For 20 points it was enough to carefully implement a brute force algorithm calculating values of all pirate codes.

Let's introduce some notation:

• ~C(k)~ - number of pirate entries of length ~k~.
• ~S(k)~ - sum of values of all pirate entries of length ~k~.
• ~f(n)~ - sum of values of all pirate codes of length ~n~.

We can easily notice that ~f(n) = \sum_{k=1}^n C(k) \times f(n-k) + 2^{n-k} \times S(k)~. It only remains to determine ~C(k)~ and ~S(k)~.

Number of pirate entries of length ~k~ is ~C(k) = Fib[k-1]~ because we see that from a pirate entry of length ~k-2~, by changing the last digit to ~0~ and concatenating two ones at the end we get a pirate entry of length ~k~ that ends with ~1011~. From a pirate entry of length ~k-1~, by changing second to last digit to ~0~ and concatenating a single ~1~ at the end, we get a pirate entry of length ~k~ that ends with ~0011~. From this analysis we get a recursion of Fibonacci numbers and we can easily check the basis for ~k = 1, 2~.

Pirate entries of length ~k~ represent numbers from ~Fib[k]~ to ~Fib[k+1]-1~ so the following holds:

$$\displaystyle S(k) = \frac{Fib[k+1](Fib[k+1]-1)-Fib[k](Fib[k]+1)}{2}.$$

Now we can finish the task with dynamic programming in time complexity ~\mathcal{O}(n^2)~.