Prove that $a ( n ) = b( n + 2)$ 
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that  $a(n) = b(n+2)$. for $n=1,2,3...$

My approach:
$a(1) = 1$ (only) , $b(3) = 3$
$a(2) = 2$, because $2=1+1=2$ ,and  $b(4)=2$, because $4=2+2=4$
$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$
$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$
$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$
By this way i am able to show that  $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which  i can  understand.
Background:-This problem is from pathfinder for Olympiad mathematics.
 A: Hint: Observe that the sequence $1, 2, 3, 5, 8, \ldots $ looks like the Fibonacci sequence, offset by 1
Hint: Show that $a_n$ is indeed the Fibonacci sequence (offset by 1), because $ a_{n+2} = a_{n+1 } + a_{n} $, with initial conditions $ a_1 = 1, a_2 = 2$.
Hint: Show that $b_n, n \geq 2$ is indeed the Fibonacci sequence, because $b_{n} = b_{n-2} + b_{n-3} + b_{n-4} + \ldots + b_3 + b_2 + 1$, and $ b_2 = 1, b_3 = 1$.
Hence, $ a_n, n \geq 1 $ and $ b_n, n \geq 3$ are the Fibonacci sequence offset by 1.
A: For compositions into ones and twos we have the OGF
$$A(z) = \sum_{q\ge 1} (z+z^2)^q =
\frac{z+z^2}{1-z-z^2}
= -1 + \frac{1}{1-z-z^2}.$$
Compositions into parts at least two have OGF
$$B(z) = \sum_{q\ge 1} (z^2+z^3+\cdots)^q
= \frac{z^2/(1-z)}{1-z^2/(1-z)}
= \frac{z^2}{1-z-z^2}.$$
We then have for $n\ge 1$
$$[z^n] A(z) = [z^n] \left(-1 + \frac{1}{1-z-z^2}\right)
= [z^n] \frac{1}{1-z-z^2}
\\ = [z^{n+2}] \frac{z^2}{1-z-z^2} = [z^{n+2}] B(z)$$
as claimed.
A: It is also possible to exhibit a bijection between compositions of $n$ using only $1$ and $2$ and compositions of $n+2$ that do not use $1$.
Suppose that $n=c_1+c_2+\ldots+c_k$, where $c_i\in\{1,2\}$ for $i=1,\ldots,k$. Let $c_0=2$; then $n+2=c_0+c_1+\ldots+c_k$. Let $i_1<\ldots<i_\ell$ be the set of indices $i$ such that $c_i=2$, and let $i_{\ell+1}=k+1$. For $j=1,\ldots,\ell$ let
$$d_j=\sum_{i=i_j}^{i_{j+1}-1}c_i\;;$$
then $n+2=d_1+\ldots+d_\ell$ is a composition of $n+2$ that does not use $1$.

Example: The composition $1+2+2+1+1+2$ of $9$ is first extended to the sum $2+1+\color{blue}2+\color{red}{2+1+1}+\color{green}2$, which is then reduced to the composition $3+\color{blue}2+\color{red}4+\color{green}2$ of $11$.

This mapping is clearly injective, and it’s not hard to see that it is surjective as well. Let $n+2=d_1+\ldots+d_\ell$ be a composition of $n+2$. Replace each $d_i$ by its composition
$$2+\underbrace{1+\ldots+1}_{d_i-2}$$
and drop the leading $2$ to get the composition
$$n=\underbrace{1+\ldots+1}_{d_1-2}+2+\underbrace{1+\ldots+1}_{d_2-2}+\ldots+2+\underbrace{1+\ldots+1}_{d_\ell-2}\;;$$
applying the original procedure to this composition of $n$ recovers the composition $n+2=d_1+\ldots+d_\ell$ of $n+2$.
