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I have n functions with n variables: $x_1, \dots , x_n \in \mathbb{N}/0$ that follow a nice pattern:

$f_1(x_1) = \frac{2^{x_1}}{3} - \frac{1}{3}$

$f_2(x_1,x_2) = \frac{2^{x_1+x_2}}{9} - \frac{2^{x_1} \; + \; 3}{9}$

$f_3(x_1,x_2,x_3) = \frac{2^{x_1+x_2+x_3}}{27} - \frac{2^{x_1+x_2} \; + \; 3 \times 2^{x_1} \; + \; 9}{27}$

$f_4(x_1,x_2,x_3,x_4) = \frac{2^{x_1+x_2+x_3+x_4}}{81} - \frac{2^{x_1+x_2+x_3} \; + \; 3 \times 2^{x_1+x_2} \; + \; 9\times2^{x_1} \; + \;27}{81}$

The pattern being that if $x_n = 2$ then $f_n(x_1, \dots ,x_{n-1},2)$ is equal to $f_{n-1}(x_1, \dots ,x_{n-1})$

Because of this it is easy to see that $R(f_{n-1}) \subset R(f_n)$ where $R(f_n)$ is the range of that function.

  • I am after some notation that allows me to figure out what the range of the "limiting" function is (by which I mean, find $R(f_n)$ as $n \rightarrow \infty^+$).

If the range of the limiting function contains all odd numbers then the Collatz conjecture is true. If it does not, then the Collatz conjecture is false.

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    You may find it easier to examine $g_n(y_1, ..., y_n) = f_n(y_1, y_2 - y_1, y_3 - y_2, ..., y_n - y_{n-1})$. Then $$g_{n+1}(...) = 3^{-n-1}\left(2^{y_{n+1}} - \sum_{k=1}^n 3^{n-k}2^{y_k}\right)$$ ${g_n}$ has the same image as $f_n$ ("range" is deprecated because of ambiguity as to its meaning). – 2017-02-09
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    Thanks, I'll have a play and see what I can figure out. – 2017-02-09

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