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[math-fun] crossword puzzle clue, 29 Dec 2006 New York Times
by Thane Plambeck 17 Jan '07

17 Jan '07

52 Across: Pioneer in the math of Sudoku (5 letter answer) I won't spoil it by giving the answer, which you can have by clicking this URL http://tinyurl.com/yqvedr My question is this: is there something interesting that Mr. X had to do with the "math of Sudoku"? I'm aware of general things, but is there something specific that comes to mind? -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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[math-fun] two kinds of partitions
by Schroeppel, Richard 17 Jan '07

17 Jan '07

As RKGuy would say, "probably well known to those who know it", but I found this cute: There are two kinds of partitions in common use: (a) Partitions of integers. p(5) = 7 partitions of 5, which are: 5, 41, 32, 311, 221, 2111, 11111. There's a complicated formula for p(n). (b) Partitions of sets. The 5 partitions of the set {abc} are abc, ab/c, ac/b, a/bc, a/b/c. There's a different complicated formula for these. The objects being partitioned are important in case B, while only the sizes of the pieces matter in case A. The other day, I realized that these two concepts could be joined together, with partitions of bags (multisets, sets with repeated elements). Case A is the partitioning of a bag with only one kind of element, {x,x,x,x,x}, while case B is the partitioning of a bag with all elements distinct {a,b,c}. The first nontrivial bag that's in between cases A and B is {a,a,b}, whose partitions are aab, aa/b, ab/a, a/a/b. So {aaa} has 3 partitions, {aab} has 4, and {abc} has 5. Case A has some interesting generalizations (planar partitions, cubic partitions, etc.). I don't know of similar notions for case B, or for bags. The generating functions for case A and case B are (of course) different, but have some similarities. Perhaps there's an in-between generating function for bag partitions. Rich

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Re: [math-fun] two kinds of partitions
by Daniel Asimov 17 Jan '07

17 Jan '07

The only thing I notice worth mentioning is that (clearly) in counting the partitions of a multiset, the count depends only on the "shape" of the multiset, and such shapes are in one-to-one correspondence with the set of integer partitions. So this defines an interesting function from the set of integer partitions to the integers. --Dan ---------------------------------------------------------------- Rich writes: << There are two kinds of partitions in common use: (a) Partitions of integers. p(5) = 7 partitions of 5, which are: 5, 41, 32, 311, 221, 2111, 11111. There's a complicated formula for p(n). (b) Partitions of sets. The 5 partitions of the set {abc} are abc, ab/c, ac/b, a/bc, a/b/c. There's a different complicated formula for these. The objects being partitioned are important in case B, while only the sizes of the pieces matter in case A. The other day, I realized that these two concepts could be joined together, with partitions of bags (multisets, sets with repeated elements). Case A is the partitioning of a bag with only one kind of element, {x,x,x,x,x}, while case B is the partitioning of a bag with all elements distinct {a,b,c}. The first nontrivial bag that's in between cases A and B is {a,a,b}, whose partitions are aab, aa/b, ab/a, a/a/b. So {aaa} has 3 partitions, {aab} has 4, and {abc} has 5. Case A has some interesting generalizations (planar partitions, cubic partitions, etc.). I don't know of similar notions for case B, or for bags. The generating functions for case A and case B are (of course) different, but have some similarities. Perhaps there's an in-between generating function for bag partitions.

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[math-fun] Taxicab and prime numbers
by Christian Boyer 15 Jan '07

15 Jan '07

It seems that nobody knows a solution of this Taxicab equation with prime numbers (p, q, r, s, t, u): (1) p^3 + q^3 = r^3 + s^3 = t^3 + u^3 I tried also the equivalent Cabtaxi problem. This new problem should be easier, because allowing both sums and differences. But, I am also unable to get a solution of: (2) p^3 +/- q^3 = r^3 +/- s^3 = t^3 +/- u^3 It is "irritating", because the standard solutions of (2) using prime and non-prime are small and easy to find, the smallest being: 728 = 6^3 + 8^3 = 9^3 - 1^3 = 12^3 - 10^3 Why??? Are (1) and (2) impossible? Any idea? With only (p, q, r, s), the equations have solutions, the smallest being: (3) p^3 + q^3 = r^3 + s^3 6507811154 = 31^3 + 1867^3 = 397^3 + 1861^3 12906787894 = 593^3 + 2333^3 = 1787^3 + 1931^3 (4) p^3 +/- q^3 = r^3 +/- s^3 62540982 = 397^3 - 31^3 = 1867^3 - 1861^3 (based on the first above solution of (3)!) 105161238 = 193^3 + 461^3 = 709^3 - 631^3 Christian.

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Re: AW: [math-fun] amusing math puzzle
by Daniel Asimov 13 Jan '07

13 Jan '07

<< [On the other hand, all 3 could communicate by choosing a very obvious code, e.g. 1=a, 2=b,... (all 3 must speak the same language) and deal out how to play. Since you play afterwards infinite rounds this communication would not reduce your expectation value.] >> According to the rules of the game, no communication is possible. This could be ensured by simply not revealing to the players the plays of any of the other players, or even the extent to which each of them won -- until the game is over. (Also, just as a motivation to have an opportunity to spend your winnings -- if any -- the game could be in principle played at the moments 1 - 1/n hours, n = 1,2,3,.... Then after one hour the game is over.) The question asks for the optimal strategy. I think if there is any optimal strategy, it must be 1,1,1,1,1,1,.... FOR: If you play any other strategy at all, you risk the possibility that exactly one opponent is playing the 1,1,1,1,... strategy -- guaranteeing that you will end up losing money. Whereas if you play 1,1,1,1,... you are guaranteed to do no worse than break even. --Dan

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[math-fun] amusing math puzzle
by Dave Dyer 13 Jan '07

13 Jan '07

The current puzzle at The Grey Labrynth is an amusing math puzzle. http://www.greylabyrinth.com/puzzles/puzzle.php?puzzle_id=206

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[math-fun] Piano-Hinged Dissections
by Eric Angelini 11 Jan '07

11 Jan '07

... the book by Greg Frederickson, is a delight -- I recommend the free DVD at the end and, for instance, the Ellipse-Heart dissection -- what a marvel ! Thanks, Greg, you're a good cameraman, director, actor and scriptwriter ! Best, É.

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Re: [math-fun] Re: Car Talk and prime numbers
by Phil Carmody 09 Jan '07

09 Jan '07

From: "Michael Kleber" <michael.kleber(a)gmail.com> > > > I asked this list on Saturday about deletable primes. > > Thanks to everyone who replied. But no one answered > > my specific question about the largest known deletable > > prime, so I'm adding these two sequences to the OEIS. > > Maybe someone would like to extend them! > > > %S A125589 2,13,103,1013,10039 > > %N A125589 Smallest n-digit base-10 deletable prime. > > 100103, 1000193, 10000931 100001903 1000009321 10000019321 100000124963 1000000129693 10000000370219 Worth double-checking, of course. Phil () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani] __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] CORRECTION Re: Question about primes
by Daniel Asimov 09 Jan '07

09 Jan '07

MAKE THAT INSTEAD: Is it known whether for each odd prime p there exists at least one odd prime q such that pq + 2 is prime? --Dan

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[math-fun] Reflection from a moving mirror
by Eugene Salamin 09 Jan '07

09 Jan '07

This is a "test your physics intuition" puzzle. When light reflects from a stationary mirror, we know that (1) the angle of reflection equals the angle of incidence, and (2) the frequency of the reflected light equals the frequency of the incident light, i.e. there is no Doppler shift. Does any of this change if the mirror is moving with a velocity parallel to its surface? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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