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\section*{Author contributions} \subsubsection{The accretion law}  In order to give appropriate credit to each author literature the universe's expansion is regarded as the expansion  of an article, a spherical surface.\\  We see galaxies immersed in a 3-dim space. increasing volume, however,  the individual contributions cosmological principle imposes galaxies placed above a surface. If we add the time at every point of the universe, we obtain a 4-dim spherical surface.\\  If $n$ is the increasing number  of steps for  each author to axis,  the manuscript should expansion will  be detailed obtained through the relation $N = n^2$, where $N$ are the points on the \emph{increasing} spherical surface that rises with the growing number of points on the radius $R$, following the law $S = 4\pi R^2$.\\  Therefore a universe that \emph{would increases  in this section. We recommend using author initials space}, according to the law $N = n^2$, with $N$ the number of particles  and $n$ the number of \emph{time-steps}, would increase \emph{surface} as happens to an expanding spherical surface.\\  However, in an isotropic universe any local observer $O$ puts himself at the center of a sphere with radius $R_i$ and places on its surface the distant galaxies.\\  (figure)\\  An observer sees galaxies move away radially in a 3d space, but describes the isotropic expansion as dilatation of a 2d surface.\\  In a universe with increasing space, the radius $R_i$ of any lattice-universe $U_i$ would increase as $R_i (n) = n \lambda_i$. It follows:  \begin{equation}  S_n = 4 \pi n^2 \lambda^2  \end{equation}  $S_n$ varies with increasing $R_n$;  then stating briefly how they contributed. the index $n$ will be a function of time, and so we can calculate the following derivatives:  \begin{align}  \frac{\text{d} S_n}{\text{d} n} = & \frac{\text{d}}{\text{d} n} \left ( 4 \pi n^2 \lambda^2 \right ) = 8 \pi n \lambda^2 \label{surface01}\\  \frac{\text{d} S_n}{\text{d} t} = & \frac{\text{d}}{\text{d} t} \left ( 4 \pi R_n^2 \right ) = 8 \pi R_n v_n \label{surface02}  \end{align}  Where $v_n$ is the velocity of recession of a galaxy placed on the spherical surface of radius $R_n$.\\  It's clear that the number $n$ of steps in the space expansion could be related to the time by the relation  \begin{displaymath}  n (t) = \frac{t}{\tau}  \end{displaymath}  and so:  \begin{equation}\label{surface03}  \frac{\text{d} S_n}{\text{d} t} = \frac{\text{d}}{\text{d} t} \left ( 4 \pi n^2 \lambda^2 \right ) = 8 \pi n \lambda^2 \frac{1}{\tau}  \end{equation}  Combining (\ref{surface02}) with (\ref{surface03}) we obtain the velocity of recession $v_n$:  \begin{equation}  v_n = \frac{n \lambda^2}{R_n \tau} = \frac{n \frac{R_n^2}{n^2}}{R_n \tau} = \frac{R_n}{n \tau} = \frac{R_n}{\tau_n}  \end{equation}  Now, if we assume that $H = \frac{1}{\tau_n}$, we obtain:  \begin{equation}  v_n = H R_n  \end{equation}  finding again the Hubble's law.