Generating-functions: is there a relationship between a generating function and the corresponding squared generating function Let's say we have a sequence $T(n)$ with the corresponding generating function
$$A(t) = \sum_{n = 0}^\infty T(n) t^n$$
Is there some relationship between the two functions $A(t)$ and $A(t^2)$? And for that matter is there some generalization for any integer power or $t$?
Edit: I'm actually trying to solve for the generating function $A(t)$ in the equation
$$A(t) + (1+t)A(t^2) = t/(1-t^2)$$
this is what inspired my question. My intuition suggested to me that I should look for some kind of relationship between $A(t^2)$ and $A(t)$, hence the vagueness of my question.
 A: I believe you're interested in this sequence.
I generated the series of coefficients directly from your functional equation in A using a couple of lines of Haskell:
sq (a:as) = a : 0 : sq as
a2 = sq a
a = 0 : 1 : tail (tail (zipWith (-) (cycle [0,1]) (zipWith (+) a2 (0:a2))))

I then looked up the series in the sequence database.
A: Ah, that's a much more specific question.  In that case, you should do one of two things:


*

*Rewrite the given condition in the form A(t^2) = (something that involves A(t)) and iterate it to see what you get.

*Compute the first few terms of the series and guess how they continue, then prove your guess.
A: Well, considering the operator 
$\Omega(A)=A(t)+(1+t)A(t^2)$
one sees that $\Omega(A)[0]=2A[0]$. 
So, an equation $\Omega(A)=B$ 
 with $B[0]=0$ 
implies that $A[0]=0$.
Now the operator $\Omega$ 
acts on series with zero constant term as $\Omega=I+N$ 
with  $I$ identity and $N(A)=(1+t)A(t^2)$ which is 
topologically nilpotent. Then 
$$
\Omega^{-1}=I-N+N^2-N^3+\ldots
$$
In this case $\Omega(A)=B$ (in case $B[0]=0$ 
which is your case) has only one solution which is  
\begin{eqnarray}
B-(1+t)B(t^2)+(1+t)(1+t^2)B(t^4)+\ldots +\cr 
(-1)^{k}\Big((1+t)\ldots 
(1+t^{2^{k-1}})\Big)B(t^{2^k})+\ldots  
\end{eqnarray}
(infinite sum). 
This is easy to program and gives all asymptotic expansions of equations of type 
$$
A(t)+(1+t)A(t^2)=B\ ;\ B[0]=0
$$
I tried it for $B(t)=\frac{t}{1-t^2}$ (your question) and $B(t)=sin(t)$.    
A: Alright, so on the one side, you have this:
$$A(t)+(1+t)A(t^{2})=\sum_{n=0}^{\infty}T(n)t^{n}+\sum_{n=0}^{\infty}T(n)t^{2n}+\sum_{n=0}^{\infty}T(n)t^{2n+1}$$
On the other side, you have:
$$\frac{t}{1-t^{2}}=\sum_{n=0}^{\infty}t^{2n+1}$$
Equating the coefficients of $x^{2k}$, you have the relation: $T(2k)+T(k)=0$.
Equating the coefficients of $x^{2k+1}$, you have the relation: $T(2k+1)+T(k)=1$.
Now you can start computing the coefficients: $T(0)=0$, $T(1)=1$, $T(2)=-1$, $T(3)=0$, etc.
sigfpe correctly identified the sequence. You can even see these recurrences mentioned in the formula section.
A: I think what you're looking for is a relationship between the coefficients of A(t) and the coefficients of A(t2).  There is one:
A(t) = a0 + a1 t + a2 t2 + a3 t3 + ...
and
A(t2) = a0 + a1 t2 + a2 t4 + a3 t6 + ...
so the coefficient of tn in A(t2) is the coefficient of tn/2 in A(t) if n is even, and 0 if n is odd.
