May 2005 - mathfun@mailman.xmission.com

[math-fun] hexagons, sequences, omino
by Schroeppel, Richard 17 May '05

17 May '05

Dan asks ... > Let T denote the torus obtained by identifying > each pair of opposite edges of a regular hexagon. How does this work? I can't get the edges to match up. Scott asks ... > Is there a nice general formula or a good approximation for the number > of nondecreasing sequences in N^k, for integers N and k ? For an exact formula: The 1D case is already answered, a quotient of factorials. Based on experiments done long long ago, at an Institute far far away, the 2D case seems to be a product of something like the 1D answers. Strict vs. non-strict vs. semi-strict perturbs the formulas slightly. This also worked for rectangles, MxN. But I couldn't find a similar answer for the 3D case. I always figured there was a tie-in with linear and planar partitions having a nice generating function, while cubic partitions don't seem to have one. Dave Wilson asks ... > What is the > probability that a randomly chosen square is in a unomino (I find only one > page on the web with that word!)? I think Gardner's columns used "monomino". Rich

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OOPS RE: [math-fun] nondecreasing sequence count?
by dasimov＠earthlink.net 17 May '05

17 May '05

The second paragraph below (using the separators to count the ways that I increments can be binned into k-1 bins) is flawed -- it overcounts the ways by I'm not sure what factor at this hour. Oh, well. --Dan << Think of the I increments as dots in a row. The number of ways they can be distributed into k-1 bins is the same as the way we can separate the I dots using k-2 vertical lines each before, between, or after the I dots. (Two or more separators are allowed to have 0 dots between them.) The number of ways to arrange the separators, and equally to interpolate a nondecreasing sequence between X1 and Xk, is thus (I+1)^(k-2). >>

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[math-fun] Mysognist math
by Guy Haworth 17 May '05

17 May '05

It is an old joke, but does make some points about 'proof', the testing of proof and the nature of 'belief', at least for an interested amateur like me: a) instead of 'women', substitute 'men', 'work' etc - i.e., change a parameter of the theorem - carry out the same deductive steps - do you still believe the result, or - were you believing what you wanted to believe? b) the use of '=' on line 2 is wrong - an '==>' might have been better c) 'root' has more than one meaning - the notation used is not well-defined - 'informal' languages depend on context-sensitive semantics - hence the need for designer-formal notation Otherwise, the proof is excellent: I trust the son will revise his opinion in time. g

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RE: [math-fun] nondecreasing sequence count?
by dasimov＠earthlink.net 17 May '05

17 May '05

Scott Huddleston asks: << Is there a nice general formula or a good approximation for the number of nondecreasing sequences in N^k, for integers N and k ? >> Let me give this a shot. I suspect N here is shorthand for {1,2,. . .,N} ? Assuming this, let X1 <= 2 <= . . . <=Xk with 1 <= Xj <= N for all j, 1 <= j <= k. Given the two integers X1 and Xk with 1 <= X1 <= Xk <= N, the above sequence is determined by the placement of the I = Xk - X1 increments among the k-1 locations the increments can occur: going from X1 to X2, or X2 to X3, etc. Think of the I increments as dots in a row. The number of ways they can be distributed into k-1 bins is the same as the way we can separate the I dots using k-2 vertical lines each before, between, or after the I dots. (Two or more separators are allowed to have 0 dots between them.) The number of ways to arrange the separators, and equally to interpolate a nondecreasing sequence between X1 and Xk, is thus (I+1)^(k-2). The totality of nondecreasing sequences includes N ways that I = 0, N-1 ways that I = 1, . . . , 2 ways that I = N-2, and finally 1 way that I = N-1. This means the # of nondecr. seqs. is ------------------------------------------------------------------------------------------------------------------------- nondec(N,k) = N*1 + (N-1)*(2^(k-2)) + . . . + 2*(N-1)^(k-2) + 1*N^(k-2). ------------------------------------------------------------------------------------------------------------------------- If we divide nondec(N,k) by N^k, we get nondec(N,k))/N^k = (1/N)*[(N/N)*(1/N)^(k-2)+((N-1)/N)*(2/N)^(k-2)+ . . . +(2/N)*((N-1)/N)^(k-2)+(1/N)*(N/N)^(k-2)] which, for N >> 0, is approx. the integral from x=0 to x=1 of (1-x)*x^(k-2), i.e., nondec(N,k)/N^k ~= 1/(k-1) - 1/k = 1/(k^2 - k). So asymptotically, nondec(N,k) ~ N^k / (k^2 - k). (Since it's 1:47 am, I think I'll let someone else check my work.) --Dan

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[math-fun] nondecreasing sequence count?
by Scott Huddleston 16 May '05

16 May '05

Is there a nice general formula or a good approximation for the number of nondecreasing sequences in N^k, for integers N and k ? Thanks, - Scott

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[math-fun] Hexagon puzzle
by dasimov＠earthlink.net 16 May '05

16 May '05

Puzzle: Let T denote the torus obtained by identifying each pair of opposite edges of a regular hexagon. For which positive integers N is it possible to cut T into N congruent regular hexagons? --Dan

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[math-fun] polyomino shelving
by Thane Plambeck 15 May '05

15 May '05

Pretty expensive, but if you've always wanted restackable, modular Tetris shelving here is your chance. http://www.inhabitat.com/entry_23.php -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] Fermat divisor question
by David Wilson 15 May '05

15 May '05

It is believed that the set of n >= 0 such that the Fermat number F(n) = 2^2^n+1 is composite is cofinite. Do we even know if it is infinite? - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] Misogynist math
by David Wilson 15 May '05

15 May '05

Disclaimer: My 15-year-old son has developed a rather dim view of girls which may in part be a backlash against the "Girlz rule, boyz drool" garbage that's going around in schools these days. At any rate, he showed me the following proof that women are evil. Though, I question the conclusion, I had to chuckle at the derivation. My apologies if this is an old joke in new guise. Theorem: Women are evil. Proof: Women require time and money: [1] women = time x money. But time is money: [2] time = money. and so [3] women = money^2. Now money is the root of all evil, [4] money = sqrt(evil), and so we conclude [5] women = evil. QED - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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RE: [math-fun] Re: Lotto & fortune cookies, from today's NYT
by dasimov＠earthlink.net 14 May '05

14 May '05

Rich wrote: << Thane contributes: << By Jennifer 8. Lee ... [Jennifer 8 Lee is no typo. That's her middle name: "8"] >> Clearly, 8 is her middle initial. Her middle name could be 88 or 8765 or even 8tricia. >> Or 8rocious. Because she wrote an incredibly stupid article recently (about how there's a new trend: hetero males getting together in a group of two in order to socialize!!!!!), I looked up some stuff about her, and she explains her adopting the 8. to distinguish herself from the many other Jennifer Lees. (And 8 because her background is Chinese and, she says, 8 is a very lucky number in Chinese culture.) So what do you call an "initial" that doesn't stand for anything? (Btw, such were once quite common in U.S. male given names.) --Dan

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