Footnote clarifying meaning of low-order 11 bits

Paper.tex

@@ -353,7 +353,7 @@ \subsubsection{Transaction Receipt}
 M(O) \equiv \bigvee_{t \in \{O_a\} \cup O_\mathbf{t}} \big( M_{3:2048}(t) \big)
 \end{equation}
 
-where $M_{3:2048}$ is a specialised Bloom filter that sets three bits out of 2048, given an arbitrary byte sequence. It does this through taking the low-order 11 bits of each of the first three pairs of bytes in a Keccak-256 hash of the byte sequence. Formally:
+where $M_{3:2048}$ is a specialised Bloom filter that sets three bits out of 2048, given an arbitrary byte sequence. It does this through taking the low-order 11 bits of each of the first three pairs of bytes in a Keccak-256 hash of the byte sequence./footnote{11 bits $= 2^2048$, and the low-order 11 bits is the modulo 2048 of the operand, which is in this case is "each of the first three pairs of bytes in a Keccak-256 hash of the byte sequence."} Formally:
 \begin{eqnarray}
 M_{3:2048}(\mathbf{x}: \mathbf{x} \in \mathbb{B}) & \equiv & \mathbf{y}: \mathbf{y} \in \mathbb{B}_{256} \quad \text{where:}\\
 \mathbf{y} & = & (0, 0, ..., 0) \quad \text{except:}\\