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Andras Gilyen - 8 years ago 2017-09-07 19:58:34
gilyen@clayoquot.swat.cwi.nl
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    We observe that for the initial $\rho$ we have that $\text{rank}(\rho)=n$, therefore $\text{rank}(\rho*(M_{(n)}^k))\geq n+k$, and so $\rho*(M_{(n)}^k)*\mathbbm{1}$ is obviously divisible by $p^{k}$. This implies that $a_k^{(n)}$ can be calculated by only looking at $\rho*(M_{(n)}^1)*\mathbbm{1}, \ldots, \rho*(M_{(n)}^k)*\mathbbm{1}$.
 
    
 
\newpage
 
\section{Proving the strong cancellation claim}
 
\section{Proving that $a_k^{(k+1)}=a_k^{(n)}$ for all $n>k$}
 
It is useful to introduce some new notation. We will consider variations of the Markov Chains:
 
\begin{itemize}
 
    \item $\P^{(n)}$ refers to the original process on the length-$n$ cycle.
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