= Five-qubit concatenation threshold
{title2=$p_{\ell+1}=10p_\ell^2-20p_\ell^3+15p_\ell^4-4p_\ell^5$}
Assume independent physical-error <probability> $p$, correction of zero or one errors, and failure on every block with at least two errors. The failure map is $f(p)=1-(1-p)^5-5p(1-p)^4$. Factoring $f(p)-p$ gives $p(1-p)(4p^3-11p^2+9p-1)$. The unique interior threshold is the root of $4p^3-11p^2+9p-1=0$. Below it concatenation decreases the failure <probability> toward zero. Two levels give $f(f(p))=1000p^4+O(p^5)$, compared with $10p^2+O(p^3)$ at one level.
// Target: quantum-information-theory.bigb
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