Bits - symbols mapping in M-PAM beginner question
$2^k$-ary PAM maps $k$ bits (which can be expressed as integers in the range  $[0, 2^k-1]$ to $2^k$  levels that are equally-spaced and symmetric about $0$. ($0$ is NOT one of the levels).. The levels are odd integers, and so for $k=2$, the levels are $-3, -1, +1, +3$ which are equally spaced and symmetric about $0$.  As to which integer in $[0,2^k-1]$, there are several common conventions such as:
\begin{align}
00 &amp;\to -3\\
01 &amp;\to -1\\
11 &amp;\to +1\\
10 &amp;\to +3
\end{align}
or
\begin{align}
00 &amp;\to -3\\
10 &amp;\to -1\\
11 &amp;\to +1\\
01 &amp;\to +3
\end{align}
or
\begin{align}
10 &amp;\to -3\\
11 &amp;\to -1\\
01 &amp;\to +1\\
00 &amp;\to +3
\end{align}
etc.
In all cases, there is a one-bit change in going from one level to the next. Note also that your formula $2m-M-1, m = 1,2,\ldots, M$ corresponds to the first list if you assume that $m-1$ is expressed as a $\log_2M$-bit integer and expressed in Gray code so that adjacent symbols differ in only one of the $\log_2M$ bits.