September 2006 - mathfun@mailman.xmission.com

[math-fun] Skolem-like primes sequence
by Eric Angelini 18 Sep '06

18 Sep '06

Hello SeqFans and Math-Fun, is this of interest? best, É. http://www.cetteadressecomportecinquantesignes.com/SkolemPrimes.htm

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RE: [math-fun] Collatz 3N+1 problems
by Christian Boyer 18 Sep '06

18 Sep '06

As well remarked by Franklin, (a) and (b) can be combined in a single function with negative integers in the sequences. However... at least because my program ran as I defined previously... with the initial definitions (a) and (b) using positive integers, in the range of odd integers from 1 to 10,000,000,000: 1) All integers fall in (a) 1 or 17 families (b) 1 or 13 families 2) Longest ways to get 1: (a) 212 steps with 8,316,337,203 (b) 206 steps with 6,337,579,165 3) Longest ways to get 17(a) or 13(b): (a) 190 steps with 7,575,090,961 (b) 204 steps with 8,774,370,049 4) Percentage P1 of numbers giving 1, from 1 to n (a) 78.45% < P1 < 79.15% (b) same 5) Percentage P1' of numbers giving 1, from n to n+100,000,000 (a) 62% < P1' < 94% (b) same 6) Biggest reached term within sequences (a) 32,705,624,140,216,093,177 with n = 5,376,042,513 (and same term for some others bigger n) (b) 24,529,218,105,162,069,883 with n = 4,032,031,885 Examples on 5) >From 6,400,000,000 to 6,500,000,000: P1'(a) = 93.1% Rrom 6,800,000,000 to 6,900,000,000: P1'(a) = 62.4% Supplemental remarks on 4) and 5) The oscillations of (a) and (b) are complementary (when P1'(a) is big, P1'(b) is small). Periods increasing with n. Maybe some other percentages or oscillations that I can't see, within n to n+100,000,000. Christian. -----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : dimanche 17 septembre 2006 22:49 À : 'math-fun' Objet : RE: [math-fun] Collatz 3N+1 problems After Dan's (a), I propose to define a second function (b): (a) M=(3N+1)/2^K then P=(3M-1)/2^K. (b) M=(3N-1)/2^K then P=(3N+1)/2^K. Interesting to see very similar properties: (a) two families, 1 and 17 (-> 19 -> 43 -> 97 -> 109 -> 61 -> 17...) (b) two families, 1 and 13 (-> 29 -> 65 -> 73 -> 41 -> 23 -> 13...) Similar percentages 1:17 for (a) and 1:13 for (b)??? Soon a report on the 2 functions from 1 to 10,000,000,000. Christian.

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Re: [math-fun] Collatz 3N+1 problems
by Daniel Asimov 17 Sep '06

17 Sep '06

Just for fun, tried a new algorithm in this vein recently, a mapping f: Odds+ -> Odds+ defined by: first doing N -> M=(3N+1)/2^K (where 2^K | 3N+1, but 2^(K+1) doesn't) and then doing M -> P=(3M-1)/2^L (where 2^L | 3M-1, but 2^(L+1) doesn't) We then say P = f(N). Trying each odd from 1 to 10,000, found the trivial cycle, a fixed point of f: C_1: 1 (-> 1) and the lone interesting cycle C_17: 17 -> 19 -> 43 -> 97 -> 109 -> 61 (-> 17). An interesting question (IF all positive integers fall into one or the other of these cycles) is: What fraction of pos. ints. fall into C_1, and what fraction fall into C_17 (assuming these sets have densities). --Dan

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Re: [math-fun] some graphs for distributions modulo 1.
by Daniel Asimov 17 Sep '06

17 Sep '06

Simon writes: << I have been collecting some old ideas with graphs, especially the ones related to distributions mod 1, this is used to be my favorite hobby, . . . >> . . . and then you hit puberty? Just kidding. These are fascinating graphics. Let me make sure I get it: Using various criteria, you get a whole bunch of empirical distributions mod 1, i.e. largish finite subsets of [0,1) -- and then you plot these on a circle by x -> exp(2pi i x), and finally you connect all pairs of these points on the circle to get a (largish) union of chords of the circle. (Or is something else going on?) Sometime in the past year I recall discussing (here?) the limiting *2D* distribution of [n points chosen from a circle at random, yielding all the points *inside* the circle determined by *intersections* of pairs of chords connecting the n points] -- as n -> oo. I think I see a general way to use the answer to this relatively simple question in order to describe the limiting 2D distributions suggesed by Simon's graphics . . . but more specificallly the limiting distributions of the *intersection points* of the chords (rather than those of the chords themselves, which looks a bit harder). --Dan

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RE: [math-fun] Lost in fraction
by Eric Angelini 17 Sep '06

17 Sep '06

Waow, so quick, thanks, Christoph ! (and forgive my poor math skills !) This fraction has a nice self-referential smell (in french) : 13847/1111 = 12.4635 4635 4635 4635 (a) ... take any digit of (a); when written in french, this digit has as much letters as described by the next digit : UN (2 letters) DEUX (4 letters) QUATRE (6 letters) SIX (3 letters) TROIS (5 letters) CINQ (4 letters) QUATRE (6 letters) ... What could be a similar english fraction ? Best and thanks, É. -----Message d'origine----- De : math-fun-bounces+eric.angelini=kntv.be(a)mailman.xmission.com [mailto:math-fun-bounces+eric.angelini=kntv.be@mailman.xmission.com] De la part de Pacher Christoph Envoyé : mardi 5 septembre 2006 18:00 À : math-fun Objet : RE: [math-fun] Lost in fraction Hello, >is there a (simple) fraction whose result is : >12.4635 4635 4635 4635 ... ? 12+4635/9999= 13847/1111, isn't it? since it is unique (after canceling), simpler is not possible... best wishes Christoph

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[math-fun] some graphs for distributions modulo 1.
by Simon Plouffe 17 Sep '06

17 Sep '06

Hello, I have been collecting some old ideas with graphs, especially the ones related to distributions mod 1, this is used to be my favorite hobby, http://www.lacim.uqam.ca:16080/~plouffe/distributions%20modulo%201/ some others are even older here : http://www.lacim.uqam.ca:16080/~plouffe/1229/ I believe there are some intriguing graphics for this one : http://www.lacim.uqam.ca:16080/~plouffe/distributions%20modulo%201/target33… http://www.lacim.uqam.ca:16080/~plouffe/distributions%20modulo%201/target40… and http://www.lacim.uqam.ca:16080/~plouffe/distributions%20modulo%201/target36… If you have any explanations, feel free to post them, i would be glad to hear any comment. Simon Plouffe

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Re: [math-fun] Collatz 3N+1 problems
by Daniel Asimov 16 Sep '06

16 Sep '06

Thanks, Neil. I always feel honored to receive a mention in your wonderful OEIS. Regards, Dan

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Re: [math-fun] Collatz 3N+1 problems
by N. J. A. Sloane 15 Sep '06

15 Sep '06

Dan, as a result of your message i am adding these two sequences to the OEIS. Comments or corrections welcomed! Neil %I A102421 %S A102421 1,7,1,1,5,25,7,17,19,43,1,13,7,61,1,35,37,79,5,11,23,97,25,53,55,115,7, %T A102421 31,1,133,17,71,73,151,19,5,41,169,43,89,91,187,1,49,25,205,13,107,109, %U A102421 223,7,29,59,241,61,125,127,259,1,67,17,277,35,143,145,295,37,19,77,313 %N A102421 Start with 2n+1, multiply by 3 and add 1 and divide out any power of 2; then mult\ iply by 3 and subtract 1 and divide out any power of 2. %C A102421 When a(x) is iterated, what are the limit cycles? Are there any besides (1) and (\ 17 -> 19 -> 43 -> 97 -> 109 -> 61)? %e A102421 n=1, 2n+1 = 3 -> 10 -> 5; 5 -> 14 ->7 = a(1). %O A102421 0,2 %K A102421 nonn %Y A102421 Cf. A102423. %p A102421 f:=proc(n) local j; j:=3*n+1; while j mod 2 = 0 do j:=j/2; od: j:=3*j-1; while j \ mod 2 = 0 do j:=j/2; od: j; end; %A A102421 njas, based on email from Dan Asimov (dasimov(AT)earthlink.net),Sep 15 2006 %I A102423 %S A102423 1,1,1,1,1,1,1,17,17,17,1 %N A102423 Start at 2n+1, iterate the map x -> A102421(x); sequence gives smallest number in\ resulting cycle, or -1 if the trajectory never cycles. %O A102423 0,8 %K A102423 nonn,more %A A102423 njas, based on email from Dan Asimov (dasimov(AT)earthlink.net) Sep 15 2006

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Re: [math-fun] 1,2,3,5,18 puzzle
by Daniel Asimov 15 Sep '06

15 Sep '06

<< Puzzle: What do the positive integers 1, 2, 3, 5, 18 have in common with each other, but with no other positive integer (conjecturally) >> To clarify: There is no "catch", and the answer is purely mathematical. --Dan

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[math-fun] Collatz 3N+1 problems
by Schroeppel, Richard 14 Sep '06

14 Sep '06

I looked a little at the 3N+5 and 3N+7 problems. My impression, based on some hand calculations, is that 3N+5 has more loops than 3N+-1, and 3N+7 has even more. [Only A = +-1(mod 6) are interesting as 3N+A problems.] If we use a prime divisor, branching for example on N (mod 5), then there's an obvious procedure for "deciding" if the overall average drift of the stepping function tends to increase or decrease the numbers, leading to either everything heading toward small numbers, or almost everything diverging. The procedure is to examine the output multipliers. The Collatz problem can be written as 2N -> N 2N+1 -> 6N+4 Any multipliers that are divisible by the input's dispatch prime (2 for Collatz) must have the rule applied to the output: 6N+4 -> 3N+2. (This can lead to infinite loops in some cases, or to fixed numbers like ->5.) In the ordinary case, we wind up with all the multipliers being indivisible by the dispatch prime. Collatz becomes 2N -> N 2N+1 --> 3N+2 Now we abandon rigor and guess that inputs are uniformly distributed mod P (the dispatch prime), and so are outputs. To measure average growth or shrinkage, just multiply together all the output multipliers, and compare with the product of the input multipliers, P^P. For Collatz, this is 2^2 = 4 on the input side, and 1*3 = 3 on the output side. So we expect Collatz to, on average, shrink the input, matching observations. If we do (odd N)->3N+5 instead of 3N+1, it doesn't matter; we still get shrinkage. But 3N+2 is rather different; the preprocessing step diverges. And 5N+1, instead of 3N+1, also diverges on average. (Again matching observation; unfortunately, very little of what I'm saying can be proven.) For composite dispatch moduli such as 4, 6, or 30, things are more complicated. If the output multipliers all happen to be relatively prime to the dispatch modulus, then the same product business applies. But in the more interesting case, some of the output multipliers share a common factor with the input modulus. If the preprocessing step can be completed to eliminate the common factors, all well and good. But sometimes it can't, and this analysis doesn't work. Two-Counter Machines are the natural setting for Collatz puzzles. I learned about CMs from Minsky's automata theory course that evolved into his book Finite & Infinite Machines. I don't know if he gets credit for inventing Counter Machines, but that's where I first encountered them. There's some small multiple of 30, perhaps 1500, which can be used as a dispatch modulus and will simulate a universal turing machine (via a two-counter machine implementation). I've played around some with dispatch moduli 4, 6, and 8, but have no results other than that there's a lot of variety in behavior. It might be possible to come up with a better preprocessing step that works with prime power dispatchers like 4 and 8, but I think 6 needs a new idea. Rich

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