February 2009 - mathfun@mailman.xmission.com

[math-fun] Isodiametric problem
by victor miller 16 Sep '10

16 Sep '10

A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter. He reduced the problem to finding univariate polynomials whose non-zero coefficients alternate between +/- 1 and are divisible by a particular cyclotomic polynomial.. He was able to calculate all such up to about 500. Each such polynomial is compactly encoded by th length of the gaps between the non-zero terms. For small n almost all were periodic. He showed how a certain family of periodic constrctions give solutions, but as n got bigger the "sporadic" ones swamped them. He conjectued that these are eventually all of them. He had lots of pictures. The paper is not available yet. Victor

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Re: [math-fun] Fast factoring ?? (really denesting)
by rwg＠sdf.lonestar.org 08 May '09

08 May '09

> On Fri, Dec 19, 2008 at 12:20 PM, <rwg(a)sdf.lonestar.org> wrote: > >> Amen. I've now tried ~ 10^8 cases of (a^(1/5)-b^(1/5))^(1/3) >> without further success, and for sqrt, only >> >> 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 >> 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 >> sqrt(4 - 3 ) = ---------------------------------------------------. >> 5 >> >> And, of course, these can only give pentanomials. >> --rwg >> ALGORISMIC MICROGLIAS > > I have no technical knowledge on denesting, but here is how I look at > the problem: > > Since sqrt( a^(1/5) + b^(1/5) ) = a^(-2/5) * sqrt( a + (a^4*b)^(1/5) > ), I'll consider only the form: sqrt( a + b^(1/5) ). If sqrt( a + > b^(1/5) ) could be denested, I'd expect it to be denested into the > form: > > x0 + x1*b^(1/5) + x2*b^(2/5) + x3*b^(3/5) + x4*b^(4/5). Notice that in the (comparatively easy) 4th root case, your conjecture catches 1/4 3/4 1/4 sqrt(161 - 12 5 ) = 2 5 - 3 sqrt(5) + 4 5 + 6 but not 1/4 1/4 1/4 3/4 3/4 1/4 sqrt(2) sqrt(31 - 4 3 5 ) = 3 5 - 2 sqrt(5) + 3 5 + 2 sqrt(3). > That's why I don't think you could find anything longer than > pentanomial. OTOH, sqrt( a^(1/7) + b^(1/7) ) is a better bet for > finding hexanomials or longer. Indeed, but the only one I've found so far is perversely pentanomial: 1/7 Sqrt(7 (2 - 2 )) 1/7 3/7 5/7 6/7 ------------------ = -1 + 2 2 + 2 + 2 - 2 . 1/14 2 > Just my 2 cents. > > Warut > You didn't propose 6th roots. Did you have reason to suspect that they are as infertile as they seem to be? --rwg PRECHRISTMAS PETRARCHISMS ALGORISMIC MICROGLIAS

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Re: [math-fun] muffin problem
by Michael Reid 13 Mar '09

13 Mar '09

> I've calcuated U(4,13) = 11/6 (using Veit's notation) using CPLEX > (industrial strength MIP) with the MIP that Veit suggested. i can now determine T(4, n) in general. recall that my normalization is that T(m, n) equals the maximum c such that there is a partition of 1 with all parts >= c , which is simultaneously a refinement of 1/m + 1/m + ... + 1/m = 1 and also of 1/n + 1/n + ... + 1/n = 1 . T(4, 4n) = 1/4n , with unique partition 4n * 1/(4n) . this is obvious. T(4, 4n+2) = 1/(8n+4) . 4 * [(2n+1) * 1/(8n+4)] <---> (4n+2) * [1/(8n+4) + 1/(8n+4)] achieves this bound. moreover, at least one 1/(4n+2) must be split, so it has a part at most 1/(8n+4) . the partition is not unique; at least two 1/(4n+2) 's must split into 1/(8n+4) + 1/(8n+4) . the others either split in the same way, or do not split. T(4, 2n+1) = (2n-1)/(4n(2n+1)) if n >= 1 . the case n = 1 is T(4, 3) = T(3, 4) = 1/12 , which is easy to handle separately. if n > 1 , there is a unique optimal partition, which is 2 * [n * (2n-1)/(4n(2n+1)) + 1/(4n+2)] + 2 * [n * 1/(4n)] <---> 2n * [(2n-1)/(4n(2n+1)) + 1/(4n)] + [1/(4n+2) + 1/(4n+2)] suppose that 1/4 splits as 1/(2n+1) + 1/(2n+1) + ... + 1/(2n+1) + ( p smaller parts ) , where there are k > 0 parts of size 1/(2n+1) . if p >= n - 2k + 1 , then the smallest of the parts is at most (1/4 - k/(2n+1))/p <= ((2n-4k+1)/(4(2n+1)))/(n - 2k + 1) = (2 - 1/(n - 2k + 1))/(4(2n+1)) < (2 - 1/n)/(4(2n+1)) = (2n-1)/(4n(2n+1)) , so this part is too small. on the other hand, if p <= n - 2k , then the largest of the parts not equal to 1/(2n+1) is at least (1/4 - k/(2n+1))/p . since this occurs as a part of 1/(2n+1) , it has another part at most 1/(2n+1) - (1/4 - k/(2n+1))/p <= 1/(2n+1) - (1/4 - k/(2n+1))/(n - 2k) = (2 - 1/(n - 2k))/(4n(2n+1)) < (2 - 1/n)/(4(2n+1)) = (2n-1)/(4n(2n+1)) , so again there is a part which is too small. this shows that no 1/4 can contain any parts of size 1/(2n+1) , and therefore every 1/(2n+1) must split. now suppose that 1/4 splits into p parts, which we know must all be less than 1/(2n+1) . if p >= n+2 , then the smallest of these parts is <= 1/(4(n+2)) < (2n-1)/(4n(2n+1)) , so too small. (here we need n > 1 .) if p <= n-1 , then the largest of these parts is at least 1/(4(n-1)) , and since it occurs as part of 1/(2n+1) , there must be another part of size <= 1/(2n+1) - 1/(4(n-1)) < (2n-1)/(4n(2n+1)) , which is again too small. therefore, each 1/4 splits into either n or n+1 parts. if 1/(2n+1) splits into 3 or more parts, then some part has size <= 1/(3(2n+1)) , which is less than (2n-1)/(4n(2n+1)) if n > 1 . thus each 1/(2n+1) splits into exactly 2 parts, so there are 4n+2 parts in total. now two of the 1/4 's must split into n parts, and the other two must split into n+1 parts. moreover, the largest part in either of the 1/4 's that split into n parts has size >= 1/(4n) . since it occurs as part of 1/(2n+1) , the other part has size <= 1/(2n+1) - 1/(4n) = (2n-1)/(4n(2n+1)) . thus, to have optimality, both of the 1/4 's that split into 1/(4n) + 1/(4n) + ... + 1/(4n) , and now n of the 1/(2n+1) 's must split as (2n-1)/(4n(2n+1)) + 1/(4n) . finally, there are 2n parts of the minimal size (2n-1)/(4n(2n+1)) , and neither of the remaining 1/4 's can contain n+1 of them , so both contain n of them, and thus split as n * (2n-1)/(4n(2n+1)) + 1/(4n+2) , which determines the partition. T(1, n) and T(2, n) are very easy to determine in general. it looks like there is probably also pattern in T(3, n) . T(6, n) might also be possible to determine in general. mike

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[math-fun] theta[s](pi/6,q), or eta as theta again
by rwg＠sdf.lonestar.org 07 Mar '09

07 Mar '09

A couple of years ago I restated an almost forgotten identity equivalent to %pi %pi 6 theta (---, q) = theta (---, q) = sqrt(3) eta(q ) 2 6 1 3 3 2 6 theta (0, q ) theta (0, q ) 2 4 1/3 = sqrt(3) (---------------------------) , 2 in this form as special values of Theta(pi/n) in terms of eta or equivalently the "theta constants" theta[s](0,q^k). What is the analogous value with the lhs 1 and 2 switched? 2 2 12 %pi %pi eta (q ) eta(q ) theta (---, q) = theta (---, q) = ----------------- 2 3 1 6 4 6 eta(q ) eta(q ) 3 9 theta (0, q ) theta (0, q) - 3 theta (0, q ) 2 2 1/3 2 2 = theta (0, q ) (---------------) = ------------------------------. 4 6 2 2 theta (0, q ) 4 This is a rich area of nonobvious(?) identities, e.g., 2 6 6 theta (0, q) theta (0, q ) theta (0, q ) 2 2 3 2 2 2 3 = theta (0, q ) theta (0, q ) theta (0, q ), 2 3 2 or equivalently 2 6 theta (0, q) theta (0, q) theta (0, q ) 3 4 4 2 2 3 3 = theta (0, q ) theta (0, q ) theta (0, q ) 4 3 4 --rwg

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Re: [math-fun] Chocolate bar puzzle
by James Propp 01 Mar '09

01 Mar '09

Dan Asimov writes: >Speaking of muffinsectors, here is a problem about chocosectors: > >You are given a rectangular chocolate bar that is scored in the >usual way into KxL small squares. Repeat this step: Choose one >of the pieces (at first, there's only the whole bar) and break >it all the way along one of its scorings . . . until all the >pieces are the small squares. There is a fun way to present this problem so that it has the surprising property that mathematically sophisticated undergrads take LONGER to see the answer than less sophisticated undergrads. (It'd be fun to make a whole collection of problems that smart and/or knowledgeable people tend to do poorly on: an anti-intelligence or anti-aptitude test if you will.) Specifically, present the candy bar problem as a problem in combinatorial game theory, with players taking turns, and ask WHO WINS?, without revealing that the parity of the duration of the game, and indeed the duration of the game itself, isn't affected by the strategies the players adopt. (This is in fact the way I invented this problem twenty years ago: I like to call the game "Impartial Cutcake" to disguise the fact that there's no real game-theory going on at all.) Lead the students to prove that an even-by-even or even-by-odd bar is a win for the first player (strategy: the first player cuts the bar into two identical pieces), and a fancier argument to prove that an odd-by-odd bar is a win for the second player (imagine that the pieces are still in place, and copy whatever your opponent just did but rotated 180 degrees). Now challenge the students, working alone or in small groups, to figure out who wins under optimal play when the initial configuration is a pair of chocolate bars, one a-by-b and one c-by-d. It is vital to keep a straight face, and not to hint that there's lurking triviality. My experience is that students who know combinatorial game theory jump in and compute Grundy values and get really distracted from the heart of the matter. Students who don't know the theory of impartial games but who have some mathemtical taste and a fondness for proof take longer to find the slick solution than the others. Why? I think it's partly because they value the work that the group has already done in finding strategies for the players, so they emotionally resist making a discovery that would render the earlier arguments pointless. Also, they're used to the idea that in math, you keep building on what you already know, so they want to use what the group has already proved about single rectangles as a tool for analyzing the two-rectangle case. Also, they're inclined to use a trick that's worked before on new problems to see how far the trick can be pushed. The less sophisticated students have a kind of blankness that on the one hand makes it hard for them to get started on the problem but on the other hand keeps them from getting distracted once they do start. I did meet one sophisticate who put his sophistication to good use by imediately asking "What are the invariants here?", but this kind of reaction was not the norm. Jim Propp

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[math-fun] common refinements of partitions (was: muffins)
by Michael Reid 01 Mar '09

01 Mar '09

i'm glad to see that there is still some interest in the muffin/common refinement of partitions problem. some time ago, veit elser gave the upper bound T(m, n) <= 1/(3m) for m < n < 3m/2 , and noted that some of the values he found achieved this bound. i looked at this range more closely and found that i could improve the upper bound as follows. here t denotes a positive integer. if ((3t+1)/(3t))m < n <= ((3t)/(3t-1))m , then T(m, n) <= 1/(3n) . if ((3t+2)/(3t+1))m < n <= ((3t+1)/(3t))m , then T(m, n) <= (t+1)/(2n) - t/(2m) . if ((3t+3)/(3t+2))m < n < ((3t+2)/(3t+1))m , then T(m, n) <= (t+1)/(4m) - t/(4n) . if ((3t+3)/(3t+2))m = n , then T(m, n) <= 1/(3m) . (here, as in previous messages, i am using T(m, n) = maximum c such that there is a common refinement of 1/m + 1/m + ... + 1/m = 1 and of 1/n + 1/n + ... + 1/n = 1 , with all parts >= c .) the first three cases are all improvements over veit's bound. in the fourth case, where there is no improvement, equality is always attained by veit's construction for T(3t+2, 3t+3) , and by obvious extension to T((3t+2)k, (3t+3)k) . however, the above shows that this is the exception, rather than the rule. i find the first case interesting, because i have not seen any cases (with m <= n ) where T(m, n) is strictly less than 1/(3n) . if we allow parts of size 1/(3n) , we may split some of the 1/n 's into 3 parts, and that seems to allow a lot of flexibility. in fact, the only occurrences of T(m, n) = 1/(3n) i've seen are those from the first case above. so i conjecture that equality always holds in the first case. on the other hand, the bounds in the other cases cannot always be attained. they can be attained if n = m+1 , by veit's and moulton's constructions. they can also be attained if n = m+2 by similar constructions. however, if n >= m + 3 , there are cases that require more careful analysis and for which the upper bound can be lowered; the first such case is (13, 16) . when n = m + 3 , it seems that the upper bound can be attained in all cases except when m = 4 mod 9 ; in that case the exact value is T(9k+4, 9k+7) = (9k+5)/(3(9k+4)(9k+7)) for k >= 1 . below are some data i calculated, all by pencil and paper, so subject to a small possibility of error. obviously this amount of data starts to become unwieldy; corrections and/or confirmations are welcome! i believe i have a proof for each of these values, but may have given short shrift to some subcases i thought couldn't give optimal partitions. mike T(2, 3) = 1/6 T(2, 4) = 1/4 T(3, 4) = 1/12 T(2, 5) = 1/10 T(3, 5) = 1/12 T(4, 5) = 3/40 T(2, 6) = 1/6 T(3, 6) = 1/6 T(4, 6) = 1/12 T(5, 6) = 1/15 T(2, 7) = 1/14 T(3, 7) = 5/84 T(4, 7) = 5/84 T(5, 7) = 1/21 T(6, 7) = 1/21 T(2, 8) = 1/8 T(3, 8) = 1/18 T(4, 8) = 1/8 T(5, 8) = 1/20 T(6, 8) = 1/24 T(7, 8) = 5/112 T(2, 9) = 1/18 T(3, 9) = 1/9 T(4, 9) = 7/144 T(5, 9) = 2/45 T(6, 9) = 1/18 T(7, 9) = 5/126 T(8, 9) = 1/24 T(2, 10) = 1/10 T(3, 10) = 2/45 T(4, 10) = 1/20 T(5, 10) = 1/10 T(6, 10) = 1/24 T(7, 10) = 1/30 T(8, 10) = 3/80 T(9, 10) = 1/30 T(2, 11) = 1/22 T(3, 11) = 1/24 T(4, 11) = 9/220 T(5, 11) = 13/330 T(6, 11) = 7/198 T(7, 11) = 1/28 T(8, 11) = 1/33 T(9, 11) = 13/396 T(10, 11) = 7/220 T(2, 12) = 1/12 T(3, 12) = 1/12 T(4, 12) = 1/12 T(5, 12) = 1/30 T(6, 12) = 1/12 T(7, 12) = 1/28 T(8, 12) = 1/24 T(9, 12) = 1/36 T(10, 12) = 1/30 T(11, 12) = 1/33 T(2, 13) = 1/26 T(3, 13) = 11/312 T(4, 13) = 11/312 T(5, 13) = 1/30 T(6, 13) = 1/30 T(7, 13) = 8/273 T(8, 13) = 1/32 T(9, 13) = 1/39 T(10, 13) = 7/260 T(11, 13) = 1/39 T(12, 13) = 1/39 T(2, 14) = 1/14 T(3, 14) = 1/30 T(4, 14) = 1/28 T(5, 14) = 11/350 T(6, 14) = 5/168 T(7, 14) = 1/14 T(8, 14) = 5/168 T(9, 14) = 1/36 T(10, 14) = 1/42 T(11, 14) = 2/77 T(12, 14) = 1/42 T(13, 14) = 9/364 T(2, 15) = 1/30 T(3, 15) = 1/15 T(4, 15) = 13/420 T(5, 15) = 1/15 T(6, 15) = 1/30 T(7, 15) = 1/35 T(8, 15) = 1/40 T(9, 15) = 1/36 T(10, 15) = 1/30 T(11, 15) = 1/45 T(12, 15) = 1/40 T(13, 15) = 3/130 T(14, 15) = 1/42 T(2, 16) = 1/16 T(3, 16) = 7/240 T(4, 16) = 1/16 T(5, 16) = 1/35 T(6, 16) = 1/36 T(7, 16) = 3/112 T(8, 16) = 1/16 T(9, 16) = 11/432 T(10, 16) = 1/40 T(11, 16) = 1/48 T(12, 16) = 1/48 T(13, 16) = 7/312 T(14, 16) = 5/224 T(15, 16) = 1/48 T(2, 17) = 1/34 T(3, 17) = 1/36 T(4, 17) = 15/544 T(5, 17) = 13/510 T(6, 17) = 13/510 T(7, 17) = 11/476 T(8, 17) = 1/40 T(9, 17) = 10/459 T(10, 17) = 2/85 T(11, 17) = 1/44 T(12, 17) = 1/51 T(13, 17) = 9/442 T(14, 17) = 5/238 T(15, 17) = 7/340 T(16, 17) = 11/544 T(2, 18) = 1/18 T(3, 18) = 1/18 T(4, 18) = 1/36 T(5, 18) = 1/40 T(6, 18) = 1/18 T(7, 18) = 1/42 T(8, 18) = 7/288 T(9, 18) = 1/18 T(10, 18) = 1/45 T(11, 18) = 1/44 T(12, 18) = 1/36 T(13, 18) = 1/54 T(14, 18) = 5/252 T(15, 18) = 1/45 T(16, 18) = 1/48 T(17, 18) = 1/51 T(2, 19) = 1/38 T(3, 19) = 17/684 T(4, 19) = 17/684 T(5, 19) = 16/665 T(6, 19) = 1/42 T(7, 19) = 25/1064 T(8, 19) = 13/608 T(9, 19) = 1/45 T(10, 19) = 11/570 T(11, 19) = 9/418 T(12, 19) = 1/48 T(13, 19) = 1/57 T(14, 19) = 1/57 T(15, 19) = 7/380 T(16, 19) = 1/57 T(17, 19) = 1/57 T(18, 19) = 1/57 T(2, 20) = 1/20 T(3, 20) = 1/42 T(4, 20) = 1/20 T(5, 20) = 1/20 T(6, 20) = 1/45 T(7, 20) = 3/140 T(8, 20) = 1/40 T(9, 20) = 1/45 T(10, 20) = 1/20 T(11, 20) = 13/660 T(12, 20) = 1/48 T(13, 20) = 1/52 T(14, 20) = 1/60 T(15, 20) = 1/60 T(16, 20) = 3/160 T(17, 20) = 1/60 T(18, 20) = 1/60 T(19, 20) = 13/760 T(2, 21) = 1/42 T(3, 21) = 1/21 T(4, 21) = 19/840 T(5, 21) = 1/45 T(6, 21) = 1/42 T(7, 21) = 1/21 T(8, 21) = 1/48 T(9, 21) = 5/252 T(10, 21) = 1/50 T(11, 21) = 4/231 T(12, 21) = 5/252 T(13, 21) = 1/52 T(14, 21) = 1/42 T(15, 21) = 1/63 T(16, 21) = 11/672 T(17, 21) = 2/119 T(18, 21) = 1/63 T(19, 21) = 13/798 T(20, 21) = 1/60 T(2, 22) = 1/22 T(3, 22) = 5/231 T(4, 22) = 1/44 T(5, 22) = 9/440 T(6, 22) = 1/48 T(7, 22) = 1/49 T(8, 22) = 9/440 T(9, 22) = 7/396 T(10, 22) = 13/660 T(11, 22) = 1/22 T(12, 22) = 7/396 T(13, 22) = 21/1144 T(14, 22) = 1/56 T(15, 22) = 1/66 T(16, 22) = 1/66 T(17, 22) = 3/187 T(18, 22) = 13/792 T(19, 22) = 13/836 T(20, 22) = 7/440 T(21, 22) = 1/66 T(2, 23) = 1/46 T(3, 23) = 1/48 T(4, 23) = 21/1012 T(5, 23) = 1/50 T(6, 23) = 19/966 T(7, 23) = 19/966 T(8, 23) = 17/920 T(9, 23) = 1/54 T(10, 23) = 17/920 T(11, 23) = 1/55 T(12, 23) = 13/828 T(13, 23) = 53/2990 T(14, 23) = 17/966 T(15, 23) = 1/60 T(16, 23) = 1/69 T(17, 23) = 1/69 T(18, 23) = 19/1242 T(19, 23) = 27/1748 T(20, 23) = 7/460 T(21, 23) = 29/1932 T(22, 23) = 15/1012 T(2, 24) = 1/24 T(3, 24) = 1/24 T(4, 24) = 1/24 T(5, 24) = 7/360 T(6, 24) = 1/24 T(7, 24) = 1/56 T(8, 24) = 1/24 T(9, 24) = 1/54 T(10, 24) = 1/60 T(11, 24) = 19/1056 T(12, 24) = 1/24 T(13, 24) = 5/312 T(14, 24) = 1/56 T(15, 24) = 1/60 T(16, 24) = 1/48 T(17, 24) = 1/72 T(18, 24) = 1/72 T(19, 24) = 9/608 T(20, 24) = 1/60 T(21, 24) = 5/336 T(22, 24) = 1/66 T(23, 24) = 1/69 T(2, 25) = 1/50 T(3, 25) = 23/1200 T(4, 25) = 23/1200 T(5, 25) = 1/25 T(6, 25) = 1/54 T(7, 25) = 1/56 T(8, 25) = 1/56 T(9, 25) = 4/225 T(10, 25) = 1/50 T(11, 25) = 19/1100 T(12, 25) = 1/60 T(13, 25) = 14/975 T(14, 25) = 17/1050 T(15, 25) = 1/60 T(16, 25) = 1/64 T(17, 25) = 1/75 T(18, 25) = 1/75 T(19, 25) = 13/950 T(20, 25) = 3/200 T(21, 25) = 1/75 T(22, 25) = 23/1650 T(23, 25) = 1/75 T(24, 25) = 1/75 T(2, 26) = 1/26 T(3, 26) = 1/54 T(4, 26) = 1/52 T(5, 26) = 1/55 T(6, 26) = 11/624 T(7, 26) = 1/56 T(8, 26) = 11/624 T(9, 26) = 19/1170 T(10, 26) = 1/60 T(11, 26) = 9/572 T(12, 26) = 1/60 T(13, 26) = 1/26 T(14, 26) = 4/273 T(15, 26) = 37/2340 T(16, 26) = 1/64 T(17, 26) = 1/68 T(18, 26) = 1/78 T(19, 26) = 1/78 T(20, 26) = 7/520 T(21, 26) = 11/819 T(22, 26) = 1/78 T(23, 26) = 4/299 T(24, 26) = 1/78 T(25, 26) = 17/1300 T(2, 27) = 1/54 T(3, 27) = 1/27 T(4, 27) = 25/1404 T(5, 27) = 23/1350 T(6, 27) = 1/54 T(7, 27) = 22/1323 T(8, 27) = 7/432 T(9, 27) = 1/27 T(10, 27) = 1/60 T(11, 27) = 17/1188 T(12, 27) = 7/432 T(13, 27) = 1/65 T(14, 27) = 5/378 T(15, 27) = 2/135 T(16, 27) = 13/864 T(17, 27) = 1/68 T(18, 27) = 1/54 T(19, 27) = 1/81 T(20, 27) = 1/81 T(21, 27) = 5/378 T(22, 27) = 4/297 T(23, 27) = 1/81 T(24, 27) = 1/72 T(25, 27) = 17/1350 T(26, 27) = 1/78 T(2, 28) = 1/28 T(3, 28) = 13/756 T(4, 28) = 1/28 T(5, 28) = 1/60 T(6, 28) = 1/60 T(7, 28) = 1/28 T(8, 28) = 1/56 T(9, 28) = 1/63 T(10, 28) = 11/700 T(11, 28) = 1/66 T(12, 28) = 5/336 T(13, 28) = 1/65 T(14, 28) = 1/28 T(15, 28) = 17/1260 T(16, 28) = 5/336 T(17, 28) = 1/68 T(18, 28) = 1/72 T(19, 28) = 1/84 T(20, 28) = 1/84 T(21, 28) = 1/84 T(22, 28) = 1/77 T(23, 28) = 33/2576 T(24, 28) = 1/84 T(25, 28) = 1/84 T(26, 28) = 9/728 T(27, 28) = 1/84 T(2, 29) = 1/58 T(3, 29) = 1/60 T(4, 29) = 27/1624 T(5, 29) = 26/1595 T(6, 29) = 25/1566 T(7, 29) = 1/63 T(8, 29) = 1/64 T(9, 29) = 41/2610 T(10, 29) = 21/1450 T(11, 29) = 1/66 T(12, 29) = 19/1392 T(13, 29) = 45/3016 T(14, 29) = 1/70 T(15, 29) = 16/1305 T(16, 29) = 19/1392 T(17, 29) = 27/1972 T(18, 29) = 1/72 T(19, 29) = 1/76 T(20, 29) = 1/87 T(21, 29) = 1/87 T(22, 29) = 15/1276 T(23, 29) = 49/4002 T(24, 29) = 17/1392 T(25, 29) = 17/1450 T(26, 29) = 1/87 T(27, 29) = 37/3132 T(28, 29) = 19/1624 T(2, 30) = 1/30 T(3, 30) = 1/30 T(4, 30) = 1/60 T(5, 30) = 1/30 T(6, 30) = 1/30 T(7, 30) = 13/840 T(8, 30) = 13/840 T(9, 30) = 2/135 T(10, 30) = 1/30 T(11, 30) = 1/66 T(12, 30) = 1/60 T(13, 30) = 11/780 T(14, 30) = 1/70 T(15, 30) = 1/30 T(16, 30) = 1/80 T(17, 30) = 7/510 T(18, 30) = 1/72 T(19, 30) = 1/76 T(20, 30) = 1/60 T(21, 30) = 1/90 T(22, 30) = 1/90 T(23, 30) = 4/345 T(24, 30) = 1/80 T(25, 30) = 1/75 T(26, 30) = 3/260 T(27, 30) = 1/90 T(28, 30) = 1/84 T(29, 30) = 1/87 T(2, 31) = 1/62 T(3, 31) = 29/1860 T(4, 31) = 29/1860 T(5, 31) = 1/65 T(6, 31) = 1/66 T(7, 31) = 25/1736 T(8, 31) = 25/1736 T(9, 31) = 23/1674 T(10, 31) = 1/70 T(11, 31) = 24/1705 T(12, 31) = 1/72 T(13, 31) = 21/1612 T(14, 31) = 3/217 T(15, 31) = 1/75 T(16, 31) = 17/1488 T(17, 31) = 20/1581 T(18, 31) = 11/837 T(19, 31) = 54/4123 T(20, 31) = 1/80 T(21, 31) = 1/93 T(22, 31) = 1/93 T(23, 31) = 1/93 T(24, 31) = 25/2232 T(25, 31) = 26/2325 T(26, 31) = 1/93 T(27, 31) = 37/3348 T(28, 31) = 19/1736 T(29, 31) = 1/93 T(30, 31) = 1/93 T(2, 32) = 1/32 T(3, 32) = 1/66 T(4, 32) = 1/32 T(5, 32) = 7/480 T(6, 32) = 7/480 T(7, 32) = 1/70 T(8, 32) = 1/32 T(9, 32) = 1/72 T(10, 32) = 1/70 T(11, 32) = 23/1760 T(12, 32) = 1/72 T(13, 32) = 5/416 T(14, 32) = 3/224 T(15, 32) = 1/75 T(16, 32) = 1/32 T(17, 32) = 19/1632 T(18, 32) = 11/864 T(19, 32) = 31/2432 T(20, 32) = 1/80 T(21, 32) = 1/84 T(22, 32) = 1/96 T(23, 32) = 1/96 T(24, 32) = 1/96 T(25, 32) = 11/1000 T(26, 32) = 7/624 T(27, 32) = 1/96 T(28, 32) = 5/448 T(29, 32) = 5/464 T(30, 32) = 1/96 T(31, 32) = 21/1984 T(2, 33) = 1/66 T(3, 33) = 1/33 T(4, 33) = 31/2112 T(5, 33) = 1/70 T(6, 33) = 1/66 T(7, 33) = 1/70 T(8, 33) = 1/72 T(9, 33) = 1/72 T(10, 33) = 3/220 T(11, 33) = 1/33 T(12, 33) = 3/220 T(13, 33) = 1/78 T(14, 33) = 23/1848 T(15, 33) = 13/990 T(16, 33) = 1/80 T(17, 33) = 2/187 T(18, 33) = 7/594 T(19, 33) = 47/3762 T(20, 33) = 49/3960 T(21, 33) = 1/84 T(22, 33) = 1/66 T(23, 33) = 1/99 T(24, 33) = 1/99 T(25, 33) = 17/1650 T(26, 33) = 3/286 T(27, 33) = 13/1188 T(28, 33) = 1/99 T(29, 33) = 10/957 T(30, 33) = 7/660 T(31, 33) = 7/682 T(32, 33) = 1/96 T(2, 34) = 1/34 T(3, 34) = 8/561 T(4, 34) = 1/68 T(5, 34) = 31/2210 T(6, 34) = 1/72 T(7, 34) = 29/2142 T(8, 34) = 15/1088 T(9, 34) = 29/2142 T(10, 34) = 13/1020 T(11, 34) = 1/77 T(12, 34) = 13/1020 T(13, 34) = 1/78 T(14, 34) = 11/952 T(15, 34) = 13/1020 T(16, 34) = 1/80 T(17, 34) = 1/34 T(18, 34) = 5/459 T(19, 34) = 23/1938 T(20, 34) = 1/85 T(21, 34) = 1/84 T(22, 34) = 1/88 T(23, 34) = 1/102 T(24, 34) = 1/102 T(25, 34) = 1/102 T(26, 34) = 9/884 T(27, 34) = 8/765 T(28, 34) = 5/476 T(29, 34) = 1/102 T(30, 34) = 7/680 T(31, 34) = 16/1581 T(32, 34) = 11/1088 T(33, 34) = 1/102 T(2, 35) = 1/70 T(3, 35) = 1/72 T(4, 35) = 33/2380 T(5, 35) = 1/35 T(6, 35) = 31/2310 T(7, 35) = 1/35 T(8, 35) = 29/2240 T(9, 35) = 4/315 T(10, 35) = 1/70 T(11, 35) = 1/77 T(12, 35) = 1/84 T(13, 35) = 64/5005 T(14, 35) = 1/70 T(15, 35) = 1/84 T(16, 35) = 1/80 T(17, 35) = 1/85 T(18, 35) = 19/1890 T(19, 35) = 22/1995 T(20, 35) = 1/84 T(21, 35) = 1/84 T(22, 35) = 1/88 T(23, 35) = 1/92 T(24, 35) = 1/105 T(25, 35) = 1/105 T(26, 35) = 1/105 T(27, 35) = 5/504 T(28, 35) = 3/280 T(29, 35) = 41/4060 T(30, 35) = 1/105 T(31, 35) = 43/4340 T(32, 35) = 11/1120 T(33, 35) = 3/308 T(34, 35) = 23/2380 T(2, 36) = 1/36 T(3, 36) = 1/36 T(4, 36) = 1/36 T(5, 36) = 1/75 T(6, 36) = 1/36 T(7, 36) = 1/77 T(8, 36) = 1/72 T(9, 36) = 1/36 T(10, 36) = 1/80 T(11, 36) = 5/396 T(12, 36) = 1/36 T(13, 36) = 29/2340 T(14, 36) = 1/84 T(15, 36) = 1/90 T(16, 36) = 7/576 T(17, 36) = 1/85 T(18, 36) = 1/36 T(19, 36) = 7/684 T(20, 36) = 1/90 T(21, 36) = 1/84 T(22, 36) = 1/88 T(23, 36) = 1/92 T(24, 36) = 1/72 T(25, 36) = 1/108 T(26, 36) = 1/108 T(27, 36) = 1/108 T(28, 36) = 5/504 T(29, 36) = 5/522 T(30, 36) = 1/90 T(31, 36) = 7/744 T(32, 36) = 1/96 T(33, 36) = 1/99 T(34, 36) = 1/102 T(35, 36) = 1/105 T(2, 37) = 1/74 T(3, 37) = 35/2664 T(4, 37) = 35/2664 T(5, 37) = 33/2590 T(6, 37) = 1/78 T(7, 37) = 33/2590 T(8, 37) = 1/80 T(9, 37) = 1/81 T(10, 37) = 1/80 T(11, 37) = 29/2442 T(12, 37) = 1/84 T(13, 37) = 28/2405 T(14, 37) = 1/84 T(15, 37) = 23/2220 T(16, 37) = 27/2368 T(17, 37) = 59/5032 T(18, 37) = 1/90 T(19, 37) = 20/2109 T(20, 37) = 23/2220 T(21, 37) = 103/9324 T(22, 37) = 9/814 T(23, 37) = 1/92 T(24, 37) = 1/96 T(25, 37) = 1/111 T(26, 37) = 1/111 T(27, 37) = 1/111 T(28, 37) = 19/2072 T(29, 37) = 61/6438 T(30, 37) = 16/1665 T(31, 37) = 1/111 T(32, 37) = 11/1184 T(33, 37) = 1/111 T(34, 37) = 1/111 T(35, 37) = 1/111 T(36, 37) = 1/111

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Re: [math-fun] Chocolate bar puzzle
by Dan Asimov 28 Feb '09

28 Feb '09

This passage << (It'd be fun to make a whole collection of problems that smart and/or knowledgeable people tend to do poorly on: an anti-intelligence or anti-aptitude test if you will.) >> was erroneously attributed to me. (Possibly because in the previous post where it appeared, it was hard to tell who wrote it.) In general I think we in math-fun could be a little clearer about whose passage we're quoting. Also on this subject, I think many posts would be easier to read if they contained quoted material *only* to the extent that it is relevant to what's being posted. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Chocolate bar puzzle
by Dan Asimov 27 Feb '09

27 Feb '09

Or restated clearly, that the sum of a pair of primes that differ by 2 and are no smaller than 5, 7 is divisible by 12. << Or slightly restated, that the sum of a prime pair no smaller than 5,7 is divisible by 12. On Fri, Feb 27, 2009 at 12:56 PM, Stephen Gray <stevebg(a)roadrunner.com> wrote: > Try giving the following trivial problem to engineers or other fairly > math-savvy people who have no exposure to number-theory thinking. > > Show that the number between a prime pair no smaller than 5,7 is divisible > by 6 . >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Only two operations inserted between digits of N
by Eric Angelini 26 Feb '09

26 Feb '09

Hello MathFun, 736 = 7+3^6 2502 = 2+50^2 34425 = 3^4*425 39343 = 39+34^3 Are there many more integers N which give back N if we insert only two "operations" somewhere between the digits of N? The "operations" are +, -, *, /, ^ and the expression at the right of the equal sign must be unambiguous, of course. Best, É.

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[math-fun] Re Wired: The Formula That Killed Wall Street
by Axel Vogt 25 Feb '09

25 Feb '09

> > Message: 5 > Date: Wed, 25 Feb 2009 11:12:29 -0800 > From: Henry Baker <hbaker1(a)pipeline.com> > Subject: [math-fun] Wired: The Formula That Killed Wall Street > To: math-fun(a)mailman.xmission.com > Message-ID: <E1LcPBI-00011G-AG(a)elasmtp-banded.atl.sa.earthlink.net> > Content-Type: text/plain; charset="iso-8859-1" > > FYI -- > > http://www.wired.com/print/techbiz/it/magazine/17-03/wp_quant ... > [ ... see the link above for the rest of the article] and here how involved guys look at such articles: http://www.nuclearphynance.com/Show%20Post.aspx?PostIDKey=127185

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