the node with degree $k_i$ coincides for all nodes with the same degree. The node value itself enters only multiplicatively into this expression. Hence, the denominator of the sum is equal for all nodes.

Crossing from the summation over all nodes to sums with the same degrees similar transformations as above are made. The different degrees $k_i$ are again divided into classes of units with the same degree $k$ and with occupation number $N_k$. Obviously, $\sum_{k=k_{\mathrm{min}}}^{k_{\mathrm{max}}}N_k=N$ has to be satisfied. The denominator can be rewritten $$\sum_{j=1}^{N}k_j=\sum_{k=k_{\mathrm{min}}}^{k_{\mathrm{max}}}N_k k.  
  \label{eq:rewrite}$$ In the limit of large number of nodes ${N\to \infty}$ this expression becomes $ N \langle k \rangle$ where the symbol $\langle \cdot
\rangle = \sum_k \;\cdot\;p(k)$ assigns averaging over the degree distribution.

Hence, the signal function of the $i$th node becomes $$f_i(t)=\frac{k_i}{N}r(t). 
  \label{eq:signal}$$ Therein, the mean-field amplitude $r(t)$ is defined by $$\begin{aligned}
  r(t) = \frac{1}{\langle k \rangle} \,
  \sum_{k=k_{\mathrm{min}}}^{k_{\mathrm{max}}}\, \frac{N_k}{N}\, k\,
  \frac{1}{N_k} \sum_{j=1}^{N} \, \delta_{k_, k_j} \, {s}_{k_j}(t).
  \label{eq:mfA2}\end{aligned}$$

It is assumed that, after forgetting initial conditions, units with the same degree share stochastic pulse sequences which are statistically identical. Therefore, the same pulse sequence can be assigned to units of the same degree class by taking the average over the 