When I was learning the transformation from the Schrodinger picture to the interaction picture in quantum mechanics, I have to use the Baker-Campbell-Hausdorf Lemma, \begin{align} e^ABe^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+\ldots. \end{align} I was wondering what if there are two operators needed to be transformed, namely \begin{align} e^ABCe^{-A}=BC+[A,BC]+\frac{1}{2!}[A,[A,BC]]+\ldots. \end{align} I expand each term in $BC$ as \begin{align} e^ABCe^{-A}=BC+[A,B]C+B[A,C]+\frac{1}{2!}\left\{[A,[A,B]]C+B[A,[A,C]]+2[A,B][A,C]\right\}+\ldots. \end{align} Now I use the shorthand notation that \begin{align} [A,X]=[X]^1,\ \ [A,[A,X]]=[X]^2,\ \ [A,[A,[A,X]]]=[X]^3,\ldots, \end{align} for $X$ is operator $B$ and $C$. Then, it shows \begin{align} e^ABCe^{-A}=&BC+[B]^1C+B[C]^1+\frac{1}{2!}([B]^2C+2[B]^1[C]^1+B[C]^2)\\ &+\frac{1}{3!}([B]^3C+3[B]^2[C]^1+3[B]^1[C]^2+B[C]^3)\\ &+\frac{1}{4!}([B]^4C+4[B]^3[C]^1+6[B]^2[C]^2+4[B]^1[C]^3+B[C]^4)+\ldots. \end{align} I notice that the coefficients of these terms actually follow the rows of the Pascal's triangle. For practical purposes, I want to find an expression of this expansion (I primarily assume that $[B,C]=0$ $[A,B]\neq0$ and $[A,C]\neq0$).
Is this already investigated in mathematics? If it has been, could you share that with me?
Thank you!