March 2009 - mathfun@mailman.xmission.com

Re: [math-fun] prime ladders
by Warut Roonguthai 13 Mar '09

13 Mar '09

I think 929573 is missing from http://mathworld.wolfram.com/WeaklyPrime.html because 029573 is prime. Warut On Fri, Mar 13, 2009 at 6:42 AM, Andrew Trevorrow <andrew(a)trevorrow.com> wrote: > Edwin wrote: > >> Even relaxing the adjacency definition to allow 3-->31 and 3-->13, etc, >> the graph of all primes will not be connected: >> >> If you change any digit of the prime 971767 it becomes composite. ... > > Yep, according to http://mathworld.wolfram.com/WeaklyPrime.html such > numbers are said to be "weakly" prime. According to my code, the > first 6 such numbers are: > > 294001 > 505447 > 584141 > 604171 > 929573 > 971767 > > Note that 929573 is missing from the above MathWorld page. > > Some more info about prime ladders: > > no. of digits: 2 3 4 5 6 > longest ladder: 4 7 9 11 13 > example: 23 389 2441 88259 440497 > 43 383 5441 88289 410497 > 47 883 5741 88789 416497 > 37 863 5701 88799 416477 > 463 3701 82799 416473 > 461 3709 12799 416873 > 761 3109 12791 406873 > 9109 92791 406883 > 9199 92761 706883 > 99761 705883 > 99721 705833 > 905833 > 995833 > > Rich gave the challenge of turning 1009 into 9973, and Warut > found a few 6-number solutions. A slightly longer solution > is needed to turn 1009 into 9227: > > 1009 > 1109 > 7109 > 7129 > 7127 > 9127 > 9227 > > Andrew > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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Re: [math-fun] mildly interesting cycloid note
by Dan Asimov 13 Mar '09

13 Mar '09

Kewl! Thanks! << > http://gosper.org/cycloid.html > I added the Fourier series, at the behest of DanA. The derivation was fairly slick and is probably how Lagrange got his series solution to Kepler's problem. (http://www.tweedledum.com/rwg/pizza.htm, which I am redoing.) --rwg >> _____________________________________________________________________ "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] Remez Principle for lexically optimal muffin solutions
by Huddleston, Scott 10 Mar '09

10 Mar '09

David Moulton described an appealing criterion for a unique optimal solution to muffin problems -- order the pieces of the corefinement in nondecreasing size, and choose the lexicographically largest solution. I'll call this the "lexically optimal" solution. Any solution to the original problem, which only requires maximizing the smallest piece, I call a "weak optimum". In this note I present some weak optimal solutions together with (what I believe are) their lexically optimal solutions. These examples are intended to illustrate a principle I now formulate and conjecture: Muffin Remez Principle: The lexically optimal solution to a muffin problem is found by distributing non-unifomity with maximal evenness at every stage of solution construction. I chose this name by imperfect analogy with the Remez algorithm in numerical analysis. Example 1: Instances 3/2 < r < 2 are usually given with either 3 or 4 pieces per person. But S(15/8) = 3/8 = 1-r/3 = r/5 admits a weak optimum with either 3 or 5 pieces per person: 1/8 * { [3,5]*15; [3*5]*3; [5*3]*5 } The lexical optimum is 1/8 * { [3,5]*6, [4*2]*9; [3,4*3]*6; [5*3]*2 } which exceeds the weak optimum at the 7th piece, 4/8 > 3/8. Example 2: Lexical optima in 3/2 < r < 2 minimize the number of muffins with both halves in 3-parts or 4-parts. S(17/10) = 2/5 = (5-2r)/4 = (2r-1)/6 admits a weak optimum with muffins concentrated in both 3-parts and 4-parts: 1/10 * { [4,6]*12, [5*2]*5; [4*3,5]*4; [6*2,5]*6 } The lexical optimum concentrates all unshared muffins in 3-parts: 1/30 * { [12,18]*4, [13,7]*12, [15*2]; [12,13*3]*4; [18*2,15]*2, [17*3]*4 } and exceeds the weak optimum at the 5th piece, 13/30 > 4/10. Example 3: All 9/5 < r < 2 admit optimal solutions using unsplit muffins, but these are usually weak. Consider S(11/6) = 7/18 = 1-r/3 A weak optimum using unsplit muffins is 1/18 * { [7,11]*6, [8,10]*2, [9*2], [18]*2; [7*2,9,10]*2, [7,8,18]*2; [11*3]*2 } Its lexical optimum is 1/36 * { [14,22]*6, [17,19]*4, [18*2]; [14*2,19*2]*2, [14,17*2,18]*2; [22*3]*2 } which exceeds the weak optimum at the 7th piece, 17/36 > 8/18.

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Re: [math-fun] muffin problem
by James Propp 09 Mar '09

09 Mar '09

Scott Huddleston writes: >The muffin problem should really be considered in projective space >r = m/p, and I'll henceforth usually write S(r) instead of S(m,p). I don't understand; if we haven't proved that S(km,kp) = S(m,p) for all k, then how can this be right? Jim

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[math-fun] working on a recipe/formula
by Michael Reid 09 Mar '09

09 Mar '09

i'm not sure i fully understood scott's post, but it is probably similar to something i'm working on here. i'm trying to find a general "formula" for T(m, n) . perhaps i should say *bound* on T(m, n) , but my intent is to get it exact in as many cases as possible. i will always have m <= n , and here i will focus on the range n >= 3m/2 . this is very much "in progress" as we speak, so to speak. firstly, there is the trivial case, when m divides 2n . i call this "case I", which splits into case I A if m divides n , and case I B if m divides 2n but not n . we have the exact values T(m, n) = 1/n and 1/(2n) in these cases, respectively. case II is if m does not divide 2n and 2n >= 3m . divide 2n by m to get 2n = mq + r , where 0 < r < m , and now compare r(q + 1) to (m - r)q . there are three possibilities: case II A: r(q + 1) > (m - r)q . here we have the easy bound T(m, n) <= 1/n - 1/(qm) . case II B: r(q + 1) < (m - r)q . here we have the easy bound T(m, n) <= 1/((q+1)m) . case II C: r(q + 1) = (m - r)q . here we have the exact value T(m, n) = 1/((q+1)m) , which also equals 1/n - 1/(qm) . (in case II C, we have (m, n) = ((2k+1)j, k(k+1)j) for some k >= 3 . i had previously identified the family T(2k+1, k(k+1)) = 1/((k+1)(2k+1)) in an earlier message, and this is just a "multiple" of that.) when can the bounds in II A and II B be attained? here's a guess. in case II A, divide r(q + 1) - (m - r)q by r to get r(q + 1) - (m - r)q = rs + t where 0 <= t < r . it seems plausible to believe that if s >= 2 , then equality can be attained. (i was initially skeptical of this, thinking there might be some behavior like that in the case m < n < 3m/2 , but i have not encountered such.) case II A.1 is when s >= 2 , so there's no improvement in the bound. case II A.2 is when s = 1 and r - t divides r + t (equivalently, divides 2t ). in this case, T(m, n) = 1/n - 1/(qm) can be achieved. case II A.3 is when s = 1 and r - t does not divide r + t . so divide r + t by r - t to get r + t = (r - t)u + v , where 0 < v < r - t . then we have the improved bound T(m, n) <= ((u+2)/m - (u+1)/n)/((q-1)(u+2)+2) . case II A.4 is when s = 0 . in this case, the bound can be improved to T(m, n) <= (1/m - 1/(2n))/q . it is unlikely that this bound can always be attained. here's the corresponding refinement in case II B: divide (m - r)q - r(q + 1) by m - r to get (m - r)q - r(q + 1) = (m - r)s + t where 0 <= t < m - r . as in case II A, it is reasonable to believe that the bound 1/((q+1)m) can be attained if s >= 2 . so we have the cases, similar to those in II A: case II B.1: s >= 2 . in this case there's no improvement in the bound. case II B.2: s = 1 and m - r - t divides m - r + t (equivalently, divides 2t ). in this case, T(m, n) = 1/((q+1)m) can be achieved. case II B.3: s = 1 and m - r - t does not divide m - r + t . so divide m - r + t by m - r - t to get m - r + t = (m - r - t)u + v , where 0 < v < m - r - t . we then have the improvement T(m, n) <= 1/n - ((u+2)/m - (u+1)/n)/((q-2)(u+2)+2) . case II B.4: s = 0 . here the bound can be improved to T(m, n) <= 1/n - (1/m - 1/(2n))/(q - 1) . it is likely that cases II A.3, A.4, B.3 and B.4 all need further refinement in subcases where the bound can be improved. i speculate that these subcases will be similar to the splitting of cases II A and B , and there will be further splittings into sub-sub-...-subcases, probably with arbitrarily deep splitting. some examples: (m, n) = (13, 23) , which is one of the complex examples we've already seen. here we have q = 3 and r = 7 . then r(q + 1) > (m - r)q , so we have case II A. next, s = 1 and t = 3 , and since r - t does not divide r + t , we have case II A.3. now u = 2 and v = 2 , so our improved bound is T(13, 23) <= (4/13 - 3/23)/10 = 53/2990 , which is the exact value, as i previously showed. (m, n) = (25, 41) . again we have q = 3 and r = 7 , but this time r(q + 1) < (m - r)q , so we're in case II B. next, s = 1 and t = 8 , and since m - r - t does not divide m - r + t , we're in case II B.3. now u = 2 and v = 6 , so the improved bound is T(m, n) <= 1/41 - (4/25 - 3/41)/6 = 61/6150 . indeed, this bound can be attained via 4 * [3 * 61 + 63] + 3 * [2 * 61 + 2 * 62] + 6 * [68 + 2 * 89] + 6 * [75 + 82 + 89] + 4 * [71 + 87 + 88] + 2 * [2 * 79 + 88] <---> 18 * [61 + 89] + 6 * [62 + 88] + 4 * [63 + 87] + 6 * [68 + 82] + 4 * [71 + 79] + 3 * [2 * 75] where i've scaled the parts by 6150 for those who don't like fractions. the first case where the above is not sharp is (m, n) = (20, 33) , where the bound from the "formula" is 41/3300 , but the exact value is 49/3960 . among the long list of values i previously sent, the only other pair (in case II) where the formula is not sharp is (21, 37) . i also have some refinements of case III ( n < (3/2)m ) beyond what i previously sent, that work for all but 1 of the values on that long list. but it'll have to wait for another message ... mike

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

09 Mar '09

> Veit Elser kindly noted errors in: > > On the other hand, we've already seen examples S(ka,kb,kc) < > > S(a,b,c) in the > > The correct statement: We've seen an example > S(ka,kb,kc,kd) > S(a,b,c,d), namely Michael Reid's example > 1/3 = S(4k,5k,6k,7k) > S(4,5,6,7) for k > 1. don't worry, i'll soon have a 3 parameter example. :-) on another note: ) lowest terms. But I expect the more interesting and "universal" projective ) quantity is ) ) S^*(r_1,r_2) = infimum_{k>0} S(k*m, k*m*r_1, k*m*r_2). ) ) It's surely the case, for given a<b<c, that S^*(b/a,c/a) = S(ka,kb,kc) for ) some finite k. I conjecture there's a single finite K_3 which achieves certainly it's plausible that S(ka, kb, kc) is (eventually) periodic in k , with period larger than 1 . whether or not this is *likely* i do not know. but i cannot rule it out. perhaps you are predicting this does not happen; but if it does happen then your S^*(r, s) function does not tell the whole story. mike

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[math-fun] FW: muffin problem
by Huddleston, Scott 09 Mar '09

09 Mar '09

Veit Elser kindly noted errors in: > On the other hand, we've already seen examples S(ka,kb,kc) < > S(a,b,c) in the The correct statement: We've seen an example S(ka,kb,kc,kd) > S(a,b,c,d), namely Michael Reid's example 1/3 = S(4k,5k,6k,7k) > S(4,5,6,7) for k > 1.

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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] muffin problem
by James Propp 07 Mar '09

07 Mar '09

Here are some minuses of "outing" math-fun in print: 1) Having a flood of new subscribers could be a drain on Rich's time for a week or two. 2) There is the risk that a few people who don't fit in with the group and who post too much could lower the overall quality of the posts and take away some of the value of math-fun for the rest of us (remember what happened to sci.math?). 3) I'd hate to put Rich in the position of having to ban people from the group (temporarily or permanently), especially in cases where the "offense" isn't a matter of rudeness but a matter of the content of the person's posts. But I think that the big plus of publicizing the group (in a low-key way) is that we'd be opening up the group to people who should be part of math-fun but don't have the good fortune to be part of the informal network that brought us all into math-fun in the first place. As things stand now, we run the risk of turning the group into an old boy's network, not by design but by default. (And yes, when I say "boy" I say it advisedly. There aren't many women in the group, and they seldom post. This is one sign that we aren't casting our net wide enough to bring in the people who should be in the group but who don't know about it.) Jim

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Re: [math-fun] muffin problem
by James Propp 04 Mar '09

04 Mar '09

>Leaving aside the problems of normalization, I have a great title: > >"Do you know the Muffin Man?" (perhaps we might even use the cartoon >from punch -- see the wikipedia reference below). > >http://en.wikipedia.org/wiki/The_Muffin_Man I've been leaning toward "I Got Plenty o' Muffin" myself. Jim

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