March 2009 - mathfun@mailman.xmission.com

[math-fun] (exp(Pi)-ln(3))/ln(2)
by Mike Stay 17 Mar '09

17 Mar '09

31.80000000332526 = (exp(Pi)-ln(3))/ln(2) Does this run of zeros have something to do with the j function? -- Mike Stay - metaweta(a)gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com

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[math-fun] mathematicaster fun
by Huddleston, Scott 16 Mar '09

16 Mar '09

I'm forwarding this math fun (definition of mathematicaster) from A.Word.A.Day. The theme last week was 15 letter words, and there was a side competition to give the best 15 letter definition for them. Munsters might enjoy a free email subscription at http://wordsmith.org/awad/subscribe.html . Enjoy, - Scott Mathematicaster<http://wordsmith.org/words/mathematicaster.html> Winner: 2nd rate math geek. -Greg Foster (lokesman gmail.com) Honorable mentions: Ramanujan manque. -Eric Towne (etowne bates.edu) A minor math major. -Howard Distelzweig (howard_distelzweig pall.com) He divides by zero. -Peirce Hammond (Peirce.Hammond ed.gov) Rounds pi to three. -Joselyne Gonzalez (joselyne yahoo.com) Uneven math maven. -Catherine Masters (cmasters schiffhardin.com) He's no Pythagoras. -Jacquie L Lowell (jlowell.improv juno.com) Dave Zobel (dzobel alumni.caltech.edu) Two plus two's five. -R. Ganesh (r.ganesh iflexsolutions.com) No Euclid or Gauss. -Devika Nair (devikanair1979 hotmail.com) Digits: all thumbs. -Jason Morgan (aeelectra yahoo.com) In-add-equatician. -Brendon L. Etter (better carleton.edu) Can do two plus two. -Matt Schmidt (mschmidt hussoninc.com) Math challenged. (I could have made it "maths challenged" but I prefer the irony of the above.) -Alan Broom (alan.broom macquarie.com) Numbers Are Not Us. -Colleen (argonauta4 aol.com) IQ's greater than pi. -Bill Ward (bill wards.net) Bank executive, e.g. (not a timeless definition, but it works currently). -Aaron Long (aarondavidlong hotmail.com) Just a calculator. -Tal Cohen (tal forum2.org) One does not add up. -Mike Freedman (mike freedthinkers.com) A nonplussed soul. -K. F. Turtletaub (doctorkf verizon.net) Calculably unfit. -John Hudspeth (johnhudspeth windstream.net) Big sum, small mind. -Caroline Murphy (ckmurphy54 hotmail.com) An amateur at math. -John P. Marhin (john.p.marhin mainroads.qld.gov.au) 0 < x < a math savant. -Greg Foster (lokesman gmail.com) 99 percent of folk. -Kassy Daggett (kdaggett efn.org) rkathleendillon. -R Kathleen Dillon (rkdillon verizon.net) Math isn't my forte. -Kevin Ogburn (kevin.ogburn co.hennepin.mn.us) Calculating Risk. -David Lehner-Smith (david.lehner-smith target.com) Small-time mathmo<http://www.mathmo.org/>. -Owen Biesel (owenbiesel gmail.com) Many readers sent: Number numbskull. Math geek wannabe. No good at numbers. A numbers bumbler. A numbers fumbler.

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Re: [math-fun] prime ladders
by N. J. A. Sloane 16 Mar '09

16 Mar '09

I've seen two different versions of how you can move from one prime to another: Version 1: change a single digit Version 2: change a single digit and then permute the digits subject in both cases to the restriction that the new prime cannot begin with 0; the number of digits must remain constant. Call two primes equivalent if you can go from one to the other in this way. For each version of equivalence, there are four obvious sequences: a(n) = number of equivalence classes of primes with n digits. Arrange the equivalence classes by the size of the smallest member. b(k) = size of the k-th equivalence class c(k) = smallest member of the k-th equivalence class d(k) = largest member of the k-th equivalence class Presumably the two a-sequences will begin with a bunch of 1's, the two b-sequences will start like A006879, the two c-sequences will start like A003617, and the two d-sequences will start like A003618. There are potentially eight (new?) sequences here - could someone compute them? Thanks! Neil

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[math-fun] $500 pi approximation contest
by Jason 15 Mar '09

15 Mar '09

From: http://www.kaidy.com/PiReward.htm Alas, someone already won the contest with a remarkably elegant submission. rwg, you'll just have to calculate faster next time. Magic of 355 / 113 355 / 113 is a 99.9999% accurate approximation of pi. 355 / 113 = 3.1415929203 pi = 3.1415926535 The difference = 0.0000002668 Can you find a BETTER fraction to more closely approximate the value of pi? The first one to do so will earn a reward of $500. (NOTE: The fraction you submit must be simplified to lowest terms, so not exactly equivalent to 355 / 113, and the numerator and denominator must each be whole numbers.) Click here to see if someone has come up with a BETTER solution. And from the "winner" page: We have a winner! John McKiernan was the earliest to submit 6,283,185,307/2,000,000,000. Thank you to all of our other participants.

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[math-fun] corefinement for a muffin subproblem
by Huddleston, Scott 14 Mar '09

14 Mar '09

I found an interesting corefinement family that arises in finding the second smallest piece size for certain muffin problems with r = m/p < 3/2. I'll show how it arises later, and proffer the corefinement problem now in case anyone wants to chew on it. Find the corefinement of { A, B } with maximal lexicographic order of piece sizes, where A = { [1]*(u+v)] } B = { [(u+v)/2v]*v, [(u+v)/2u]*u } This notation means B consists of v parts of size (u+v)/2v and u parts of size (u+v)/2u.

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[math-fun] Patterson's 1944 triangle
by N. J. A. Sloane 14 Mar '09

14 Mar '09

Dear SeqFans, I came across a paper, Phys. Rev. 65, 195 - 201 (1944) Ambiguities in the X-Ray Analysis of Crystal Structures by A L Patterson http://prola.aps.org/abstract/PR/v65/i5-6/p195_1 is the URL, if your institution subscribes (mine does not) In Table I there is a triangle of numbers, which the author admits he is not confident about (which is probably why I left them out of the 1973 book) I know that there have been several followup papers. It might be an interesting project for someone to look into this, determine the correct triangle(s), and submit it to the OEIS, The subject is the ways to place n points around a circle so that the set of n-choose-2 distances do/do not determine the points [Related to the classical problem, worked on by Petard, as I recall, who got it wrong - hoisted with your own Petard? - of when the distances between n points anywhere in the plane determine the pints.] Might be an interesting problem to work on (the Patterson paper, I mean). Neil

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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] analysis of case m < n < 3m/2
by Michael Reid 13 Mar '09

13 Mar '09

continuing my quest for a "formula" for T(m, n) for the "muffin" problem. here's some further analysis in the case m < n < 3m/2 . as in the case n > 3m/2 , this is still under construction. i previously sent the cases > 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) . david moulton pointed out that something is wrong with these. it seems that i mistakenly included the wrong endpoint in the first two cases. indeed, the four cases should cover the entire range m < n < 3m/2 as t runs through positive integers, without overlap between the cases. i have already used case I to denote the case that m divides 2n , and case II is when n > 3m/2 (and m does not divide 2n ), so case III will be when m < n < 3m/2 , which splits into 4 subcases, as above, but corrected: case III A: ((3t+1)/(3t))m <= n < ((3t)/(3t-1))m here we have the bound T(m, n) <= 1/(3n) , and i believe this is the exact value. (moreover, it seems that T(m, n) > 1/(3n) in all other cases.) the bound T(m, n) <= 1/(3n) for 4m/3 <= n < 3m/n , which is the t = 1 version of case III A, had already been observed by moulton and others. case III B: ((3t+2)/(3t+1))m <= n < ((3t+1)/(3t))m here we have T(m, n) <= (t+1)/(2n) - t/(2m) . this bound cannot always be attained, and i will refine this case below. case III C: ((3t+3)/(3t+2))m < n < ((3t+2)/(3t+1))m here we have the bound T(m, n) <= (t+1)/(4m) - t/(4n) . again, this cannot always be attained, so this case needs refinement as well. case III D: if ((3t+3)/(3t+2))m = n , then T(m, n) = 1/(3m) . this was veit's original bound, and he showed that it is exact in this case. now i refine case III B: divide n by 3(n - m) to get n = 3(n - m)t + u , where n - m < u <= 2(n - m) . (the quotient t is the same integer in the bounds defining case III B.) case III B.1 is if u - (n - m) divides 3(n - m) - u (equivalently, divides 2(n - m) , equivalently, divides 2u ). in this case there is no improvement, and the bound (t+1)/(2n) - t/(2m) can be attained. so now divide 3(n - m) - u by u - (n - m) to get 3(n - m) - u = (u - (n - m))y + z , where 0 < z < u - (n - m) , and compare z(y + 4) to (u - (n - m) - z)(y + 3) . case III B.2 is if z(y + 4) < (u - (n - m) - z)(y + 3) . here we have the improved bound T(m, n) <= ((ty + t + 1)n - t(y + 1)m)/(mn(y + 4)) . case III B.3 is if z(y + 4) > (u - (n - m) - z)(y + 3) . here we have the improved bound T(m, n) <= ((2ty + 3t + y + 3)m - (2ty + 3t + 1)n)/(mn(y + 3)) . case III B.4 is if z(y + 4) = (u - (n - m) - z)(y + 3) . here the bounds in III B.2 and III B.3 coincide, and the common value, ((ty + t + 1)n - t(y + 1)m)/(mn(y + 4)) can be attained. here's my refinement for case III C. i am not entirely satisfied with it, so it is subject to change/reorganization. divide n by 3(n - m) to get n = 3(n - m)t + u , where 2(n - m) < u < 3(n - m) . case III C.1 is if (12/5)(n - m) <= u . at the moment there is no improvement on the bound, although i know it can't always be attained. case III C.2 is if (12/5)(n - m) > u . here we have the improved bound T(m, n) <= ((3t + 3)m - (3t + 1)n)/(3mn) . the first example where the bound in III C.1 is not exact is (31, 38) . it seems that i should split off the case (12/5)(n - m) <= u < (5/2)(n - m) where the bound T(m, n) <= (5t+5)/(4n) - (5t+2)/(4m) holds. i need to double-check this. maybe case III C should be split up in a different way. it's also possible that case III D should be considered as a subcase of case III C. these cases, III A, III B.1,2,3,4, and III C.1,2 give the exact value of T(m, n) in all the examples (where m < n < 3m/2 ) on the long list i previously sent, except for (23, 29) , which goes into case III B.2. so i developed a refinement of that case: case III B.2i is if (u - (n - m) - z)(y + 3) - z(y + 4) >= u - (n - m) - z . here there's no improvement on the bound. case III B.2ii is if (u - (n - m) - z)(y + 3) - z(y + 4) < u - (n - m) - z . here we have the improved bound T(m, n) <= ((4ty + 5t + 2y + 5)/n - (4ty + 5t + 2)/m)/(2(y + 2)) . now (23, 29) goes into case III B.2ii, (with t = 1 and y = 1 ) where the bound is 49/4002 , and this is the exact value of T(23, 29) . presently, i'm not certain if we should expect the bound to always give the exact value in case III B.2i. if not, there is another refinement waiting to be discovered. there's an analogous refinement of case III B.3: case III B.3i is if z(y + 4) - (u - (n - m) - z)(y + 3) >= z . here there's no improvement on the bound. case III B.3ii is if z(y + 4) - (u - (n - m) - z)(y + 3) < z . here we have the improved bound T(m, n) <= ((2ty + 3t + 2)/m - (2ty + 3t + 1)/n)/(2(y + 3)) . some examples: (m, n) = (42, 53) . this is the smallest example that goes into case III B.4. here we have t = 1 and y = 1 (and u = 20 , z = 4 ). the formula gives T(m, n) = 5/742 , and this is attained by the partition 8 * [2 * 15 + 23] + 4 * [15 + 2 * 19] + 10 * [2 * 16 + 21] + 20 * [26 + 27] where i have scaled by 2226 to give integers. i am not giving the rearrangement as a refinement of 53 * 1/53 , since that is automatically determined. i believe that this is the unique optimal partition, and that in case III B.4, there is always a unique optimal partition. (m, n) = (61, 77) goes into case III B.3ii, with t = 1 , u = 29 , y = 1 and z = 6 . the bound in this case becomes 173/37576 , and this is attained by the partition 4 * [2 * 173 + 270] + 4 * [173 + 175 + 268] + 4 * [2 * 175 + 266] + 4 * [174 + 220 + 222] + 2 * [180 + 2 * 218] + 12 * [185 + 187 + 244] + 2 * [2 * 186 + 244] + 12 * [301 + 315] + 4 * [302 + 314] + 12 * [303 + 313] + [2 * 308] where i've scaled by 37576 . (m, n) = (17, 19) goes into case III A with t = 3 . then the bound is T(17, 19) <= 1/57 , and as predicted, is attained by the partition 3 * [17 + 2 * 20] + 2 * [3 * 19] + 6 * [25 + 32] + 6 * [26 + 31] (scaled by 969 ). (m, n) = (32, 37) goes into case III B with t = 2 and u = 7 . here we have u - (n - m) = 2 , which divides 3(n - m) - u = 8 , so we are in case III B.1. the formula gives T(32, 37) = 11/1184 , which is attained by the partition 4 * [2 * 11 + 15] + 4 * [11 + 12 + 14] + 2 * [11 + 2 * 13] + 14 * [16 + 21] + 4 * [17 + 20] + 4 * [18 + 19] . (m, n) = (37, 42) goes into case III C with t = 2 and u = 12 . here we have (12/5)(n - m) = u , so we are in case III C.1. the bound is T(m, n) <= 13/1554 , and this can be attained with the partition 6 * [2 * 13 + 16] + 4 * [3 * 14] + 12 * [18 + 24] + 12 * [19 + 23] + 3 * [2 * 21] . mike

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[math-fun] Space-filling curves
by Henry Baker 13 Mar '09

13 Mar '09

Space-filling curves, anyone? http://tinyurl.com/c9bno3 It looks like there may be an opportunity for a web site to generate Euler-like paths/circuits for runners/cyclists/... Navteq must already have such programs for their "drive-bys".

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Re: [math-fun] prime ladders
by Steve Witham 13 Mar '09

13 Mar '09

>From: rcs(a)xmission.com > >3 -> 13: >My notion was that leading 0s are ruled out, so each size is a separate >problem space. With this restriction, I think all the two digit primes >are connected, and all the three digit primes are connected. > >From: Warut Roonguthai <warut822(a)gmail.com> > >I found that all the 4-digit primes are connected even under the >restriction, and this is also true for the 5-digit case. I think all the one-digit primes are connected!! --Steve

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