I have the following recurrence $$A_m = -(1/2)^m A_{m-1} - (1/2)^{2m}A_{m-2},$$ where $A_0 = 0, A_1 = -1$. I am hoping to explicitly solve for $A_n$.
I know that explicitly solving recurrences is often not possible. The only technique I have really seen is the use of generating functions, so by defining $$F(x) = \sum_{k=2}^{\infty} A_k x^k, $$ but in this case I cannot express the LHS in terms of $F$ and derivatives of $F$, except if I allow $F(x)$ to be expressed in terms of $F(x/2)$ and $F(x/4)$, but I don't see how to use that.
Searching online, I only found information on linear recursions, but that could be caused by the fact that I don't know what the type of recursion that I have is called. Still, I have the feeling that it can be solved explicitly, since Mathematica easily finds an answer (and it does not for most recursions).
Could anyone help me solve this recursion or provide information on how hard it is to solve by hand?
Let $\,B_m = 2^{m(m+1)/2}A_m = 2^{1+2+\ldots+m}A_m\,$, then multiplying the recurrence by $\,2^{1+2+\ldots+m}\,$:
$$ \begin{align} 2^{1+2+\ldots+m}A_m &= -2^{1+2+\ldots+m}\frac{A_{m-1}}{2^m} - 2^{1+2+\ldots+m}\frac{A_{m-2}}{2^{2m}} \\ 2^{1+2+\ldots+m}A_m &= - 2^{1+2+\ldots+(m-1)}A_{m-1} - \frac{1}{2} \cdot 2^{1+2+\ldots+(m-2)}A_{m-2} \\ B_m &= - B_{m-1} - \frac{1}{2}\,B_{m-2} \end{align} $$
The latter is a linear homogeneous recurrence with constant coefficients, which can be solved for $\,B_m\,$ using the standard techniques, then reverting the substitution $\,A_m = 2^{-m(m+1)/2} B_m\,$.