algorithms - When not to assume $n$ is a power of 2, while solving recurrences? In Udi Manber - Introduction to Algorithms. A Creative Approach, exercise 3.29, p.59

Although in general it is sufficient to evaluate recurrence relations only for powers of 2, that is not always the case. Consider the following relation:
$$T\left(n\right)=\begin{cases}
T\left(\frac{n}{2}\right)+1 & \text{if $n$ is even}\\
2T\left(\frac{n-1}{2}\right) & \text{if $n$ is odd}
\end{cases}$$
a. Prove that the solution of this recurrence for powers of $2$ is $T\left(2^{k}\right)=k+1$ (namely, for powers of $2$, $T\left(n\right)=O\left(\log n\right)$)
b.  Show that, for an infinite number of values of $n$, $T\left(n\right)=\Omega\left(n\right)$. Discuss why the usual assumption about the relative behavior for powers of $2$ and nonpowers of $2$ breaks down for this recurrence.

I have no problem solving this problem, but there an important section of the solution provided in the book make me confused, and I can't search for explanations on the Internet or books about that:

We can restrict ourselves to powers of $2$ if the solution of the recurrence relation is a function that behaves "nicely." In particular, the expressions for the running times of most algorithms are monotonically increasing functions. If the difference between the value of $T\left(n\right)$ and $T\left(2n\right)$ is no more than a constant and if $T\left(n\right)$ is monotonically increasing, then $T\left(m\right)$ for $n < m < 2n$ is no more than a constant times $T\left(n\right)$. Therefore, the analysis for powers of $2$ is sufficient in that case.


Here is what I understand about the bold-italic part above:
Consider, 
$$\begin{align*}
T\left(n\right)&=cT\left(\frac{n}{2}\right)+O\left(n\right)\\T\left(2n\right)&=cT\left(n\right)+O\left(2n\right)\\&=c^{2}T\left(\frac{n}{2}\right)+O\left(n\right)\\
\end{align*}$$
The difference between $T\left(n\right)$ and $T\left(2n\right)$ is a constant $c$, and it does not affect the analyzing of $O\left(\right)$ of the recurrence if we chose $2n$ as the input value or any $m$, $n<m<2n$.
Now what I confused is what is the case of not to assume that $n$ is a power of $2$. According to the book, it's not applicable for the case, such as $T\left(n\right)=nT\left(\frac{n}{2}\right)$, when $T\left(2n\right)=2n^{2}T\left(\frac{n}{2}\right)$.
But I see some examples were still solved by this way, such as exercise 8.2.53, p.527 K. Rosen's Discrete Mathematics:
$$T\left(n\right)=nT^{2}\left(\frac{n}{2}\right)$$
or this one.
So, my questions are:


*

*Am I wrong?

*Could you provide simple example(s) of monotonic $T\left(n\right)$ where replacing $n$ by a power of 2 is not applicable? (I desire $T\left(n\right)$ have unified formula, without different formulas for specified values of $n$, such as the example provided in Udi Manber's book above).
Thanks!

For the case of Kenneth Rosen's exercise, I posted an answer's here. As the result of solution 2, I choose $n$ such that $2^{m-1}<n<2^{m}$, in particular, $n=2^{m-1}\cdot2^{\frac{1}{2}}=2^{m-\frac{1}{2}}$.
$$\begin{align*}
T\left(n\right)&=\frac{n^{2^{\left\lfloor \log_{2}n\right\rfloor }-1}\cdot n^{2^{\log_{2}n-\left\lfloor \log_{2}n\right\rfloor }}}{2^{\left\lfloor \log_{2}n-2\right\rfloor 2^{\left\lfloor \log_{2}n\right\rfloor }+2}\cdot2^{\left(\log_{2}n-\left\lfloor \log_{2}n\right\rfloor \right)2^{\log_{2}n-\left\lfloor \log_{2}n\right\rfloor }}}6^{n}\\&=\frac{n^{2^{m-1}-1}\cdot n^{2^{m-\frac{1}{2}-\left(m-1\right)}}}{2^{\left\lfloor m-\frac{1}{2}-2\right\rfloor 2^{m-1}+2}\cdot2^{\left(m-\frac{1}{2}-\left(m-1\right)\right)2^{m-\frac{1}{2}-\left(m-1\right)}}}6^{n}\\&=\frac{n^{2^{m-1}-1}\cdot n^{\sqrt{2}}}{2^{\left(m-3\right)2^{m-1}+2}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&\\&=\frac{n^{2^{m-1}-1}\cdot n^{\sqrt{2}}}{2^{\left(m-\frac{1}{2}-3+\frac{1}{2}\right)2^{m-1}+2}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{n^{2^{m-1}}\cdot n^{\sqrt{2}-1}}{4\cdot\left[n\cdot2^{-\frac{5}{2}}\right]^{2^{m-1}}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{n^{2^{m-1}}\cdot n^{\sqrt{2}-1}}{4\cdot n^{2^{m-1}}\cdot2^{-5\cdot2^{m-2}}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{n^{\sqrt{2}-1}}{4\cdot2^{-5\cdot n\cdot2^{-1-\frac{1}{2}}}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{2^{\frac{5n}{2\sqrt{2}}}\cdot n^{\sqrt{2}-1}}{4\cdot2^{\frac{1}{\sqrt{2}}}}6^{n}
\end{align*}$$
When, if $n=2^{m}$, $T\left(n\right)=\frac{6^{n}\cdot4^{n-1}}{n}$.
So, I think Udi Manber's statement is right. While I don't exactly know why some recurrences would be solved by replace $n=2^{m}$, but my guess is that it's easier to do, and the results are acceptable (??) (Hope someones would correct it if I was wrong).
