How to prove this inequality: $f(2h-1)≤\frac{3h-1}{2}$

Question

Let $I$ denote the set of odd integers. If $k \in I$, then $3k+1$ is even, so $3k+1=2^a k'$ with $k' \in I$ and $a \ge 1$. The Syracuse function is the function $f:I \to I$, for which $f(k) = k'$ (sequence A075677 in the OEIS).

Some properties of the Syracuse function are:

• $\forall k \in I, f(4k+1) = f(k)$. (Because $3(4k+1) + 1 = 12k+4 = 4(3k+1)$.)
• In more generality: For all $p \ge 1$ and $g \in I$, $f^{p-1}\left(2^ph-1\right) = 2 \cdot 3^{p-1}h - 1$. (Here, $f^{p-1}$ is the function iteration notation.)
• $$\forall h\in I, f(2h-1) \le \frac{3h-1}{2}$$

The Collatz conjecture is equivalent to the statement that, $\forall k \in I$, there exists and integer $n \ge 1$ such that $f^n(k)=1$.

I dont understand why it must be

$$f(2h-1)≤\frac{3h-1}{2}??$$

I cannot prove this inequality.

Answer 1

Consider $f(\color{blue}{2h-1})$; how many factors of two can we divide $3(\color{blue}{2h-1})+1$ by ? \begin{eqnarray*} 3(2h-1)+1=6h-2=2(3h-1) \end{eqnarray*} and $3h-1$ is even, so there is at least another factor of $2$. So \begin{eqnarray*} f(2h-1) \leq \frac{3h-1}{2}. \end{eqnarray*}

Answer 2

Note that you can generalize a step further:

from $h\cdot2^p-1$ you always go up till $h\cdot3^p-1$ (which is even if $h$ is odd, so is not the result of any $f(k)$)

Same way, you always go from $h\cdot4^p+1$ down to $h\cdot3^p+1$.

Or from $h\cdot8^p-5$ up to $h\cdot9^p-5$

Or from $h\cdot2048^{p}-17$ to $h\cdot2187^{p}-17$