: http://www.cameronius.com/games/mutton/ Basically, it's a deduction game where the wolf hides among the sheep and periodically eats one. The farmer is trying to shoot the "sheep" he suspects are actually wolves in disguise. My two guiding thoughts are (1) that any identifiable strategy by either player leaks a lot of information, which leads to that strategy being ineffective. For example, it's clearly advantageous for the wolf be be adjacent to as many sheep as possible, to make himself harder to identify - but if the wolf always runs to hide among the largest number of sheep, he's very easy to identify. (2) which leads to the thought that a completely random strategy might be best. But some actions are so clearly non-optimum that they ought to be excluded completely. For example, eating the only adjacent sheep unequivocally identifies the wolf. It seems like some sort of Nash equilibrium must exist, but there probably are too many possibilities to make a complete analysis practical.

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[math-fun] Visualizing Prime Numbers
by Ray Tayek 14 Dec '09

14 Dec '09

http://www.sievesofchaos.com/ pretty pictures and videos --- co-chair http://ocjug.org/

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