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

Solution 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*}

Solution 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 $h\cdot2187^{p}-17$