The top voted answer in Is it possible to describe the Collatz function in one formula? looks like a really complicated function at first sight.
Let me restate the function in it's original form here; $$f(n)=\frac74n+\frac12+(-1)^{n+1} \left(\frac54n+\frac12\right)$$
I do not understand what is so special about this function when there are simpler non-modular functions than that.
Of course one author of another answer says, quote
..the way the Collatz function is written definitively influences how it's studied
is an acceptable (and somewhat good) thing to state, but the former author does not describe fully what the formula is all about, why are the fractions the way they are. When so many have accepted it as the "best" answer, then there must be something special about it or was it just a fad of the moment? Maybe someone could explain to me in deeper detail why this function is of interest.
- Has anyone studied it in more detail?
- What can be said about each term and their long term behaviour?
- Why is it so popular?
https://math.stackexchange.com/q/1449997
Edit: Here are a few simpler looking functions:
- $C(n)=n-\lceil \frac{n(-1)^n}{2}\rceil$
- $f(n)=\frac{\lceil\frac{3n}{2}\rceil}{2+(-1)^n}$
- $f(n)=\lceil{2+(-1)^{n-1}\over2}n\rceil$
I think I can answer the question "Why is it so popular?". People like formulas that involve only elementary operations: addition, multiplication, exponentiation.
When you have a modulo 2 expression in your formula, there is a trick to remove it and replace it with an exponential term: namely, $(-1)^n$. In this case, since the piecewise expressions are rather simple, we can use a simple strategy. We write our expression as: $$ f(n)= an+b + (-1)^n (cn+d) $$ Then for $n$ even, we get $f(n)=(a+c)n+(b+d)$, and for $n$ odd, we get $f(n)=(a-c)n+(b-d)$. Then we just have to solve for $a,b,c,d$. In the case of the Collatz function, we want $a+c=1/2$, $b+d=0$, $a-c=3$, $b-d=1$. It is easy to see that this gives the coefficients in the formula you refer to.