Can one do anything useful with a functional equation like $g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$? I got 
$$g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$$
as a functional equation for a generating function.  Is there a way to get a closed form or some asymptotic information about the Taylor coefficients from such an equation?
Here g(0) = 1, g'(0) = 0, and g''(0) = 4.
Edit: Thanks, everyone; as has been pointed out , I've just made a mistake in obtaining the recurrence relation.  I'll leave this as-is instead of editing because I think some of the comments will be helpful to others and if I change it now they won't make sense.
 A: Suppose you know that $g(x) = f(x) g(x^2)$ where $f(0) = 1$ (this assumption is crucial). Then applying this recurrence infinitely many times gives
$$g(x) = \prod_{n \ge 0} f(x^{2^n}).$$
The hypothesis that $f(0) = 1$ guarantees that this product makes sense as a formal power series. For example, if $g(x) = \frac{1}{1 - x}$, then $f(x) = 1 + x$, and we obtain the infinite product
$$\frac{1}{1 - x} = \prod_{n \ge 0} (1 + x^{2^n}).$$
This infinite product encodes the existence and uniqueness of the binary repesentation of a nonnegative integer. 
A: I get that
the only solution is
$g(x) = 0$
if we can write
$g(x)
=\sum_{n=0}^{\infty} a_n x^n
$.
Here is my proof:
We have
$g(x^2) 
= \frac{4x^2-1}{2x^2+1}g(x)
$
or
$(2x^2+1)g(x^2) 
= (4x^2-1)g(x)
$.
From this,
as copper.hat pointed out,
$g(0) = 0$.
If
$g(x)
=\sum_{n=1}^{\infty} a_n x^n
$,
(since
$g(0) = 0$)
the left side is
$\begin{align*}
\sum_{n=1}^{\infty} (2x^2+1)a_nx^{2n}
&=\sum_{n=1}^{\infty} 2a_nx^{2n+2}+\sum_{n=1}^{\infty} a_nx^{2n}\\
&=\sum_{n=2}^{\infty} 2a_{n-1}x^{2n}+\sum_{n=1}^{\infty} a_nx^{2n}\\
&=a_1x^2+\sum_{n=2}^{\infty} (2a_{n-1}+a_n)x^{2n}\\
\end{align*}
$
and the right side is
$\begin{align*}
\sum_{n=1}^{\infty} (4x^2-1)a_nx^{n}
&=\sum_{n=1}^{\infty} 4a_nx^{n+2}-\sum_{n=1}^{\infty} a_nx^{n}\\
&=\sum_{n=3}^{\infty} 4a_{n-2}x^{n}-\sum_{n=1}^{\infty} a_nx^{n}\\
&=-a_1x-a_2x^2+\sum_{n=3}^{\infty} (4a_{n-2}-a_n)x^{n}\\
\end{align*}
$
Equating coefficients,
$-a_1x = 0$,
$-a_2x^2 = a_1x^2$,
and
$2a_{n-2}+a_n
=4a_{2n-2}-a_{2n}
$
and
$0
=4a_{2n-1}-a_{2n+1}
$.
Therefore
$a_1 = a_2 = 0$
and
$a_{2n}
=4a_{2n-2}-2a_{n-2}-a_n
$
and
$a_{2n+1}
=4a_{2n-1}
$.
From this second recurrence,
since $a_1 = 0$,
$a_{2n+1} = 0$
for all $n$.
From
$a_{2n}
=4a_{2n-2}-2a_{n-2}-a_n
$,
for 
$
n=2,
a_4
=4a_2-2a_0-a_2
=0
$,
for 
$
n=3,
a_6
=4a_4-2a_2-a_3
=0
$.
By strong induction,
all the
$a_{2n} = 0$,
so the only solution is
$g(x) = 0$.
