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Re: [math-fun] bad pun
by Daniel Asimov 18 Mar '07

18 Mar '07

Not sure if I'd ever heard of the Szilassi polyhedron before, but (for those in the same boat) it has 7 (nonconvex) polygonal faces each of meets the other 6, is topologically a torus, and it embeds in 3-space. My favorite torus polyhedron has 7 regular hexagons as faces, any two of which meet each other along an edge.* This embeds isometrically in R^6 (so that its 84 symmetries are all in the orthogonal group O(6)). Does anyone know for sure whether 6 is the lowest possible dimension for this? --Dan _______________________________________________________ * Which is of course the nicest way to 7-color a torus.

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[math-fun] ..and of course
by wouter meeussen 18 Mar '07

18 Mar '07

k=1;NestList[1+1/((k++)#)&,1,12] gives {1,2,5/4,19/15,91/76, ..., A052169(n+1)/A002467(n) } if I got my offsets correct. grin, W.

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[math-fun] Mousetrap and "Kurepa hypothesis for left factorial"
by wouter meeussen 18 Mar '07

18 Mar '07

looking at Sloane's A052169 (Kurepa hypothesis for left factorial) and A002467 (game of Mousetrap), is it self-evident that [[1]] a(n)=1+1/(x (k++) a(n-1)) , a(1)=x, for x->1 produces the same sequence as [[2]] a(2) = 1, a(3) = 2, a(n) = (n-2)*a(n-1) + (n-3)*a(n-2) ? Expanding [[1]], with the usual meaning of k++ as post-increment, we get: k=1;NestList[1+1/((k++)x #)&,x,3] {x, 1 + 1/x^2, 1 + 1/(2*(1 + 1/x^2)*x), 1 + 1/(3*(1 + 1/(2*(1 + 1/x^2)*x))*x)} Taking only the numerators gives: {x, 1 + x^2, 2 + x + 2*x^2, 2 + 6*x + 5*x^2 + 6*x^3, 14 + 27*x + 26*x^2 + 24*x^3, 8 + 94*x + 155*x^2 + 154*x^3 + 120*x^4, 118 + 699*x + 1060*x^2 + 1044*x^3 + 720*x^4, 48 + 1390*x + 5823*x^2 + 8344*x^3 + 8028*x^4 + 5040*x^5, ... set x to 1 and you get A052169 in a different way. As a 'limping' triangular table, the coefficient of the highest power in x stands out as n!, and the definition of A052169 as pairwise sums of A002467 (itself equal to n!-!n) is curious. {0,1}, {1,0,1}, {2,1,2}, {2,6,5,6}, {14,27,26,24}, {8,94,155,154,120}, {118,699,1060,1044,720}, {48,1390,5823,8344,8028,5040}, {1210,16013,54004,74060,69264,40320}, {384,22010,190701,552788,730764,663696,362880}, {14730,364217,2393046,6194420,7931016,6999840,3628800}, {3840,382130,5913397,31851386,75446260,93878136,80627040,39916800}, {208110,8591947,97284270,450355252,992596296,1203535872,1007441280,479001600 } maybe not math-funny to all, but quite amusing to at least one, W.

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[math-fun] counting set systems
by N. J. A. Sloane 17 Mar '07

17 Mar '07

Dear Mathfun, Seqfans: Numbers of set systems of various types --------------------------------------- There are some sequences that may be missing from the OEIS, in case somebody would be interested in calculating them. This was suggested by reading pages 1 and 2 of the book by Heinz Koenig on Measure and Integration (Springer, 1997) We are considering a "set system" S, which is a nonempty collection of subsets A of a set X, where X has n points. There will be two versions of each sequence, depending on whether the points are labeled or unlabeled. We define several properties which S might have: U : A, B in S => A U B in S I : A, B in S => A intersect B in S \ : A, B in S with A contained in B => B\A in S - : A in S => complement of A in X is in S 0 : S contains the empty set 1 : S contains the universe set X We also define a new operation on subsets of X: |P|A|Q| = (P intersect complement of A) union (Q intersect A) [points in P but not in A plus points in both Q and A] Now Koenig defines 4 types of set systems: a lattice : properties U, I an oval : P,A,Q in S => |P|A|Q| in S a ring : properies U, I, \ an algebra : properties U, I, - The question is, how many lattices, ovals, rings, and algebras are there on n labeled/unlabeled points? (In fact one might consider any of the 2^7 combinations of the above 7 properies!) There are of course many sequences like these in the OEIS already, but usually there is no precise definition (this should be corrected!) so it is hard to tell which of these may be new. For example: posets: A000112 (unlabeled), A001035 (labeled) Lattices: A006966 (unlabeled); A055512 (labeled) Boolean lattices: A005493 Set systems of various types, studied recently by Mitch Harris and Don Knuth: A102894-A102897. E.g.: %I A102895 %S A102895 2,2,8,90,4542,2747402,151930948472 %N A102895 Number of ACI algebras or semilattices on n generators, with no identity element. %C A102895 An ACI algebra or semilattice is a system with a single binary, idempotent, commutative and associative operation. %C A102895 Or, number of families of subsets of {1, ..., n} that are closed under intersection and contain the empty set. - this seems to count subsets with properties I, 0 in the labeled case %I A102897 %S A102897 2,4,14,122,4960,2771104,151947502948 %N A102897 Number of ACI algebras (or semilattices) on n generators. %C A102897 Also counts Horn functions on n variables, boolean functions whose set of truth assignments are closed under 'and', or equivalently, the boolean functions that can be written as a conjunction of Horn clauses, clauses with at most one negative literal. %C A102897 Also, number of families of subsets of {1,...,n} that are closed under intersection (because we can throw in the universe, or take it out, without affecting anything else). %C A102897 An ACI algebra or semilattice is a system with a single binary, idempotent, commutative and associative operation. - this seems to count subsets with property I in the labeled case ? Likewise A102896 seems to count subsets with properties I, 1 in the labeled case and A102894 seems to count subsets with properties I, 0, 1 in the labeled case (see the full entries for more details) Neil

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[math-fun] today's Jeopardy! outcome: nonzero three-way tie
by Thane Plambeck 17 Mar '07

17 Mar '07

The first three-way nonzero tie in the history of the program occurred in the program aired today. I copied this interesting (if slightly flawed) analysis from http://newsfromme.com/ * * * **snip * * * ** No Winner, No Losers I know we've been paying too much attention to this one Jeopardy! episode but I wanted to write one more post about how it's unlikely to happen again. The three-way tie was an anomaly in large part because one player didn't play the game to win. Going into Final Jeopardy!, here's how the totals stood... * Scott: $13,400 * James: $8,000 * Anders: $8,000 So how much do you wager if you're one of those contestants? If you're James or Anders, you'd presume the following: That you can't win unless you're right and Scott is wrong. In theory, you could win the game with the wrong answer if Scott was also wrong and he wagered so much that his total dropped to below where you wound up, but you have to operate from the assumption that he's not dumb enough to do that. James and Anders made the right wagers. Neither one could afford the risk that the other would bet more than he did. You'd also assume that the odds are that it's unlikely you will get the answer right and the other two will both get it wrong. That's possible but it's much more likely that it either stumps everyone or no one. So you'd wager as each of them did: The full eight thousand. Now, let's say you're Scott. You'd assume from the above that at least one of the other two players — and probably both — is wagering $8,000. A person who does that will beat you unless you're also right and you wind up ahead of them. Therefore, the correct bet for you is at least $2,601. That way, one or both of them could wind up with $16,000 and you'd have $16,001. You'd also win if all three of you were wrong. The only way you could lose is if you're wrong and one or both of the others is right...but that's going to be true no matter what you bet. So at least $2,601 is the proper bet here. The trouble is that Scott didn't bet that way. I'm assuming the Game Theory expert opined that this wouldn't happen again because future players will recognize that Scott blew his chance to win and they won't make that mistake. But is it really much of a mistake? If Scott had bet $2,601, he'd take home $16,001 and come back on Monday to play against two new opponents. Since he bet $2,600, he takes home $16,000 and comes back on Monday to play against the same two opponents again. Not much difference from his standpoint. Scott didn't really lose anything except being able to say he won. * * * Also, this comment from Roy Leban on the puzzlers.org list: * * * Not the greatest analysis. How could anyone think that a computer science professor who had already won twice, once betting $1401 once, wouldn't know exactly what it would take to win. I figured it was a three-way tie early on when Squonk [ie Scott] didn't sweep the board (and, wow, that first board was tough for me too! I think I might have had a $0 at the end, if I was lucky). When I saw the twin $8,000's, it seemed inevitable. Very cool! > Yeah, very exciting! Infinity noticed that the guy in the middle must > have said, "Thank you," since during Squonk's close-up he shakes the > dude's hand and says, "You're welcome." My wife and I thought Squonk also said "You're welcome" earlier. He turned to the player next to him right before his answer was unveiled and said something. And "You're welcome" is right -- he could have won and had some more money, but he essentially gave those two guys $16,000 each. Plus, it should be an interesting rematch on Monday. Squonk, I hope you'll fill us all in at some point about what went through your mind and what the other players and Alex thought about it. Is there another three-way tie on Monday?! -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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[math-fun] Centrist Groupoids, Universal Equation, Somos Sequences & Modular Theta functions, ExpMath slides
by Richard Schroeppel 15 Mar '07

15 Mar '07

Centrist Groupoids Some of you will recall Central Groupoids, which are systems with a binary operator (magmas, shudder) that satisfy the equation AB.BC=B (or, (AB)(BC)=B), and have the curious property that the finite systems have a square number of elements. I decided to look at the relaxed equation AB.BA=B, which, after 30 seconds of deep thought, I have decided to call Centrist Groupoids. (Medial Magmas?) Of course, any Central Groupoid is also Centrist. The rule is anti-commutative. How Many of Order N? There are no 2- or 3-element systems. There are 58 4-element systems, and 6634 5-element systems, and no/nada/nil/0/zilch 6-element systems. Universal Equation Is there an equation that is true for all 2-element binary operators? (Besides A=A. :-)) PDP10 aficionados will recall the 16 boolean instructions (ANDCA, etc.). There are 10 non-isomorphic tables, or 7 if we allow argument swapping. (Zero, Left, And, Xor, Nor, ~Left, Andcm.) All of them satisfy the identity A:.A.AA:A = A:.AA.A:A or A{[A(AA)]A} = A{[(AA)A]A}. Puzzle: Show that similar equations exist with more letters, and for systems with more than 2 elements, and for ternary operators (tragmas). Problem: Find some. Somos Sequences etc. Gosper has come up with some interesting determinant identities for Somos sequences. I've finally updated my web page. Gosper's results, along with my foray into defining theta functions mod 43, are on the new page. I've also put up the slides from my Experimental Math talks, which include contributions from several Funsters. (counting polyhypercubes, ...) The page is at http://www.cs.arizona.edu/~rcs Rich rcs(a)cs.arizona.edu

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[math-fun] bad pun
by R. William Gosper 15 Mar '07

15 Mar '07

http://gosper.org/rasta.png

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[math-fun] Probably dumb question about primes
by Daniel Asimov 14 Mar '07

14 Mar '07

Goldbach's conjecture is that every even number >= 4 is the sum of two primes. But what's known about this question: << Is every even number the difference of two primes? >> --Dan

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[math-fun] Today is the picentennial (correction)
by R. William Gosper 14 Mar '07

14 Mar '07

of 14-15 Jan 1693

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[math-fun] Today is the picentennial
by R. William Gosper 14 Mar '07

14 Mar '07

of 27-28 Feb 1693. Various sites claim for 1693: Newton's 2nd mental breakdown, 1st published description of fluxions, 1st publication of Wallis's Algebra, Leibniz rediscovers determinants, ..., Not clear what happened on 27-28 Feb, or 16-17 Feb if we count the missing eleven days. --rwg

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