Average difference between two odd numbers of equal length If I select two different odd numbers of binary length $l$, what is the formula that will tell me the average difference between those two numbers?
Note that the high order digit must always be $1$, so for example, qualifying numbers of length $4$ are from $9$ to $15$ inclusively, qualifying numbers of length $5$ are from $17$ to $31$ inclusively, etc.
For length $l=3$, the answer is $2$ (the only possible numbers are $5$ and $7$).  For length $l=4$, the answer is $3.3333...$ For length $l=5$, the answer is $6$.  But I'd like to be able to calculate it for much larger lengths, and explore what happens as the length $l$ gets larger.

We try to simplify the set of all differences of odd $l$-digit binary numbers. Ignoring the first and last bits, because they are always $1$, we look at the differences of all numbers from the remaining $l-2$ digits, which is to say, numbers $1,2,\ldots,2^{l-2}$. Note, because we dropped the last digit, we must multiply our new result by $2$. Now, the sum, $S$, of all possible differences is
\begin{eqnarray*}
S &=& 2\sum_{i=2}^{2^{l-2}} \sum_{j=1}^{i-1}{(i-j)} \\
&=& 2\sum_{i=2}^{2^{l-2}} \sum_{j=1}^{i-1}{j} \\
&=& \sum_{i=2}^{2^{l-2}}{(i^2 - i)} \\
&=& \dfrac{1}{6} 2^{l-2} \left(2^{l-2}+1\right) \left(2^{l-1}+1\right) - \dfrac{1}{2} 2^{l-2} \left(2^{l-2}+1\right) \\
\end{eqnarray*}
using the Sum of Squares formula.
Now, the number, $N$ of these number-pairs, is
$$N=\binom{2^{l-2}}{2} = 2^{2l-5} - 2^{l-3}.$$
So the expected value is
$$\dfrac{S}{N} = \dfrac{2^{l-1}+2}{3}\qquad\text{after some tedious simplification.}$$
