Finding general expression of series expansion $A^{-1} = \sum_{n=0}^{\infty} (-1)^n a_n x^n$, $A = \sum_{n=0}^{\infty} a_n x^n$ Let
\begin{align}
&A^{-1} = \sum_{n=0}^{\infty} (-1)^n a_n x^n \\
&A = \sum_{n=0}^{\infty} a_n x^n
\end{align}
Using the relation $A^{-1} A =1$, I want to find the general expression for $a_n$.
My assumption and initial conditions are given as follows with $a_0=1, a_1=1$.
With this I find explicitly, $a_2= \frac{1}{2}$, $a_3 = \frac{1}{4}$, $a_4 = \frac{1}{8}$, $a_5=0, \cdots$. How I can set the general expression for $a_n$?

After seeing the comment from @metamorphy,  I want additionally impose all odd powers vanishes except $a_3$. i.e., $a_{2n+1}=0$ for $n>1$. In this case, is this fix the uniqueness?
Additionally I relax the positiveness condition for $a_n$.
 A: The solution is not unique, but as some in the comments have pointed out the even coefficients in the expansion of A(x) are determined (up to a sign) by the odd coefficients, which are arbitrary.
Let $f(x)$ and $g(x)$ be respectively the even and odd parts of $A(x)$. Then
$$1 = A(x)A(-x) = f(x)^2 - g(x)^2$$
So $f(x) = \pm \sqrt{1+g(x)^2}$ and $A(x) = g(x) \pm \sqrt{1+g(x)^2}$ for any odd function $g(x)$. To expand in series you can apply the fractional binomial theorem.
A: So what you are looking for is an $f$ so that
$$
f(x)f(-x)=1\tag1
$$
and
$$
f(0)=1\tag2
$$
and so that the odd part is
$$
\frac{f(x)-f(-x)}2=x+\frac{x^3}4\tag3
$$
Solving $(3)$ for $f(-x)$ in terms of $f(x)$ and multiplying by $f(x)$ gives
$$
f(x)^2-2\left(x+\frac{x^3}4\right)f(x)-1=0\tag4
$$
Applying the quadratic formula to $(4)$ and remembering $(2)$ gives
$$
f(x)=\left(x+\frac{x^3}4\right)+\sqrt{\left(x+\frac{x^3}4\right)^2+1}\tag5
$$
At the moment, I don't see an easy expression for the coefficients of the Taylor series of $\sqrt{\left(x+\frac{x^3}4\right)^2+1}$ .
However, you say that you deduced that $a_3=\frac14$. Given your initial conditions, you can set $a_3$ to anything and get a solution (just replace the $\frac{x^3}4$ term in $(4)$ and $(5)$).
A: Starting from @robjohn's answer
$$\sqrt{1+\left(x+\frac{x^3}4\right)^2}=1+\sum_{n=1}^\infty \frac{a_n}{2^{2n-1}}\,x^{2n}$$ where the $a_n$ form the (bizarre) sequence
$$\{1,1,-1,1,0,-3,8,-11,0,44,-126,177,34,-892,2498,-3355,-1598,20732,-55374,\cdots\}$$ which does not seem to appear anywhere.
