Re: [math-fun] Collatz 3N+1 problems
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
The Dan's proposal is true for all odd integers n, from 1 to 1,000,000,000. Each fall into C_1 or C_17. In this range, the longest way is with n=458,788,881: 193 steps to get 1. The percentage F_1 of odd integers n falling into C_1 is not very stable, but is always: 78.4% < F_1 < 79.2% Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Daniel Asimov Envoyé : vendredi 15 septembre 2006 22:54 À : math-fun Objet : Re: [math-fun] Collatz 3N+1 problems 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I forgot to say that, in the same range, the longest way for the C_17 family is with n=789,829,547: 168 steps to get 17 Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : samedi 16 septembre 2006 12:31 À : 'math-fun' Objet : RE: [math-fun] Collatz 3N+1 problems The Dan's proposal is true for all odd integers n, from 1 to 1,000,000,000. Each fall into C_1 or C_17. In this range, the longest way is with n=458,788,881: 193 steps to get 1. The percentage F_1 of odd integers n falling into C_1 is not very stable, but is always: 78.4% < F_1 < 79.2% Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Daniel Asimov Envoyé : vendredi 15 septembre 2006 22:54 À : math-fun Objet : Re: [math-fun] Collatz 3N+1 problems 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Can you get good evidence of the 1:17 frequency oscillation? Does it look random, or periodic, sinusoidal, ?? Even if it's not provable, getting a description of the behavior would be very interesting. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Christian Boyer Sent: Sat 9/16/2006 5:19 AM To: 'math-fun' Subject: RE: [math-fun] Collatz 3N+1 problems I forgot to say that, in the same range, the longest way for the C_17 family is with n=789,829,547: 168 steps to get 17 Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : samedi 16 septembre 2006 12:31 À : 'math-fun' Objet : RE: [math-fun] Collatz 3N+1 problems The Dan's proposal is true for all odd integers n, from 1 to 1,000,000,000. Each fall into C_1 or C_17. In this range, the longest way is with n=458,788,881: 193 steps to get 1. The percentage F_1 of odd integers n falling into C_1 is not very stable, but is always: 78.4% < F_1 < 79.2% Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Daniel Asimov Envoyé : vendredi 15 septembre 2006 22:54 À : math-fun Objet : Re: [math-fun] Collatz 3N+1 problems 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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.
Totally to be expected. These are really the same thing. Let f(N) = (3N+1)/2^K, and g(N) = (3N-1)/2^K. Than a(N) = g(f(N)), b(N) = f(g(N)). So each family is the intermediate values from the other family. Actually, you can combine these into a single, signed function: h(N) = (1-3N)/2^K. Now, we have: 13 -> -19 -> 29 -> -43 -> 65 -> -97 -> 73 -> -109 -> 41 -> -61 -> 23 -> -17 -> 13. Franklin T. Adams-Watters -----Original Message----- From: cboyer@club-internet.fr 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. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
participants (4)
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Christian Boyer -
Daniel Asimov -
franktaw＠netscape.net -
Schroeppel, Richard