February 2006 - mathfun@mailman.xmission.com

Re: [math-fun] Kobon Triangle progress
by dasimov＠earthlink.net 08 Feb '06

08 Feb '06

Ed writes: << People curious about the Kobon triangle problem might be interested in the progress mentioned in my latest column. http://www.maa.org/editorial/mathgames/mathgames_02_08_06.html >> Very interesting question and column! I'm not certain, but I believe I've heard that Paul Erdos wrote a paper, possibly in collaboration, about this very problem. Sorry I don't have the citation. --Dan

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[math-fun] a conjecture based on Euler pol. and the Fermat's little theorem.
by Simon Plouffe 08 Feb '06

08 Feb '06

hello everybody, I had this idea one night a while ago, if k = n^2 - n + 41 and if 2^(k-1) = 1 mod k then k is prime. It seems to work all the time, I verified for n up to 500,000,000. In other words : n^2 - n + 41 is never a 2-psp, a pseudo-prime in base 2 (that is the conjecture). There is another one too : if k = n^2 + 3 and if 2^(k-1) = 1 mod k then k is prime. I tried to construct an argument about the factorization of n^2 - n + 41 and 2-psp's but failed to see any convincing details. I have other values that seems to work also like k = n^2 + 163 (only 1 fail), another counter example is 103*n^2 - 3945*n + 34831 when n = 2400371 it does satisfy the little fermat theorem but the number is composite : 5315987*111635707. and that polynomial (103*n^2..) is known to produce a lot of primes. Based on this I launched a series of computations but (as you may know), it all comes back to the fact that the little fermat theorem produces primes very often and if we combine it with a simple g.f. like the Euler polynomial, chances are that it will produce many primes. Even by knowing this, the best probable prime I could find is about 35000 digits. 1) I could not find any reference about this. 2) So far n^2 + 3 and the Euler pol. always produces primes and avoid the 2-psp's elegantly. 3) Maybe there is alink between class numbers and the little fermat theorem but I can't see it. My question is do you see any connections ? Simon Plouffe

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Re: [math-fun] parity of floor(2^n / n)
by Michael Reid 07 Feb '06

07 Feb '06

> What is known about the parity of floor of 2^n / n , > for general values of n? Some information is given a small observation: it seems that this is equivalent to knowing the remainder of 2^n modulo 2n ; if the remainder is 0, 1, 2, ... , n - 2 or n - 1 , then floor(2^n / n) is even, while if the remainder is n, n + 1, n + 2, ... , 2n - 2 or 2n - 1 , then floor(2^n / n) is odd. perhaps this formulation makes the problem more tractable; it certainly *looks* more promising. of course, the odd remainders cannot occur. more generally, if n = 2^k m , where m is odd, then the remainder modulo 2^(k+1) m is a multiple of 2^(k+1) , so it "suffices" to know the remainder modulo m , by the chinese remainder theorem. unfortunately, it does not appear immediately obvious how to determine from the remainder modulo m , whether the remainder modulo 2^(k+1) m is in {0, 1, ... , 2^k m - 1} . > The entry in the OEIS mentions that the parity is > always even when n is prime; this is a nice little > application of the binomial theorem. E.g., for > n = 5, (1+1)^5 = 1+5+10+10+5+1, so the floor of > 2^5 / 5 is 1+2+2+1, which is even. that's cute. that also holds for any pseudoprime to base 2, by the above remarks. indeed, if n is such a pseudoprime, then 2^n = 2 mod 2n . mike

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[math-fun] parity of floor(2^n / n)
by James Propp 07 Feb '06

07 Feb '06

What is known about the parity of floor of 2^n / n , for general values of n? Some information is given in the OEIS entry for the sequence A071354. Is it believed, or known, that as N approaches infinity, the number of n between 1 and N for which the floor of 2^n / n is even, divided by N, converges to 1/2 ? (This may be asking way too much: I don't even know how to prove that the floor of 2^n / n is odd for infinitely many values of n.) The entry in the OEIS mentions that the parity is always even when n is prime; this is a nice little application of the binomial theorem. E.g., for n = 5, (1+1)^5 = 1+5+10+10+5+1, so the floor of 2^5 / 5 is 1+2+2+1, which is even. Jim Propp

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[math-fun] Planar density question
by dasimov＠earthlink.net 07 Feb '06

07 Feb '06

Given integer n > 0, draw a line segment in C connecting each pair of nth roots of unity. Then as n -> oo, does the set of intersection points in C (assume each is given equal weight and the weights sum to 1) approach a continuous density on the unit disk? (Note: we care only about the intersection points, not the rest of the line segments.) Since for n even and > 4, the line segments must have at least one triple intersection (and I think usually more), it may be simpler to just consider n odd (in which case I think no triple intersections can occur, at least not in the *interior* of the unit circle). ------------------------------------------------------------------------ (((It's kind of clunky how you'd define a limiting distribution (if any), but it's done all the time in probability theory: Suppose for k = 1,2,3,... you're given a set of points S_k in R^2 -- and a corresponding discrete distribution obtained by giving each point q in S_k the probability Pr(q). Then there's a cumulative distribution F_k(a,b), defined by F_k(a,b) = sum of Pr(q) for all q = (qx, qy) in S_k such that qx <=a and qy <= b. Then the discrete distributions on the S_k's approach a "limit" as k -> oo if for almost all (a,b) in R^2, the limit as k -> oo of F_k(a,b) exists. In this case call the limiting cumulative distribution F(a,b). Then the limiting density will be given by d(q) = d^2 (F(a,b)) / (da db) assuming it exists and equals d^2 (F(a,b)) / (db da) almost evereywhere.))) There ought to be a cleaner way to define the limiting density, but I don't know what it is. So: Do these string figures' intersection points approach a limiting density, at least through n odd? --Dan

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[math-fun] Last term of the self-1-describing sequence ?
by Eric Angelini 06 Feb '06

06 Feb '06

Hello MathFun, This (partially) self-describing sequence says, for every a(n): the a(n)th *digit* of the sequence is a "1": 1,3,10,20,22,31,32,33,34,35,41,51,52,53,54,55,111,112,200,210,220, 222,231,1111... So the 1st digit of the sequence is indeed a "1", and the 3rd, and the 10th, and the 20st, and the 22nd, etc. ^^ This column is the sequence itself (which will replace A098645 in the OEIS when Neil will be back from Europe). Many such "self-1-describing" sequences can be build, for ins- tance this one (all "1"s must be described, not a few of them): 1,4,5,11,20,22,31,... Question: Find the *finite* such "self-1-describing" sequence where 16 "1"s can be seen and where the last term is the smallest one. Best, É. (I'll send a bunch of "self-n-describing" such seq to the OEIS this week or so; BTW I will change also A098670 because the concept of "slowest growingness" is ambiguous -- actually quite impossible to define for such seq.)

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RE: [math-fun] Why learn algebra?
by Eric Angelini 06 Feb '06

06 Feb '06

[David]: > One of the most basic faults of high school education is failure to teach students to stick with an unpleasant task. Inability to deal with unpleasantness will not serve students in their careers. Ergo algebra. ... I agree 100% -- though algebra has never been unpleasant for me! Best, É. (I'll send a puzzle to the list in a couple of hours for been slightly off topic with this remark)

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Re: [math-fun] Why learn algebra?
by Michael Reid 05 Feb '06

05 Feb '06

> This is not a (political) troll. Due to recent test results in LA & other > places, people are starting to ask the question "why teach algebra?" I've as usual, the pundits miss the point. the real evil is not algebra; it is *reading*, and that's what should be slated for immediate removal from the school curriculum. this may seem extreme, but consider the potential market for optical character recognition devices equipped with voice output, as well as the need for periodic software and hardware upgrades. written books are out; audio books are in. why do students need to learn to read? they can get all the infotainment, er, news, they need from tv anyway. between audio books and OCR devices for use on old written media, reading is essentially unnecessary. oh, sure perhaps reading should be taught to a select group of elite schoolchildren who receive a thorough classical education, but that's only for the privileged elite. for the masses, why don't we dumb down everything to the barest minimum? education is out; edu-tainment is in! with tongue firmly in cheek, mike

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Re: [math-fun] Why learn algebra?
by dasimov＠earthlink.net 05 Feb '06

05 Feb '06

Marc LeBrun writes: << Henry wriote: > This is not a (political) troll..."why teach algebra?" I know you sincerely do not intend it as a troll, but since "why teach X?" is ultimately a social question, it would seem impossible for this to remain a technical or even philosophical discussion. (Personally, I believe the whole issue is moot (ie arguable but of no significance). As, inevitably, the effectiveness of automation improves and the wealth of the world grows, the fraction of humanity for which it is of value to know anything about anything whatsoever is rapidly becoming asymptotically zero...it's all ultimately "enrichment exercises" for which there will uniformly be decreasingly compelling pragmatic (versus esthetic or moral) justification (which was the sort you were soliciting). I view this evolution as a good thing, but recognize others may not, and agree math-fun is not the best venue to contest such positions, so I'll leave it at that). >> Funny, Marc's comment reminds me of The Feeling of Power, a short story starring Myron Aub. What is amazing is that I read that story over 45 years ago and have not thought of it once in the interim, yet I remembered the name "Myron Aub". Go *figure*. --Dan P.S. I happen to think that questions of math education are entirely appropriate for this forum, and are important as well.

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[math-fun] Why learn algebra?
by Henry Baker 05 Feb '06

05 Feb '06

This is not a (political) troll. Due to recent test results in LA & other places, people are starting to ask the question "why teach algebra?" I've been pondering this question for years, and have never been able to come up with a simple answer that would be compelling to a non-mathematician, non-engineer. Does anyone on this list have such an argument? ------------------------- Steve Lopez of the LA Times: "One reason for the dropout problem is that kids are flunking algebra. In the fall of 2004, 44% of the district's 48,000 ninth-graders came up short, and Times reporter Duke Helfand found a Birmingham student who flunked algebra six times. This is because huge numbers of students can't do basic math, let alone algebra, which in 2003 became a requirement for graduation in all LAUSD schools. In future years, no one will get a diploma without three years of advanced math. At the very least, the district should have phased in the existing algebra requirement, beefing up basic math instruction first. But even at that, why force all students to take advanced math courses, driving out those who might never use algebra in the real world? Sure, good jobs await those who know math and science, and yes, algebraic computations are a mind-expanding exercise. But so is the reading of great books. We ought to make sure students can take all the trig and calculus they want. But for others, I'd rather see a reality-based math curriculum that prepares students for life, equipping them to balance a checkbook, understand credit card finance charges and challenge the IRS." President Bush on algebra: "You know," Bush said, "a lot of people probably think 'Math and science isn't meant for me Â it kind of seems a little hard, algebra.' I can understand that, frankly." Even now, Bush said, he too could use some expert guidance. "I'm looking for a mentor, by the way," he said to laughter, "both in math and English."

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