  A Hilbert space operator $A\in\B$ is left $(X,m)$-invertible by $B\in\B$
(resp., $B\in\B$ is an $(X,m)$-adjoint of $A\in\B$) for some operator $X\in\B$
if
$\triangle_{B,A}^m(X)=\sum_{j=0}^m(-1)^j\left(\begin{array}{clcr}m\\j\end{array}\right)B^{m-j}XA^{m-j}=0$
(resp.,
$\delta_{B,A}^m(X)=\sum_{j=0}^m(-1)^j\left(\begin{array}{clcr}m\\j\end{array}\right)B^{(m-j)}XA^j=0$).
No Drazin invertible operator $A\in\B$, with Drazin inverse $A_d$, can be left
$(I,m)$-invertible (equivalently, $m$-invertible) by its adjoint or its Drazin
inverse or the adjoint of its Drazin inverse. For Drazin inverrtible operators
$A$, it is seen that the existence of an $X$ acts as a conduit for implications
$\triangle_{B,A}(X)=0\Longrightarrow \delta^m_{C,A}(X)=0$, where the pair
$(B,C)=$ either $(A,A_d)$ or $(A_d,A)$ or $(A^*,A^*_d)$ or $(A^*_d,A^*)$.
Reverse implications fail. Assuming certain commutativity conditions, it is
seen that $\triangle_{A^*_d,A}^m(X)=0=\triangle^n_{B^*_d,B}(Y)$ implies
$\delta^{m+n-1}_{A^*B^*,AB}(XY)=0=\delta^{m+n-1}_{A^*+B^*,A+B}(XY)$.
