There's actually a reasonably good mathematical reason for doing bit replication:

First note that the n-bit string, $N$, actually represents the value $\frac{N}{2^n-1}$ and we want to produce the m-bit string, $M$, where $n<m$ and
$$\frac{N}{2^n-1}\approx\frac{M}{2^m-1}$$

We first scale numerator and denominator with
$$\frac{N.(2^n+1)}{(2^n-1)(2^n+1)}\approx \frac{M}{2^m-1}$$
and this simplifies to 
$$\frac{N.(2^n+1)}{2^{2n}-1}\approx \frac{M}{2^m-1}$$

In your case, $n\in \{5,6\}$ and $m=8$ and we can "stop" here, but but the process can be repeated, (ad nauseum), if m >> n.

We next make the approximation...
$$\frac{N.(2^n+1)}{2^{2n}}\approx \frac{M}{2^m}$$ which simplifies to 
$$\frac{N.(2^n+1)}{2^{2n-m}}\approx M $$

Note that $N.(2^n+1)$ is equivalent to repeating the n-bit string, to create a 2n-bit string, and the division shifts off the $2n-m$ LSBs to leave an M bit result. 

QED

Of course, the  'correct' calculation is $M=\lfloor(\frac{(2^m-1) N}{2^n-1}+\frac{1}{2}\rfloor$ but this approximation, in general, works *most* of the time. Of course there are times when it's inaccurate, but IIRC only by one bit and relatively infrequently.