\frac{H^{(0)}_N}{D^{(0)}_N} + 
O\left(\frac{A^0}{Q^2}\right)\ .
\label{expLTN}$$ Substituting above Eqs. (\[expLTA\]) and (\[expLTN\]) into our definition of the nuclear broadening of the transverse momentum square in Eq. (\[DLT2\]), we derive $$\Delta\langle \ell_T^2\rangle \approx 
\left[\frac{H^{(0)}_A}{D^{(0)}_A} -
      \frac{H^{(0)}_N}{D^{(0)}_N} \right] +  
\frac{H^{(2)}_A}{D^{(0)}_A} + 
O\left(\frac{A^0}{Q^2}\right)\ ,
\label{DLT2com}$$ where the power suppressed terms $O(A^0/Q^2)$ should be very small due to the cancellation between $\langle \ell_{T}^2\rangle^{eA}$ and $\langle \ell_{T}^2\rangle^{eN}$ and the fact that they are not enhanced by the $A^{1/3}$. The first term in Eq. (\[DLT2com\]) would be exactly equal to zero if the normalized effective nuclear parton distributions $\phi_{a/A}(x)=\phi_{a/N}(x)$. However, due to the well-known nuclear effects in the parton distributions, the first term in Eq. (\[DLT2com\]) does not have to vanish. But, because of very weak $A$-dependence of nuclear parton distributions shown in Fig. \[fig0\], we identify this term with the $a$ in Eq. (\[BA13\]), $$a \approx \frac{H^{(0)}_A}{D^{(0)}_A} -
      \frac{H^{(0)}_N}{D^{(0)}_N}\ .
\label{DLTa}$$ It is the second term in Eq. (\[DLT2com\]) that is responsible for the $A^{1/3}$-type dependence of the nuclear broadening.

We introduce $\Delta \langle \ell_T^2\rangle_{1/3}$ as $$\begin{aligned}
\Delta \langle \ell_T^2\rangle_{1/3} &\equiv & b\, A^{1/3} 
\approx \frac{H^{(2)}_A}{D^{(0)}_A} 
\nonumber \\
&=& \left. 
\int d\ell_T^2 \, \ell_T^2 \, 
\frac{d\sigma^{(D)}_{eA\rightarrow e\pi X}}{dx_B dQ^2 d\ell_T^2} 
\right/ \frac{d\sigma^{(0)}_{eA\rightarrow eX}}{dx_B  dQ^2} \, ,
