March 2005 - mathfun@mailman.xmission.com

RE: [math-fun] well posed problem?
by Daniel Asimov 11 Mar '05

11 Mar '05

Wouter Meeussen writes: << of all subsets of the n-th roots of -1, n= 1 to 20, how many add to zero? ?? 0, 1, 1, 3, 1, 6, 1, 11, 4, 8, 1, 50, 1, 10, 9, 107, 1, 240, 1, 316 ?? where are the 'groupies' when you need them? >> It appears you have 6 subsets of the 6th roots of unity, but I count 9 subsets (03, 14, 25, 024, 135, 0134, 2350, 3401, 012345). (In any case there could be another sequence based only on minimal such subsets,but it wouldn't be as interesting.) --Dan

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[math-fun] well posed problem?
by wouter meeussen 11 Mar '05

11 Mar '05

of all subsets of the n-th roots of -1, n= 1 to 20, how many add to zero? ?? 0, 1, 1, 3, 1, 6, 1, 11, 4, 8, 1, 50, 1, 10, 9, 107, 1, 240, 1, 316 ?? where are the 'groupies' when you need them? W.

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[math-fun] sliding numbers
by Chris Landauer 10 Mar '05

10 Mar '05

hihi, all - look at the equation eric angelini wrote for sliding numbers 1/a + 1/b = (a+b)/10^k the left-hand side is (a+b)/ab, and we can cancel out (a+b), so we are looking for all (a+b) with a and b positive integers, such that a*b is a (non-negative integral) power of 10, sorted by (a+b) it should be easy to prove some things about them (to guarantee that after a certain point, there will be no new smaller ones) for example, for a given k, the smallest (a+b) with that k is 2*sqrt(10^k) (equality when k i even, unless we also require a!=b) more soon, cal

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[math-fun] close approximations
by Schroeppel, Richard 10 Mar '05

10 Mar '05

daz> this approximation suggests that9^(4^(7*6)) and 3^(2^85) must be fairly good approximations to one another. Indeed! sin(11) is a bit of a cheat, since it takes the closeness of 11 to 7pi/2 and squares it. But sin(2) would be perfectly reasonable, so maybe instead of banning "asymptotic" functions, the problem should be recast as finding the appropriate price (in complexity units). Ed's example of cos(1+cos(1+cos(...))) shows you can square the epsilon at a constant cost, but if we charged according to 1/derivative, this could be fixed. He also hasn't provided for including function definitions, non-named infinite series/products/limits/sups, and functional inverses. MaxCF doesn't seem to be defined. If it's supposed to be the maximum CF partial quotient, some care is needed: it's likely infinite for everything except quadratic surds. If it's the sup of the average p.q., this is usually bounded and can be guessed from the first few CF terms. Rich rcs(a)cs.arizona.edu

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[math-fun] A001682
by Richard Guy 10 Mar '05

10 Mar '05

I'm collecting Murray Klamkin problems for a book, and have reached Amer Math Monthly 64(1957) 665 where Joe Lipman solves a Murray problem ``as long as tables of sufficient accuracy are available.'' Of course, we don't use tables any more. I've corrected the arithmetic in the original, and in so doing am able to make a modest addition to A001682 which currently reads 21,42,65,86,109,130,151,174,195,218,239,262,283, 304,327,348,371,392,415,436,457,480,501,524,545, 568,589,610,633,654,677,698,721,742,763 and to which may be added 786,807, 830, 851, 874, 895, 916, 939, 960, 983,1004,1027,1048,1069,1092, 1113,1136,1157,1180,1201,1222,1245, 1266,1289,1310,1333,1354,1375,1398, 1419,1442,1463,1486,1507,1528,1551, 1572,1595,1616.1639,1660,1681,1704, 1725,1748,1769,1792,1813,1834,1857, 1878,1901,1922,1945,1966,1987,2010, 2031,2054,2075,2098,2119,2140,2163, 2184,2207,2228,2249 (not 2251) where I've continued the calculation until the difference pattern 21 23 21 23 21 21 23 is broken. What's the magic number whose continued fraction expansion will tell me when to make a gear change? As always, someone should check my hand calculations! R.

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[math-fun] Re: Sliding numbers
by Eric Angelini 10 Mar '05

10 Mar '05

Thanks a lot, Kerry -- before going to bed, yesterday, I found also that the sequence had to start with 2, which is solution of: 1/a+1/b = (a+b)/10^0 I will submit this shortly to the OEIS. Best, É. __________________________________________________ http://angelink.be/?CeRaccourciFaitCinquanteSignes ----- Original Message ----- From: "Kerry Mitchell" <> To: "math-fun" <math-fun(a)mailman.xmission.com>; "math-fun" <math-fun(a)mailman.xmission.com>; <seqfan(a)ext.jussieu.fr> Cc: "Eric Angelini" <keynews.tv(a)skynet.be> Sent: Wednesday, March 09, 2005 7:32 PM Subject: Re: [math-fun] Sliding numbers > I ran an Excel spreadsheet and came up with these first 116 sliding numbers: > > 2, 7, 11, 20, 25, 29, 52, 65, 70, 101, 110, 133, 200, 205, 205, 250, 254, 290, 425, 502, 520, 641, 650, 700, 785, 925, 1001, 1010, 1100, 1258, 1330, 2000, 2005, 2050, 2050, 2225, 2500, 2504, 2540, 2900, 3157, 3445, 4025, 4250, 5002, 5020, 5200, 6266, 6325, 6410, 6500, 7000, 7850, 8125, 9250, 10001, 10010, 10100, 11000, 12508, 12580, 13300, 15689, 16265, 16625, 20000, 20005, 20050, 20500, 20500, 22025, 22250, 25000, 25004, 25040, 25400, 29000, 31282, 31570, 34450, 35125, 40025, 40250, 42500, 50002, 50020, 50200, 52000, 62516, 62660, 63250, 64100, 65000, 70000, 78253, 78500, 79405, 79625, 80125, 81250, 90925, 92500, 100001, 100010, 100100, 101000, 110000, 125008, 125080, 125800, 133000, 156314, 156890, 160625, 162650, 166250 > > They agree with what you posted and I included 2 because 1/1 + 1/1 = 2, and that shows that 20 is a multiple. I'm reasonably sure that no integer has been missed. I stopped at 116 because the next one should be 200,000 from 10^10 = 10^5 x 10^5 (I think), but I only checked exponents out to 9. > > The powers of 10 corresponding to these numbers are: > > 0, 1, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 4, 3, 4, 4, 3, 4, 4, 3, 4, 4, 5, 5, 5, 5, 3, 4, 5, 4, 5, 6, 4, 5, 6, 6, 6, 4, 5, 6, 5, 6, 5, 6, 4, 5, 6, 5, 7, 6, 7, 7, 7, 6, 7, 4, 5, 6, 7, 5, 6, 7, 6, 7, 7, 8, 5, 6, 7, 8, 8, 8, 8, 5, 6, 7, 8, 6, 7, 8, 8, 6, 7, 8, 5, 6, 7, 8, 6, 7, 9, 8, 9, 9, 7, 9, 8, 9, 7, 8, 9, 9, 5, 6, 7, 8, 9, 6, 7, 8, 9, 7, 8, 8, 9, 9 > > and I think that the next one should be 10, but it was 6 in my truncated list. > > As Eric pointed out, each sliding number leads to infinitely more, making this a fractal sequence. Here are the first several "primitive" elements, that is, those that aren't multiples of 10: > > 2, 7, 11, 25, 29, 52, 65, 101, 133, 205, 205, 254, 425, 502, 641, 785, 925, 1001, 1258, 2005, 2225, 2504, 3157, 3445, 4025, 5002, 6266, 6325, 8125, 10001, 12508, 15689, 16265, 16625, 20005, 22025, 25004, 31282, 35125, 40025, 50002, 62516, 78253, 79405, 79625, 80125, 90925, 100001, 125008, 156314, 160625 > > Perhaps it would be interesting to look at this problem with bases with more factors, like 6 or 12. > > Kerry Mitchell > > > > > -------------- Original message from "Eric Angelini" <keynews.tv(a)skynet.be>: -------------- > > > > Hello SeqFan and Math-fun, > > > > Let's start with an example of "sliding number": > > > > 1/4 + 1/25 = 0.29 --> 29 is a "sliding number" > > 1/8 + 1/125 = 0.133 --> 133 is a "sliding number" > > 1/2 + 1/5 = 0.7 --> 7 is a "sliding number" > > etc. > _______________________________________________ > 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] Keen approximations
by Daniel Asimov 09 Mar '05

09 Mar '05

Ed Pegg writes: << I recently did a column on approximations. http://www.maa.org/editorial/mathgames/mathgames_02_14_05.html >> I urge other people to look at this incredibly interesting stuff. I admit that I don't understand very well how keenness is defined. Here's one item from that site that suggests some comments: << Richard Sabey's spectacular approximation to e, (1+9-47*6)3285, gets 18457734525360901453873570 decimal digits of accuracy due to high exponents and asymptotic effects. >> Yikes, I can barely count the number of digits in the number of digits! Since this is (1 + (very small))^(very large), and since e = lim_(t -> oo) (1 + 1/t)^t, and (1 + 1/t)^t is monotonic increasing for t > 1, this approximation suggests that9^(4^(7*6)) and 3^(2^85) must be fairly good approximations to one another. So, what would happen if we just used any incredibly large value of t with its own reciprocal? E.g., how close is (1 + 3^(-2^85))^(3^(2^85)) to e ? Or (1 + 9^(-4^(7*6)))^(9^(4^(7*6))) ? Now, I'm not sure if this is the same as keenness, but for me a nice measure of an approximation would be the "surprisingness" of it. This would be defined as a probabilty, roughly as follows: Suprisingness^(-1) = the probability that if one considers *all* the numbers with the same or smaller complexity to be randomly distributed within their range [smallest, largest] (or maybe this should be done logarithmically), that one of them gets as close as or closer than the approximation actually does. --Dan

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[math-fun] New Mersenne Prime confirmed
by Schroeppel, Richard 09 Mar '05

09 Mar '05

Tony Reix reported this morning on the NMBRTHRY list that the Mersenne prime 2 ^ 25964951 - 1 is confirmed. A triple check is expected to finsh tomorrow. George Woltman has requested that the exponent be kept off "the web" until his press release is posted, and the discoverer's local newspaper has a chance to report it. This is probably M42. It is the largest known Mersenne prime. The next smaller MP exponent known is 24036583, giving modest support to the cluster theory. The GIMPS project is burning at about 16Tflops. Every prime exponent < 15 million has been tested at least once. There are still lots of untested numbers < 25 million, so there's a chance of another, smaller, MP being found. Rich rcs(a)cs.arizona.edu

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[math-fun] ortho-polygon
by David Gale 09 Mar '05

09 Mar '05

For those of you who are geometrically inclined, have you ever seen anything like, Problem. Given n lines a_1,a_2,. . ,a_n in the plane in general position, Construct a (actually, the) polygon with vertices A_1,A_2, . . ,A_n, where Vertex A_i lies on a_i, and Segment A_iA_(i+1) is perpendicular to a_(i+1) (mod n). dg

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Re: [math-fun] Sliding numbers
by lkmitch＠att.net 09 Mar '05

09 Mar '05

I ran an Excel spreadsheet and came up with these first 116 sliding numbers: 2, 7, 11, 20, 25, 29, 52, 65, 70, 101, 110, 133, 200, 205, 205, 250, 254, 290, 425, 502, 520, 641, 650, 700, 785, 925, 1001, 1010, 1100, 1258, 1330, 2000, 2005, 2050, 2050, 2225, 2500, 2504, 2540, 2900, 3157, 3445, 4025, 4250, 5002, 5020, 5200, 6266, 6325, 6410, 6500, 7000, 7850, 8125, 9250, 10001, 10010, 10100, 11000, 12508, 12580, 13300, 15689, 16265, 16625, 20000, 20005, 20050, 20500, 20500, 22025, 22250, 25000, 25004, 25040, 25400, 29000, 31282, 31570, 34450, 35125, 40025, 40250, 42500, 50002, 50020, 50200, 52000, 62516, 62660, 63250, 64100, 65000, 70000, 78253, 78500, 79405, 79625, 80125, 81250, 90925, 92500, 100001, 100010, 100100, 101000, 110000, 125008, 125080, 125800, 133000, 156314, 156890, 160625, 162650, 166250 They agree with what you posted and I included 2 because 1/1 + 1/1 = 2, and that shows that 20 is a multiple. I'm reasonably sure that no integer has been missed. I stopped at 116 because the next one should be 200,000 from 10^10 = 10^5 x 10^5 (I think), but I only checked exponents out to 9. The powers of 10 corresponding to these numbers are: 0, 1, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 4, 3, 4, 4, 3, 4, 4, 3, 4, 4, 5, 5, 5, 5, 3, 4, 5, 4, 5, 6, 4, 5, 6, 6, 6, 4, 5, 6, 5, 6, 5, 6, 4, 5, 6, 5, 7, 6, 7, 7, 7, 6, 7, 4, 5, 6, 7, 5, 6, 7, 6, 7, 7, 8, 5, 6, 7, 8, 8, 8, 8, 5, 6, 7, 8, 6, 7, 8, 8, 6, 7, 8, 5, 6, 7, 8, 6, 7, 9, 8, 9, 9, 7, 9, 8, 9, 7, 8, 9, 9, 5, 6, 7, 8, 9, 6, 7, 8, 9, 7, 8, 8, 9, 9 and I think that the next one should be 10, but it was 6 in my truncated list. As Eric pointed out, each sliding number leads to infinitely more, making this a fractal sequence. Here are the first several "primitive" elements, that is, those that aren't multiples of 10: 2, 7, 11, 25, 29, 52, 65, 101, 133, 205, 205, 254, 425, 502, 641, 785, 925, 1001, 1258, 2005, 2225, 2504, 3157, 3445, 4025, 5002, 6266, 6325, 8125, 10001, 12508, 15689, 16265, 16625, 20005, 22025, 25004, 31282, 35125, 40025, 50002, 62516, 78253, 79405, 79625, 80125, 90925, 100001, 125008, 156314, 160625 Perhaps it would be interesting to look at this problem with bases with more factors, like 6 or 12. Kerry Mitchell -------------- Original message from "Eric Angelini" <keynews.tv(a)skynet.be>: -------------- > Hello SeqFan and Math-fun, > > Let's start with an example of "sliding number": > > 1/4 + 1/25 = 0.29 --> 29 is a "sliding number" > 1/8 + 1/125 = 0.133 --> 133 is a "sliding number" > 1/2 + 1/5 = 0.7 --> 7 is a "sliding number" > etc.

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