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? 

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*}

