The question is a bit old, but I've just studied the subject and so I can provide a detailed answer.
If you just want the answer :
The minimum of nodes in an AVL tree of height h is f(h+2) - 1 where f is the Fibonacci sequence, defined as follow :
f(0) = 1, f(1) = 1, f(n+2) = f(n+1) + f(n).
Proof :
We can prove this relationship by proving the two following proposition by recursion :
Let s(n) be the minimum number of nodes in an AVL tree of height n.
We have : s(0) = 1, s(1) = 2, s(n+2) = s(n+1) + s(n) + 1.
Proposition 1 : s(n) = f(0) + f(1) + f(2) + ... + f(n)
This can quite easily be showed by recursion :
First, at rank 0 and 1, we have :
s(0) = f(0), s(1) = 2 = f(0) + f(1)
Now let's show that if the propostion is true at rank n and n+1, then it is true at rank n+2 :
s(n+2) = s(n+1) + s(n) + 1 [formula for s(n+2)]
= (f(0) + f(1) + ... + f(n+1)) + (f(0) + f(1) + ... + f(n)) + 1 [apply Proposition 1 at rank n and n+1]
= f(0) + ((f(1) + f(0)) + (f(2) + f(1)) + ... + (f(n+1) + f(n))) + 1 [rearrange the sums]
= f(0) + (f(2) + f(3) + ... + f(n+2)) + 1 [apply f(n+2) = f(n+1) + f(n)]
= f(0) + f(1) + (f(2) + f(3) + ... + f(n+2)) [f(1) = 1 and rearrange]
= f(0) + f(1) + ... + f(n+2)
Which by recursion shows that the property is true for any n greater than 0.
Proposition 2 : f(n+2) - 1 = f(0) + f(1) + f(2) + ... + f(n)
This proof is left as an exercice to the reader, but it's exactly the same principle as the previous one.
Now we have, by applying props 1 and 2 successively :
s(n) = f(0) + f(1) + ... + f(n) = f(n+2) - 1
There is a general expression for the Fibonacci sequence, so we can apply it for n+2 to very quickly compute the minimum number of nodes in an AVL tree of any height.