quite trivial from the formal perspective. The main problem lies in the fact that a bulk system is considered to consist of an infinite number of “*atoms in the system*”(AIS). Calculating the atomic charge densities for all AIS can, for a periodic system, be reduced to only the atoms in a single unit cell since all periodic copies should yield the same results.\
In addition to this, also the summation limits in Eqs. ,  and change. Where for molecules the sum over $B$ is a finite sum over all AIM, it becomes an infinite sum over all AIS. Because atomic CDDs drop exponentially, the density contribution to the “prosystem” CDD $\rho_{prosys}(\mathbf{r})=\sum^{AIS}_{B}{\rho_B^{\circ}(n_{B}^{\circ},\mathbf{r})}$ of atoms at larger distances becomes negligible. This allows us to truncate the infinite sum to include only the atoms within a certain ‘*sphere of influence*’ (SoI), *i.e.* all atoms of which the contribution to the prosystem CDD is not negligible. Within the iterative Hirshfeld-I scheme we then get: $$\label{eq:Hir_w_iterativeHirshfeldDef}
w^{H,i}_{A}(\mathbf{r}) = \frac{\rho^{i-1}_{A}(n_{A}^{i-1},\mathbf{r})}{\sum_{B}^{SoI}\rho_{B}^{i-1}(n_{B}^{i-1},\mathbf{r})} \quad \forall A\ \epsilon\ \mathrm{unit\ cell}$$ where $$\sum_{A}^{SoI}w^{H,i}_{A}(\mathbf{r})\leq \sum_{A}^{AIS}w^{H,i}_{A}(\mathbf{r})\equiv 1 \qquad \forall \mathbf{r},$$ with $i$ indicating the iteration step, and $\rho^{i}_{A}(n_{A}^{i},\mathbf{r})$ the atomic CDD for an atom $A$ with a population $n_{A}^{i}$ given by $$\label{eq:Hir_pop_int}
n_{A}^{i} 