Collatz sequence

[Ronnie Ginzburg]

Assuming the Collatz conjecture is true, each initial value V leads to a finite sequence the last member of which is 1.
Let the number of even and odd values in the sequence be E and N respectively.
I noticed that as V tends to infinity, the ratio E/N may tend to a value L close to 2.
Data suspiciously point to the exact value L = 2 but this is a dubious conjecture...
Clearly, the convergence is not uniform (is it considered pointwise?).
Can analysis find an exact limit L?

 

[Cubist]

Consider the expected value for the number of consecutive even values after a single odd value. Immediately after an odd value there is always an even number in the Collatz sequence. After that assume that the probability of an even or odd at every step is 50 50.

Total number of even values after an odd (before the next odd number turns up)	Probability that this will occur
1	0.5 (because there's a 50-50 chance that the next value after the first even will be an odd)
2	0.25 (draw a probability tree if you're not sure about this)
3	?
n	P(n)

Once you have an expression for P(n), then the expected value of E/N is given by

[math] \sum_{n=1}^{\infty} n*P(n) [/math]
Can you work this out?

 

[Ronnie Ginzburg]

Not sure I understand — actually, I'm sure I don't understand...
A simple 50-50 chance predicts the evens/odds ratio to tend to the golden ratio.
The sequence is finite, and in the end, we always meet a power of 2, 2^k, which terminates the sequence adding k evens.
??? HELP...

 

[Ronnie Ginzburg]

The "Golden Ratio" bit was a silly mistake of min:; I counted the (same generation) leaves of a biased binary tree representing Collatz possibilities.
A simple computer simulation arrives at a ratio which indeed tends to 2 as the Collatz sequence lengthens.
Sorry to have troubled you...

 

[Cubist]

Not sure I understand — actually, I'm sure I don't understand...
A simple 50-50 chance predicts the evens/odds ratio to tend to the golden ratio.

I'm not sure how you obtained the golden ratio, perhaps you could post your work? When I use the method outlined in post#2 I get an expected value for "E/N" of exactly 2. Extra hint:- use P(n) = 2^-n

The sequence is finite...

You asked about what happens when V tends to infinity, which would surely result in an infinite sequence? I guess you could argue a special case like V=2^infinity, which would still result in an infinite sequence (always dividing by two). But I thought it would be sensible to assume V is tending towards infinity in unit steps, and therefore it's a good assumption that V's infinite binary representation would contain a 50-50 ratio of '0' and '1' digits, rather than mostly zeros.

I think the slight doubt in my mind is:- would multiplying V by 3 change the distribution of binary '0' and '1' (V is a tending towards infinity random number)? I would argue that (a random number)*3 is still a random number, therefore 50-50 chance of odd/even is reasonable.

 

[Cubist]

I didn't see your message#4 before I wrote the above. It was probably awaiting moderator approval.

Sorry to have troubled you...

No trouble, I like the Collatz conjecture! I spent some time thinking about it a couple of years ago. I made some rules for navigating the sequence in reverse (starting at 1) and tried to prove that this would reach all positive integers. Needless to say I didn't succeed. I'm not sure I would recommend this approach to a possible proof, in the end it just seemed like a needless complication!