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Re: [math-fun] Origins of a problem.
by POPPY9X＠aol.com 20 Nov '02

20 Nov '02

there is an old problem in which one asks that the integers from 1 to N be listed so that the sum of adjacent integers are primes. I've never seen a solution. It is probably true that one can always do this with N at the end of the list. if one knew that for N>9 there is always a pair of twin primes betwen N and 2N all of this would follow. But there should be a real proof. anyone know anything?

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Re: [math-fun] Origins of a problem.
by POPPY9X＠aol.com 20 Nov '02

20 Nov '02

the original problem was not "circular" but just a list for example with N=12 we have: 12, 7, 10. 9. 8. 11, 6, 5, 2, 1, 4, 3, down thru 6 illustrates the twin prime connection (here the primes are 17 and 19)_ it also provides an inductive procedure for verifying the conjecture for N up to the linit of tables of primes (millions)

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RE: [math-fun] funny little limit on central binomial coefficient s
by Meeussen Wouter (bkarnd) 20 Nov '02

20 Nov '02

thanks Shel, Rich's shortish answer helped me grok it in half a day (he must have seen it in half a second): ratio of successive central binomial coefficients = 4-2/(1+n) (running sum of seq ...), seq[n], seq[n+1] x + 1 = 4 (solve for x) (* same goes for catalans *) cute no? that's why I like this group. Wouter Meeussen email: wouter.meeussen(a)vandemoortele.com =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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[math-fun] five, six, and eight (was: seven)
by James Propp 20 Nov '02

20 Nov '02

Last week, I wrote (in response to Dan Hoey's posting): >My mistake; as Dan Asimov pointed out, the sequence has period 5. (thereby showing that I can confuse one Dan with another with as much facility as I can confuse small natural numbers :-) ). I should mention that the theory of cluster algebras currently being developed by Fomin and Zelevinsky gives a context for these facts, but it doesn't really explain them in an elementary way. Here is a restatement of the "order-5 fact", along with two other facts of a similar kind: 1) Any sequence x(1),x(2),x(3),... defined by the recurrence x(n) = (x(n-1)+1)/x(n-2) (for n>2) has period 5. 2) Any sequence x(1),x(2),x(3),... defined by the recurrence { (x(n-1)+1)/x(n-2) (for n>2, n odd) x(n) = { { (x(n-1)^2+1)/x(n-2) (for n>2, n even) has period 6. 3) Any sequence x(1),x(2),x(3),... defined by the recurrence { (x(n-1)+1)/x(n-2) (for n>2, n odd) x(n) = { { (x(n-1)^3+1)/x(n-2) (for n>2, n even) has period 8. (Note that this is different from the period-8 recurrence I mentioned in my previous email.) In the theory of cluster algebras, these recurrences correspond to the Lie algebras A_2; B_2 (or C_2); and G_2 (respectively). But I'm just quoting someone else here; I don't actually understand the link between Lie algebras and cluster algebras. Nor do I have the foggiest notion of whether the x(n) = (x(n-1)+x(n-2)+1)/x(n-3) recurrence fits into the story somehow. But since there's such a nice story for the fact that the map t -> 1-1/t has order 3, it doesn't seem far-fetched to hope for something similar in these cases. Jim Propp

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[math-fun] funny little limit on central binomial coefficients
by Meeussen Wouter (bkarnd) 19 Nov '02

19 Nov '02

it seems to be 3, but why?: (2n choose n) /( Sum(k=0..n-1, (2k choose k) ) or (2n choose n) /(1+Sum(k=0..n-1, (2k choose k) ) hmmm, Mathematica 4.0.1.0 doesn't seem to handle it. Wouter. email: wouter.meeussen(a)vandemoortele.com =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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RE: [math-fun] Dominos and non-attacking rooks.
by Andy Latto 19 Nov '02

19 Nov '02

> -----Original Message----- > From: ed pegg [mailto:ed@mathpuzzle.com] > Sent: Tuesday, November 19, 2002 12:07 PM > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Dominos and non-attacking rooks. > > > > Can a 6x6 board be covered with 15 dominoes and 6 > non-attacking rooks? > > I have a nice little proof that it can't be, but I'm > wondering if it is a new > problem. Here's my proof: I'm curious if it is the same as yours. Define a function f by f(n) = rank of the rook in file n. None of the rooks attack each other, so f is a permutation on S = {1,2,3,4,5,6}. So the number of elements of f that map an even element of S to an odd element is equal to the number that map an odd element to an even one. So the total number of elements of S that are mapped to an element of opposite parity by f is even. But this is just the number of rooks on white squares, which must be odd, since the number of rooks on white and black squares must be equal (since each domino covers a white and a black square). This proof works for any square chessboard of size 4k+2. Is this the same as your proof? No idea whether it's a new problem. But it's a nice one.

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[math-fun] Domino problem
by asimovd＠aol.com 19 Nov '02

19 Nov '02

I thank Richard Guy for his answer (negative) to the converse domino question I posed. Can someone please tell me where to find the "Inquisiive Problem Solver" ? I don't even know whether it's a magazine or website or what. Somehow I don't think I associate Sol LeWitt's name with the problem I'm trying to recall. It was posted by someone whose name I don't think I recognized then and don't recall now. (If you are that person, please forgive me.) But it had to do with 4x4 arrangements of black & white squares such that each row and each column contains 2 W and 2 B squares. The question was whether all such arrangements could be obtained in a certain way, which I don't remember, either. --Dan

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Re: [math-fun] Origins of a problem.
by asimovd＠aol.com 19 Nov '02

19 Nov '02

<< >From my site, a few years ago ... At the dinner, I met Bernardo Recaman Santos of Columbia. >> Just wondering -- is the Columbia University, or Colombia, the country? --Dan

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[math-fun] Lego Escher
by Helger Lipmaa 18 Nov '02

18 Nov '02

http://www.lipsons.pwp.blueyonder.co.uk/lego.htm Nice. Especially the "Ascending and Descending" is brilliant. Helger

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Re: [math-fun] Origins of a problem.
by asimovd＠aol.com 18 Nov '02

18 Nov '02

Ed Pegg Jr. writes: << Joe Kisenwether has upped the ante. Arrange the numbers 1-32 in a circle so that any two adjacent numbers sum to a square number. The answer is unique. >> That's an interesting circle. I'd like to see a picture of the circle of numbers 1-32 in numerical order, with 32 chords indicating adjacency in the Kisenwether ordering -- especially with that ordering being unique. --Dan

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