Generating Function for $(4,9,16,25,36,.....)$ I have a sequence $(4,9,16,25,36,...)$ it is being generated by $a_n=(n+1)^2)$
I have found the generating function for $n^2$ here: 
Proving the generating function of $n^2$.
I know I can shift a sequence:
$(1,4,9,16,25,....)$ to the right $(0,1,4,9,25,...)$ via $x^1*A(x)$ with $ A(x)$ a generating function. Does this also work with a left shift? Is there a definition? $n$ was a natural numbers. Maybe something else? 
 A: Let $f(t)=t^2+t^3+t^4+\dots=\dfrac{t^2}{1-t}$, for $|t|<1$.
Then $f'(t)=2t+3t^2+4t^3+\dots=\dfrac{2t-t^2}{(1-t)^2}$
And
$$\dfrac{\mathrm d\left(tf'(t)\right)}{\mathrm dt}=4t+9t^2+16t^3\dots=\frac{t^3-3t^2+4t}{(1-t)^3}$$
Finally
$$4+9t+16t^2+\dots=\frac{t^2-3t+4}{(1-t)^3}$$
A: We consider a power series $A(x)$ with constant term $a_0=0$.
\begin{align*}
A(x)=\sum_{j=j_0}^\infty a_jx^j
\end{align*}
A power series $B(x)=\sum_{j=j_0}^\infty a_{j+1}x^j$ can be derived via
\begin{align*}
A(x)&=\sum_{j=j_0}^\infty a_j x^j=\sum_{j=j_0-1}^\infty a_{j+1} x^{j+1}\\
&=x\sum_{j=j_0}^\infty a_{j+1}x^j+a_{j_0}x^{j_0}\\
&=xB(x)+a_{j_0}x^{j_0}
\end{align*}

We conclude the left-shifted power series $B(x)$ has the representation
  \begin{align*}
\color{blue}{B(x)=\frac{1}{x}A(x)-a_{j_0}x^{j_0-1}}
\end{align*}
In the current situation we have
  \begin{align*}
A(x)&=\frac{x(1+x)}{(1-x)^3}\\
&=x+4x^2+9x^3+16x^4+\cdots\\\\
\color{blue}{B(x)}&=\frac{1}{x}A(x)-1\\
&=\frac{1+x}{(1-x)^3}-1\\
&=\frac{x(x^2-3x+4)}{(1-x)^3}\\
&\,\,\color{blue}{=4x+9x^2+16x^3+\cdots}
\end{align*}

A: An alternative way to think about it is to consider the forward differences of the sequence, $1,4,9,16,25,\ldots$ which is $3,5,7,9,\ldots$, and then forward differencing again we get constants.  We translate this into generating function language:
Knowing that $$f(x) = 4 + 9x + 16x^2 + 25x^3 + \cdots$$
We calculate that $$\tag{1} f(x) - xf(x) - 4 = 5x + 7x^2 + 9x^3 + \cdots \equiv g(x)$$
and then $$\tag{2} g(x) - xg(x) - 5x = 2x^2 + 2x^3 + 2x^4 + \cdots = \frac{2x^2}{1-x}$$
From $(2)$ we solve for $g(x)$ as $$g(x) = \frac{-3x^2 + 5x}{(1-x)^2}$$
and then substitute back into $(1)$ to find 
$f(x) = \frac{g(x) + 4}{1-x} = \frac{x^2 - 3x + 4}{(1-x)^3}$
This gives a good brute force approach for computing the generating function of any sequence whose terms are given by evaluating a polynomial at evenly spaced points.  
