Generating a System of Recurrences from a Light Bulb Word Problem Question

A row of $n$ light bulbs must be turned on. Initially they are all off. The first bulb can always be turned on or off. For $i> 1$, the $i$th bulb can be switched (turned on or off) only when $i-1$ is on and all other earlier bulbs are off. Let $a_n$ be the number of switches needed to turn all light bulbs on and let $b_n$ be the number of switches needed to turn on the $n$th bulb for the first time. Find recurrences for $a_n$ and $b_n$.

The above question is from Aigner's A course in Enumeration. 
My Attempt
The initial conditions are $a_0=b_0=0$ and computing the first couple of values gives us for example $a_1=b_1=1$, $a_2=b_2=2$, and $a_3=5;\, b_3=4$. 
We note that $a_n=b_n+a_{n-2}$ for $n\geq 2$. Indeed, to switch on all lights we must first switch on the $n$th light for the first time (and after doing so light $n-1$ and $n$ are on, all others off) and then switch on the reamining $n-2$ lights.
I was unable to come up with a recurrence for $b_n$.
My Problem
I looked in the back of the book and the following recurrence for $b_n$ is given 
$$
b_n=a_{n-1}+a_{n-2}+1\quad (n\geq 2)\tag{1}
$$
How does one explain the recurrence above (combinatorially)? My idea is that to get to the state that light $n$ is on for the first time it must be the case that we get to the state when only $n-2$ and $n-1$ are on and then we turn off $n-2$ and turn on $n$. But I am unable to relate it to the number of switches to have an initial segment of lights on. 
Any help is appreciated.
 A: The sequence of moves to reach the state where bulb $n$ is on for the first time looks something like this:
$$
00000000\to \dots \to 00000010\to 00000011
$$
However, if we could show it furthermore looked something like this:
$$
00000000\;\;\underbrace{\to \dots\to}_{a_{n-1}}\;\;  11111110 \;\;\underbrace{\to \dots\to}_{a_{n-2}}\;\; 00000010\to 00000011
$$
then the recurrence would be proved. The reason the second block takes $a_{n-2}$ moves is because of the symmetry of the problem; turning on $n-2$ takes the same time as turning off that many. 
So, we would be done if we could prove this lemma:

Lemma: Any toggling of the $n^\text{th}$ light is preceded by a time where lights $1$ through $n-1$ are all on. 

Proof: This can be proven by induction. Suppose that any toggling of light $k$ is preceded by a time where lights $1$ to $k-1$ are all on, for all $k<n$. Now, before a toggling of light $n$, we must reach state
$$
0000001*\tag{1}
$$
which is preceded at some point in the past by
$$
0000011*\tag{2}
$$
By induction, to move from $(2)$ to $(1)$, which involves a toggling of bulb $n-2$, there needs to a time where bulbs $1$ through $n-3$ are all on. But in $(2)$, we additionally have that bulbs $n-2$ and $n-1$ are on, so all of bulbs $1$ through $n-1$ are on, as desired. 
