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I was playing around with the Collatz Conjecture and came up with the following:

Take any positive integer. If it's odd, multiply by three and add one. If it's even, divide by two. Repeat indefinitely and you'll eventually get one. This is the Collatz Conjecture.

More compactly:

a = A = some positive integer
i = 0
while a != 1
    if a%2

We can alternatively say:

a = A = some positive integer
i = 0
while a != 2^n

while a != 1

This yields the pattern:

a = A           = A                 i = 0
a =(A*3+1)      = 3A + 1            i = 1
a =(A*3+1)*3+1  = 9A + 3 + 1        i = 2

Taking this and making something more math-ish we get:

$$a = 3^iA + \sum_{k=0}^{i-1} 3^k.$$

Playing around with that we can say:

$$a = 3^iA + \sum_{k=0}^{i-1} 3^k $$

$$a -3^iA = \sum_{k=0}^{i-1} 3^k $$

$$3^{-i}a -A = \sum_{k=0}^{i-1} 3^{-i}3^k = \sum_{k=0}^{i-1} 3^{k-i} = \sum_{k=0}^{i-1} \cfrac{1}{3^{i-k}} = \sum_{k=1}^{i} \cfrac{1}{3^{k}} = \cfrac{1}{2} - \cfrac{1}{2}3^{-i}$$

Which lets us say:

$$3^{-i}a -A = \cfrac{1}{2} - \cfrac{1}{2}3^{-i}$$

$$3^{-i}a + \cfrac{1}{2}3^{-i} -A = \cfrac{1}{2}$$

$$3^{-i}(a + \cfrac{1}{2}) -A = \cfrac{1}{2}$$

$$3^{-i}(a + \cfrac{1}{2}) = \cfrac{1}{2} + A$$

$$3^i = \cfrac{a + \cfrac{1}{2}}{\cfrac{1}{2} + A}$$

$$i = \log_3\cfrac{a + \cfrac{1}{2}}{\cfrac{1}{2} + A} = \log_3(a + \cfrac{1}{2}) - \log_3(\cfrac{1}{2} + A) .$$

If the conjecture applies, then $a = 2^n$ where $n$ is all positive integers. So:

$$i = \log_3(2^n + \cfrac{1}{2}) - \log_3(\cfrac{1}{2} + A).$$

Since $i$ can only be a positive integer, that means the decimal portions of the two terms on the right must match.

This would mean that for all values of $A$ and $n$ (positive integers) if $\log_3(\cfrac{1}{2} + A)$ can produce any decimal ending which $\log_3(2^n + \cfrac{1}{2})$ can't, the conjecture wouldn't apply to that value of $A$.

I feel that most likely this isn't terribly useful for proving/disproving the conjecture, mainly as my knowledge of math isn't that great and I'm the person who came up with it, but I thought someone else might be interested in the approach and either be able to take it further or point out exactly why it's useless.

Beyond this point I've thought of a couple random things which quite possibly have no basis in reality:

Working from:

$$i = \log_3(2^n + \cfrac{1}{2}) - \log_3(\cfrac{1}{2} + A).$$

As $A \rightarrow \infty$, $$\log_3(A + \cfrac{1}{2}) \rightarrow \infty,\text{ and }\cfrac{d(\log_3(A + \cfrac{1}{2}))}{dA} \rightarrow 0.$$

So clearly this term very quickly covers quite a few irrational decimal endings. (Maybe all of them? Repeatedly?)

As $n \rightarrow \infty$, $$\log_3(2^n+\cfrac{1}{2}) \rightarrow \infty, \quad\cfrac{d(\log_3(2^n+\cfrac{1}{2}))}{dn} = (\log_32)(1-\cfrac{1}{2^{n+1}+1}) \rightarrow \log_32$$

Which is to say, the larger $n$ gets, the more stably $\log_3(2^n+\cfrac{1}{2})$ moves through decimals.

The derivative has two terms:

$$\log_32\text{ and }-\cfrac{\log_32}{2^{n+1}+1}.$$


$$\ -\infty < \int_{0}^{\infty} -\cfrac{\log_32}{2^{n+1}+1} dn < \infty$$

$-\cfrac{\log_32}{2^{n+1}+1}$ would seem not to be a factor in greatly increasing the number of decimal endings.

Which just leaves $\log_32$.

So, assuming the $-\cfrac{\log_32}{2^{n+1}+1}$ term is a non-factor, then:

If the decimal endings of $n \log_32 + C$ aren't a superset of $\log_3(\cfrac{1}{2} + A)$'s decimal endings, then there should exist some number(s) the conjecture doesn't apply to. Or not, cause my logic is probably flawed.

Anyway, I thought this might be a good place to get an evaluation of this, and it's at least good for an "Oh, hrm?" moment. Hoping not to end up too embarrassed.

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It's a lot easier for us to read what you write if you use a little TeX. See… for some simple TeXniques. – Gerry Myerson May 25 '12 at 1:28
I don't understand the bit that says "we can alternatively say... while a != 2^n" - what about 6? 6 != 2^n but 6 should go to 3 not to 19? – robert king Nov 5 '12 at 23:08
chat somewhere, anyone? some new ideas on collatz – vzn Mar 4 '13 at 17:14

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