Suppose the following series
$$y=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{12}+O(x^{5}).$$
How to obtain the series reversion
$$x=A_{0}+A_{1}y+A_{2}y^{2}+A_{3}y^{3}+O(y^{4}),$$
when the series starts from $a_{0}\neq0$ and the reversed one from $A_{0}\neq0$? Unfortunately, the classical approach based on a substitution
$$y=1+(A_{0}+A_{1}y+A_{2}y^{2}+A_{3}y^{3}+O(y^{4}))+\frac{1}{2}(A_{0}+A_{1}y+A_{2}y^{2}+A_{3}y^{3}+O(y^{4}))^{2}+...,$$
does not work because of impossibility of determine $A_{0}$. I am looking for a method that can be easily implemented without additional libraries in C++...
Thanks for your help.
Updated solution:
Suppose the simplified version of the series \begin{eqnarray*} x & = & A_{1}(y-1)+A_{2}(y-1)^{2}+O((y-1)^{3}),\\ y-1 & = & x+\frac{x^{2}}{2}+O(x^{3}), \end{eqnarray*} then, after a substitution \begin{eqnarray*} y-1 & = & A_{1}(y-1)+A_{2}(y-1^{2})+\frac{1}{2}\left[A_{1}(y-1)+A_{2}(y-1^{2})\right]^{2},\\ y' & = & A_{1}y^{\prime}+A_{2}y^{\prime}y^{\prime}+\frac{1}{2}\left[A_{1}y^{\prime}+A_{2}y^{\prime}y^{\prime}\right]{}^{2},\\ & = & A_{1}y^{\prime}+(\frac{1}{2}A_{1}^{2}+A_{2})y^{\prime}y^{\prime}+O(y^{\prime3}), \end{eqnarray*} where $y^{\prime}=y-1$. Then, $A_{1}=1$, and $\frac{1}{2}+A_{2}=0\Rightarrow A_{2}=-\frac{1}{2}$, and
$$x=(y-1)-\frac{1}{2}(y-1)^{2}+O((y-1)^{3}).$$
To apply series reversion, you need to have $A_0=0$ and off-shift it accordingly:
$$y(0)=1$$
So when $y=1$, we need $x=0$:
$$x(1)=0$$
$$x=A_1(y-1)+A_2(y-1)^2+A_3(y-1)^3+A_4(y-1)^4+\mathcal O((y-1)^5)$$
$$y-1=x+\frac12x^2+\frac16x^3+\frac1{12}x^4+\mathcal O(x^5)$$
And by standard approaches, one may derive
$$x=(y-1)-\frac12(y-1)^2+\frac13(y-1)^3-\frac{23}{24}(y-1)^4+\mathcal O((y-1)^5)$$