Let $T_h$ denote the minimal number of nodes that a standard AVL tree of height $h$ could have.

If $h>0$, then there’s the root and two subtrees. One of the subtrees must have height $h-1$. The other must have height at least $h-2$, otherwise it’s not a valid AVL tree. So we have the recurrence:

$$T_h = T_{h-1} + T_{h-2} + 1$$

And we can throw in $T_0 = 0$ and $T_1 = 1$ for obvious reasons.

The easy way to solve the recurrence is to denote $F_h = T_h + 1$. Then:

$$begin{align*}F_0 & = 1 \ F_1 & = 2 \ F_h & = F_{h-1} + F_{h-2}end{align*}$$

That is, they’re Fibonacci numbers with the sequence offset by 3.

Since:

$$F_h approx frac{1}{sqrt{5}} left( frac{1 + sqrt{5}}{2} right)^{h+3}$$

Inverting this, we get:

$$h approx frac{1}{log left( frac{1 + sqrt{5}}{2} right)} log left( T_h right) + 3$$

And that is where the 1.44 comes from.

So what you need is to solve the recurrence:

$$U_h = U_{h-1} + U_{h-3} + 1$$

with suitable initial conditions.

If you can’t think of a clever way to do this, you can do it the not-clever-but-always-works way by observing that:

$$begin{bmatrix} U_h \ U_{h-1} \ U_{h-2} \ 1end{bmatrix} = begin{bmatrix} 1 & 0 & 1 & 1 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1end{bmatrix} begin{bmatrix} U_{h-1} \ U_{h-2} \ U_{h-3} \ 1end{bmatrix}$$

Therefore:

$$begin{bmatrix} U_{h+2} \ U_{h+1} \ U_{h} \ 1end{bmatrix} = begin{bmatrix} 1 & 0 & 1 & 1 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1end{bmatrix}^{h} begin{bmatrix} U_{2} \ U_{1} \ U_{0} \ 1end{bmatrix}$$

Next, perform eigendecomposition on the matrix. Let:

$$M = begin{bmatrix} 1 & 0 & 1 & 1 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1end{bmatrix}$$

Suppose you can find a matrix $Q$ and a diagonal matrix $Lambda$ such that $M = Q Lambda Q^{-1}$. Then:

$$M^h = Q Lambda^h Q^{-1}$$

Taking the power of a diagonal matrix is trivial, and this should give you a closed form. You will then need to find an approximate inverse to solve your problem.