From the previous post "If The Universe Is A Mochi...", we would expect the universe to be roughly proportioned into,
Given,
\(m_p=\cfrac{m}{2^n}\)
\(\begin{matrix} \begin{matrix} m_p & 0 \\ 2m_p & 2^{n-2} \\ 4m_p & 2^{n-4}\\8m_p &2^{n-6} \\..&..\\..&..\end{matrix} & \\ & \end{matrix}\)
The most abundant elements is of the type \(2m_p\).
If we sum all the masses,
\(0.m_p+2^{n-2}.2m_p+2^{n-4}.4m_p+2^{n-6}.8m_p+..\)
And taking the limit \(n\rightarrow\infty\),
\(\lim_{n\rightarrow\infty}\left\{2^{n-1}m_p(1+\cfrac{1}{2}+\cfrac{1}{4}+...)\right\}=\lim_{n\rightarrow\infty}\left\{2^{n}\cfrac{m}{2^n}\right\}=m\)
which is the mass the big bang started with.
It is most interesting that this model suggests a binary count of masses in the universe. That all masses are related to a constant (\(m_p\)) by some power of 2,
\(m_n=2^nm_p\),
and \(m_p\) itself being rare.
