

Note:

n(h) = 1 + n(h−1) + n(h−2) ...... (1) n(1) = 1 n(2) = 2 
>> eqn := n(h) = 1 + n(h1) + n(h2); eqn := n(h) = 1 + n(h  1) + n(h  2) >> init := n(0) = 1, n(1) = 2; init := n(0) = 1, n(2) = 2 >> z := rsolve( {eqn, init}, n ); / 1/2 \ / 1/2\h / 1/2 \ / 1/2\h 3 5   5   3 5   5   + 1/2 1/2 +  +   + 1/2 1/2    1 \ 10 / \ 2 / \ 10 / \ 2 / 1/2 / 2 \h 1/2 / 2 \h 2 5   2 5    1/2  1/2 \ 1  5 / \ 1 + 5 /   +  1/2 1/2 5 (1  5 ) 5 (1 + 5 ) ~= 2 × 1.6^{h} 
Example:
> simplify( subs(h=5, z) ); 20 > 2*1.6^5; 20.97152 
n(h) = minimum number of nodes in an AVL tree of height h ~= 2 × 1.6^{h} ....... (2) 

m(h) > min. # nodes in AVL tree of height h > n(h) > 2 × 1.6^{h} 

(This was a rough estimate, I gave you a more accurate estimate above)

