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.

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.
