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- From: steiner@andy.bgsu.edu (Ray Steiner)
- Newsgroups: sci.math
- Subject: Re: That Collatz Flu
- Message-ID: <BuKtr5.9q3@andy.bgsu.edu>
- Date: 14 Sep 92 16:28:16 GMT
- References: <BuF596.64L@news.cso.uiuc.edu>
- Organization: Bowling Green State University B.G., Oh.
- Lines: 37
-
- In article <BuF596.64L@news.cso.uiuc.edu>, levine@symcom.math.uiuc.edu (Lenore Levine) writes:
- > I've gotten interested in the Collatz function. (I guess it's something
- > every math person has to go through for a week, like the flu.)
- >
- > (Collatz function: f(n) = n/2 if n is even
- > = (3n + 1)/2 if n is odd.
- >
- > It is conjectured that for every n > 0, there is a k such that f^k(n)
- > (the kth iterate of f) = 1.)
- >
- > What I'm wondering is: Has it been shown that there are no n > 2 such
- > that f^k (n) = n?
- >
- > Also, any major results later than the 1985 survey article?
- >
- > Lenore Levine
-
- First of all, the first question is the finite cycles conjecture.
- This is still open for the 3x+1 problem. I played around with
- the Qx+1 problem(Q odd integer) some years ago
- and after a bit of computer searching for
- Q<1000 I found that only 5 and 181 yielded cycles
- (for small x, of course)!
- Are there any other Q that give cycles? For the 3x+1 problem,
- it is known that any nontrivial cycle must contain at least
- 250000 terms!
- Also consider the set of all odd n which have a divergent
- trajectory in the 3x+1 problem. Is the smallest member of
- this set congruent to 27(mod 48)? I conjectured this some
- years ago, but have never been able to prove it. Obiviously,
- the smallest member of this set cannot be of the form 4k+1,
- but how can one eliminate all 8k+7 as possible smallest
- element?
- Hope this makes the Collatz flu a bit more bearable!
- Ray Steiner
- --
- steiner@andy.bgsu.edu
-