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Talk:3x+1 problem
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Sequence of powers of 2 reached before the final downfalls
It would also be interesting to have the sequence of maximal heights, and the powers of 2 reached before the final downfalls. (message corrected: the powers of 2 reached before the final downfalls ARE NOT the maximal heights reached)
Daniel Forgues 15:41, 10 August 2010 (UTC)
- The sequence of maximal heights is already in the OEIS as A025586. However, the sequence of powers of 2 reached before the final fall does not appear to be in. I've calculated it as 1, 2, 16, 4, 16, 16, 16, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 64, 16, 16, 16, 16, ... using ColPow2[n_Integer] := Module[{i = n}, While[Not[IntegerQ[Log[2, i]]], i = f[i]]; Return[i]]. The number 16 occurs a lot, at least in the range where term visibility is concerned.
- Alonso del Arte 17:53, 10 August 2010 (UTC)
- P.S. With f of course defined thus: f[1] := 1; f[n_?OddQ] := f[n] = 3n + 1; f[n_?EvenQ] := f[n] = n/2
- P.P.S. It could also be argued that a(1) = 4 since 3 * 1 + 1 = 4.
- I'm still confused as to whether or not 16, 4, 16, 16, 16, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 64, 16, ... is the sequence you find "would also be interesting." Alonso del Arte 22:41, 13 August 2010 (UTC)
I am considering the sequence of powers of 2, i.e 2^k, that lead to the final downfalls of the hailstones. Or one may consider the sequence k of exponents of 2 instead, that makes for another sequence. Daniel Forgues 00:02, 14 August 2010 (UTC)
- If I understand you correctly, the sequence is A135282. Alonso del Arte 01:14, 14 August 2010 (UTC)
- Yes, this is it! (Largest k such that 2^k appears in the trajectory of the Collatz 3x+1 sequence started at n.) So at end of the hailstone sequence, we have the exponent of 2 go k, k-1, k-2, ..., 2, 1, 0, 2, 1, 0, 2, 1, 0, 2, 1, 0, ... the hailstone bounces forever... Daniel Forgues 03:30, 14 August 2010 (UTC)
- Alright, I'll add it to the page. Alonso del Arte 21:41, 14 August 2010 (UTC)
Further negative cycles
There exist two other negative Cycles. One is longer:
- -17, -50, -25, -74, -37, -110, -55, -164, -82, -41, -122, -61, -182, -91, -272, -136, -68, -34, -17.
The other is shorter:
- -1, -2, -1
--Karsten Meyer 00:59, 30 December 2010 (UTC)
- Ah yes, the first one you mention appears in the OEIS as A003124, albeit without the negative signs. Alonso del Arte 23:46, 30 December 2010 (UTC)
- A003124 is on 3n-1. It is, in a special way, the twin (or the mirror) of 3n+1.
- In 3n-1 is the -1, -4, -2, -1 cycle. --Karsten Meyer 00:05, 31 December 2010 (UTC)
- Say, Karsten, would you do me the honor of selecting the first Sequence of the Day for 2011? It can be about the 3n + 1 problem or about some other topic that has interested you of late. And you don't have to put it in when the ball drops, you could put it in today at Template:Sequence of the Day#SOTD for the next holiday at your leisure today and someone else takes care of moving it up at the appropriate time. Alonso del Arte 00:52, 31 December 2010 (UTC)
- Thanks very much, Karsten. Alonso del Arte 01:39, 31 December 2010 (UTC)
- Say, Karsten, would you do me the honor of selecting the first Sequence of the Day for 2011? It can be about the 3n + 1 problem or about some other topic that has interested you of late. And you don't have to put it in when the ball drops, you could put it in today at Template:Sequence of the Day#SOTD for the next holiday at your leisure today and someone else takes care of moving it up at the appropriate time. Alonso del Arte 00:52, 31 December 2010 (UTC)
Will have to mention once it's approved
I am amazed that the sequence Numbers n such that the trajectory of 3n + 1 under the '3x + 1' map reaches n. was not already in the OEIS. If the current proposal for A219696, mention of that sequence should be added to the article. - Alonso del Arte 01:08, 3 December 2012 (UTC)
The 3x+1 link of main article doesn't work
The 3x+1 link of main article doesn't work (as of now).--Bill McEachen 02:39, 1 April 2016 (UTC)
- additionally, I offer the following conjecture: Considering Collatz starting from 2 sequentially, starting from n>2, each evaluation will encounter a subsequence for a (lower start) starting value<n (which obviously ended at 1). As example, for start(3) we have 3,10,5,16,8,4,2,1. This encounters "2,1" which was seen for start(2). Starting from n=5, we immediately see the subsequence seen in start(3) of 5,16,...1. If true, Collatz conjecture would necessarily be true.--Bill McEachen 02:39, 1 April 2016 (UTC)
- The last link in the history that seems to work is
- https://oeis.org/w/index.php?title=3x%2B1_problem&oldid=1562097 [2012-02-24T07:44:01]
- It appears that my next edit
- https://oeis.org/w/index.php?title=3x%2B1_problem&oldid=1562104 [2012-02-24T08:09:59] (graph of function)
- which uses an expensive graphing template to plot the function is just excessively expensive (although it worked before since we were able to do 7 edits after). I'm afraid we might have to revert to that version (and recover the following edits by comparing versions in the history). IT DOESN'T WORK! Comparing [2012-02-24T08:09:59] with [2012-02-24T07:44:01] leaves a blank page! It looks like we won't be able to recover the edits after the last working one. May I go ahead and restore version
- https://oeis.org/w/index.php?title=3x%2B1_problem&oldid=1562097 [2012-02-24T07:44:01]
- and loose the following edits? — Daniel Forgues 16:40, 1 April 2016 (UTC)
- The following are essentially restatements of Collatz conjecture:
- What you are saying: Considering Collatz starting from 2 sequentially, starting from n > 2, each evaluation will encounter a subsequence for a (lower start) starting value < n (which obviously ended at 1).
- Each evaluation will encounter a power of 2 (actually, a power of 4), since then the hailstone just falls all the way down to 1.
- — Daniel Forgues 17:01, 1 April 2016 (UTC)
- I will respond on your Talk page.--Bill McEachen 00:05, 2 April 2016 (UTC)
- — Daniel Forgues 17:01, 1 April 2016 (UTC)
- https://oeis.org/w/index.php?title=3x%2B1_problem&oldid=1562097 [2012-02-24T07:44:01]
- Should I restore:
- https://oeis.org/w/index.php?title=3x%2B1_problem&oldid=1562097 [2012-02-24T07:44:01]
- Unfortunately, we will loose the next 8 edits which refuse to render... Look at the history:
- The last 8 edits don't render. — Daniel Forgues 01:17, 2 April 2016 (UTC)
- I will revert to the last link in the history that seems to work (the server does not send any message, e.g. "time limit exceeded" or whatever, so we we are left wondering whether the database entry of the next version was corrupted somehow, although this is not supposed to happen with modern DBMS)
- https://oeis.org/w/index.php?title=3x%2B1_problem&oldid=1562097 [2012-02-24T07:44:01] — Daniel Forgues 17:24, 3 April 2016 (UTC)
- I sent an email to Neil Sloane about the last 8 edits of 3x+1 problem failing to render, to make sure there are no other options than to lose the last 8 edits. — Daniel Forgues 17:53, 3 April 2016 (UTC)
- 3x+1 problem now renders! I restored the latest version from the raw wikitext and disabled and hid the prohibitive template {{Vertical bar graph from lists}}!
I used https://oeis.org/w/index.php?title=3x%2B1_problem&action=raw to extract the latest [2013-06-03T07:47:48] raw wikitext, then disable and hide the too expensive
{{Vertical bar graph from lists}}. — Daniel Forgues 18:35, 6 April 2016 (UTC)
- 3x+1 problem now renders! I restored the latest version from the raw wikitext and disabled and hid the prohibitive template {{Vertical bar graph from lists}}!
Path or dead end to a potential proof by induction of the Collatz conjecture?
Path or dead end to a potential proof by induction of the Collatz conjecture? — Daniel Forgues 06:21, 25 February 2012 (UTC)
If we could prove that if the iterated reduced Collatz function reaches 1 for all odd numbers from 3 to is assumed true implies that reaches 1 is also true, we would have a proof by induction of the Collatz conjecture.
Or, if we could prove that if the iterated reduced Collatz function reaches an odd number less than for all odd numbers from 3 to is assumed true implies that reaches an odd number less than is also true, we would have a proof by induction of the Collatz conjecture.