February 2005 - mathfun@mailman.xmission.com

[math-fun] RE: True So Far
by Lior Manor 24 Feb '05

24 Feb '05

Hi Eric, Looks like a(6078) = 8945 is the last term of your sequence. According to my calculations the last terms ending with 0-9 digits are: a(5914) = 5890 a(6060) = 8201 a(6064) = 8302 a(6073) = 8613 a(6075) = 8734 a(6078) = 8945 a(6059) = 7756 a(6067) = 8563 a(6051) = 6778 a(5918) = 5979 It is easy to see that there will be no additional terms. You can view the full sequence in sci.misc (http://groups-beta.google.com/group/sci.misc/msg/aa57ea84b421d2a2) Lior Manor

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[math-fun] not proven: 2^86
by Richard Schroeppel 24 Feb '05

24 Feb '05

Rich>My favorite effort in this direction is "2^86 is the last power of 2 >without a 0 digit". David Gale> Really? In my Automatic Ant book (page 43) I speculate that statements like "there are only a finite number of 7-less powers of 2" may be undecidable, in the Godel sense that there exists no proof of the statement nor of its negation from the standard axioms. Rich, you apparently have a proof with 7 replaced by 0. Could you elaborate? Whoa! Nellie! I have NOT proved 2^86 is the last power of 2 with no 0. It's a CONJECTURE. All the proof I have will fit comfortably in an arbitrarily small margin. I should have been more careful in my wording. My apologies for the confusion. The only evidence I know of is reported in Hakmem item 57. Mike Beeler wrote a program that checked a block of low-order digits, and we ran it out to exponent 30M. Every power after 2^86 had at least one 0 digit. ---- Here are some notes I made up for power-digit puzzles. The power-matching problem seems more interesting if we allow anagrams. Examples: 125/512, 169/196/961, 256/625, 1024/2401, 4096/9604, 1048576/5764801. Of course 169/961 is a cheat. And I need a rule for handling leading 0s. Here's a "stats-gone-wrong" example: Based on statistics alone, modelling numbers as random digit strings of a known size, 22nd and higher powers should all eventually have a 0 digit, since log10(10/9) = 1-.95424 = .04576 > 1/22. However, there's a parametric exception: (1,1)^22 = 1,22,231,1540,7315,26334,74613,170544,319770,497420,646646,705432 646646, ... ,22,1. (Pascal's Triangle) 1000001^22 = 1000022000231001540007315026334074613170544319770497420646646705432 646646497420319770170544074613026334007315001540000231000022000001 so (10^n+1)^22 is 8-less for n>=6. Of course, (10^n)^22 is also 8-less, and if k^22 is 8-less, so is (10k)^22. This sort of example don't work for 0-less, since 0 is the padding digit. Variations on 10001 include 10002, 20001, 20003, 100379, 23400765, 1001001, 9999, 29993, 299930001, etc. Powers of these could be 8-less in a more or less ad hoc way, and could be extensible. Is there a parametric set of 0-less squares? 777...444 seems to fail. 9999^2 = 99980001. Divide by 9 to get 3333^2 = 11108889. Close, but no cigar. [RIP WAH]. Multiply by 4 to get 6666^2 = 44435556. Bingo. Is there any way to find 0-less powers other than scanning for them? Presumably the 0-less squares thin out but never completely disappear. It might be possible to use stats to prove density upper bounds on 0-less squares or powers of 2. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] test; E. M. Wright; continued fraction convergence; Dyson's problem
by Schroeppel, Richard 24 Feb '05

24 Feb '05

We looked at the cycle of low-order digits of powers of 2 (mod 10^k). 2 is a primitive root of 5, 25, ..., so the cycle lengths are 4, 20, 100, etc. In particular, every even number not ending in 0 actually occurs as a last digit, as does every 2-digit multiple of 4 not ending in 0, and every 3-digit multiple of 8 not ending in 0, and so on. This will let you say (and prove) things like "the tens digit of powers of 2 is a 7, 10% of the time". But this idea doesn't go far enough to prove the 2^86 conjecture. It _can_ show that the density of 0-less powers of 2 is 0. But the density of primes is 0, and there are plenty of them. More interestingly, there's a unique "terminal string" of 1s and 2s for very large powers of 2: it ends with ...2112. No other terminal string (consisting solely of 1s and 2s) occurs for infinitely many powers of 2. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com on behalf of Michael Kleber Sent: Wed 2/23/2005 7:00 PM To: math-fun Subject: Re: [math-fun] test; E. M. Wright; continued fraction convergence; Dyson's problem Rich wrote, > >My favorite effort in this direction is "2^86 is the last power of 2 > >without a 0 digit". David Gale replied: > Really? In my Automatic Ant book (page 43) I speculate that statements like > "there are only a finite number of 7-less powers of 2" may be undecidable, > in the Godel sense that there exists no proof of the statement nor of its > negation from the standard axioms. > > Rich, you apparently have a proof with 7 replaced by 0. Could you elaborate? I'm just guessing, and surely Rich will fill us in soon. But for any k, the last k digits of 2^n are eventually periodic. So if you find a k where every member of the attrating cycle of "n -> 2n mod 10^k" contains a 0, you're done. I'd have to think more about whether we expect such a scheme to succeed or fail in general. As k increases, you have more digits to work with, but also probably more terms in the cycle, and if someone else wants to work out how the probabilities balance against each other, I'm as happy to read it as to do it myself... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ 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] test; E. M. Wright; continued fraction convergence; Dyson's problem
by David Gale 24 Feb '05

24 Feb '05

> Rich writes, >My favorite effort in this direction is "2^86 is the last power of 2 >without a 0 digit". Really? In my Automatic Ant book (page 43) I speculate that statements like "there are only a finite number of 7-less powers of 2" may be undecidable, in the Godel sense that there exists no proof of the statement nor of its negation from the standard axioms. Rich, you apparently have a proof with 7 replaced by 0. Could you elaborate? David >Rich rcs(a)cs.arizona.edu > > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] test; E. M. Wright; continued fraction convergence; Dyson's problem
by Schroeppel, Richard 23 Feb '05

23 Feb '05

test: If Mime conversion is working, this message should be plain text. ---- E. M. Wright, an author of Hardy & Wright's Introduction to the Theory of Numbers, died recently. ---- My memory of the rule for when CFs converge: If all partial quotients are positive, then the CF values either converge, or oscillate while approaching two limit points. If the sum of the PQs is infinite, then the CF converges. (Is this right? I think the oscillation interval drops roughly by 1/(1+PQ) for each term, and that product ->0 should be same as sum PQ -> infinity.) Of course this leaves open negative and complex partial quotients. ---- I looked at Dyson's problem. (Show that no power of 2 is the reverse of a power of 5, except 1 = 2^0 = 5^0.) There are some easy partial results: Assume a counterexample. Then both are squares and have an odd number of digits, or both are non-squares and have an even number of digits; both are cubes, or both are non-cubes, and of the opposite sense. These come from looking at the powers mod 9 and mod 11. I tried looking at additional moduli like 7, 13, 101, etc. There are restrictions on various digit-position sums, but nothing easy to state. The 1 solution means that solutions exist mod all primes (except 2,5). My favorite effort in this direction is "2^86 is the last power of 2 without a 0 digit". Rich rcs(a)cs.arizona.edu

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[math-fun] math-fun message conversion
by Richard Schroeppel 23 Feb '05

23 Feb '05

I've adjusted the filtering for the Math-Fun list. If I've got it right, the effect should be that attachments are automatically dropped, and Mimed messages are converted to plain text. Thanks to David Wilson for pointing out that the latest version of Mailman offers some useful message conversions. I'll be trying a couple of test messages. Rich rcs(a)cs.arizona.edu

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[math-fun] Long Messages
by David Wilson 23 Feb '05

23 Feb '05

Rich Schroeppel recently repeated his no-attachments warning for math-fun mail. I'm assuming I was probably the offender here, by accidentally submitting my Ulam pic to math-fun. In deference to the rules, I did not originally post the pic to math-fun, but rather offered the picture on request, and got several requests. It is not inconceivable that in satisfying those requests, I accidentally submitted the picture to math-fun. My mail tool (Outlook Express) makes this and other annoying things quite easy to do with the touch of a button. At any rate, Rich mentioned that he discourages long messages and attachments in math-fun in deference to subscribers with slow connections, which is laudable. On the other hand, I imagine that the overwhelming majority of math-fun/seqfan users have no such issues with long messages. My thought is that if math-fun/seqfan were to maintain a second mailing list or forwarding address for long messages (e.g, math-fun-long(AT)mailmain.xmission.com) I could have just posted my Ulam pic there with no hassle. -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.300 / Virus Database: 266.4.0 - Release Date: 2/22/2005

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[math-fun] Re: continued fractions
by Richard Guy 23 Feb '05

23 Feb '05

Open letter to AB, copied to sequensters & munsters, in case they can throw more light (and, in the former case, accept some new sequences). When I asked Hugh Williams, he immediately had the same reaction as I, that it didn't much matter if they were rats or reals, and that a sufficient condition was that all but a finite number of the partial quotients should be greater than one. However, I think it suffices that they be bounded away from zero. Have you looked in a large tome by H.S.Wall -- Continued Fractions ? I haven't. I'll assume they're all positive. If they're periodic, then it seems that the cf converges to the root of a quadratic, whose coefficients are in the field generated by the partial quotients. As a simple example, let's take c_0 = 0, c_n = 1/2 (n>0) See a paper by Bremner & Tzanakis. The convergents are 2/1, 2/5, 10/9, 18/29, 58/65, 130/181, where the numerators and denominators satisfy the recurrence a_n = a_n-1 + 4a_n-2 the denominators are A006131 in OEIS and the numerators twice that (tho this double, 2, 2, 10, 18, 58, 130, ..., seems not to be in OEIS). The cf converges to something with sqrt(17) in it. While not of immediate interest here, I note that c_n = n gives a sequence of convergents whose numerators are duplicated in OEIS as A001053=A103736 (and whose denominators are A001040). If the partial quotients tend to zero, then much of the classical theory still holds, though numbertheoretic aspects tend to disappear. The sequence of convergents either converges or oscillates boundedly. In the following examples I leave the reader to decide which. c_0 = 0, c_n = 1/n 1/1, 1/3, 7/9, 19/45, 159/225, 729/1575, 7407/11025, 48231/99225, ... I've cheated here and not put these in their lowest terms, so that I can say that the numerators and denominators satisfy the recurrence a_n = a_n-1 + n(n-1)a_n-2 The denominators are A000246 in OEIS, but this connexion is not mentioned. The numerators are NOT in OEIS (modulo my not being a good looker). Nor are the numerators or denominators of the convergents to the cf with c_0 = 0 and c_n = 1/2^n: 0/1, 2/1. 2/9, 66/41, 322/1193, 34114/22185, 693570/2465449, ... which satisfy a_n = a_n-1 + 2^2n-1 a_n-2 or the closely related c_n = 1/2^n-1 1/1, 1/3, 9/11, 41/107, 1193/1515, 22185/56299, 2465449/3159019, ... Best to all, R. On Thu, 17 Feb 2005, Andrew Bremner wrote: ...., where can > I find out theorems about the convergence of > > cf = c_0 + 1/c_1 + 1/c_2 +1/c_3+... > > where c_i are rationals? Specifically, when does cf > converge (to a real number?)

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[math-fun] reminder: no attachments
by Richard Schroeppel 22 Feb '05

22 Feb '05

I'd like to remind subcribers that this is a plain-text mailing list. The reason is to accommodate low-bandwidth subscribers (i.e. slow dialup). I'm (painfully) aware that some contributors are forced to use non-text mailers, so we live with it. But please avoid the temptation to include attachments, binary files, pictures, and long program listings. Rich rcs(a)cs.arizona.edu Math-Fun moderator

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Re: [math-fun] Beyond base 10
by Chris Landauer 22 Feb '05

22 Feb '05

Steve Gray wrote: > 1. When I entered computers, octal and binary were the only other bases, as > I recall. No one in the early days seemed to need anything else. ... the infamous SMSG ``New Math'', which i encountered in junior high school in pasadena in the 1960-1 school year, frequently used base 7, just because it was unfamiliar to everyone (i loved it) - but i never met anyone else who needed it (and very few who even wanted it 8-)) i think there was a machine that used T, E, ... for hex digits, but i can't remember, since i didn't use that machine (i only heard about it) one of the nice things in the (few) Russian math papers i've read is the use of certain Roman letters for variable names, so they do not need even to be italicized (they are easy to distinguish from the Cyrillic) more later, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org

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