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.

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.
