TITLE: Recurrence relation for number of bit strings of length n that contain two consecutive 1s
QUESTION [13 upvotes]: I'm pulling my hair out over a review question for my final tomorrow.

Find a recurrence relation (and its initial conditions) for the number of bit strings of length n that contain two consecutive 1s.

Here is my thought process/work, could someone please help me figure out where I'm going wrong?
Knowing that a bit string of length 0 or 1 cannot contain two consecutive bits of any kind, and that only one bit string of length 2 (the string 11) can contain two consecutive ones, I have the initial conditions a(0) = 0, a(1) = 0, and a(2) = 1. I'm now trying to find a formula for the term a(n+2).
I know there are three ways for a bit string of length n + 2 to have two consecutive 1s:

Condition X: Both n + 1 and n + 2 are 1. count(X) = 2^n because this still leaves n bits unaccounted for.
Condition Y: Both n and n + 1 are 1. count(Y) = 2^n for the same reason.
Condition Z: The first n bits contain 11 somewhere. count(Z) = 4 * a(n) because adding 2 bits to the a(n) condition quadruples the number of possible strings.

Since these three conditions are not mutually exclusive,
a(n+2) count(X) + count(Y) + count(Z) - (count(X $\cup$ Y) + count(X $\cap$ Z) + count(Y $\cap$ Z)) + count($X \cap Y \cap Z)$

or 
a(n+2) = 2^n + 2^n + 4 * a(n) - (2^(n-1) + a(n) + 2 * a(n-1)) + a(n-1)

This successfully proves the initial condition a(2) = 1 and the next one - a(3) = 3. But when I plug in 4, I get 9. Listing out the possible strings by hand and circling the ones that satisfy the condition gives only 8. What am I doing wrong?
Thanks.

REPLY [8 votes]: Here's an alternate way of solving your question.
Let $a_n$ be the number of bit strings of length $n$ containing a pair of consecutive $1$'s.
In order to construct such a bit string, one could start with

$0$ and follow with a string of length $(n-1)$ containing a pair of consecutive $1$'s
$10$ and follow with a string of length $(n-2)$ containing a pair of consecutive $1$'s
$11$ and follow with any string of length $(n-2)$

Since these three cases are mutually exclusive and exhaust the possibilities for how such a string might start, we can apply the sum rule and write down the following recurrence relation valid for all  $n \ge 2;$
$a_n = a_{n-1} + a_{n-2} + 2^{n-2}$
(Recall that there are $2^k$ bit strings on length $k$).
Now for the initial conditions:

$a_0 = 0$
$a_1 = 0$