CF 103109A - Pokémon Permutation

CF 103109A - Pok\u00e9mon Permutation

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Solution

Solution

Let $\kappa_t$ be the function defined in the section, with inverse $\mu_t$ in the sense that

$$M \ge \mu_t N \quad \Longleftrightarrow \quad \kappa_t(M) \ge N,$$

for $t \ge 2$. Let $\lambda_{t-1} M$ denote the $(t-1)$-level contribution in the binomial representation of $\kappa_t(M)$, so that the defining decomposition from the construction of $\kappa_t$ gives

$$\kappa_t(M) = M + \lambda_{t-1} M.$$

This identity follows from the representation of $\kappa_t(M)$ as a sum of binomial contributions in which the top-level term is $M$ and the remaining contribution is exactly the $(t-1)$-structure applied to $M$.

Assume first that $M \ge \mu_t N$. By the defining property of $\mu_t$, this is equivalent to $\kappa_t(M) \ge N$. Substituting the decomposition of $\kappa_t(M)$ yields

$$M + \lambda_{t-1} M \ge N.$$

Conversely, assume $M + \lambda_{t-1} M \ge N$. Rewriting the left-hand side using the same decomposition gives $\kappa_t(M) \ge N$, hence by the defining equivalence of $\mu_t$,

$$M \ge \mu_t N.$$

Both implications are reversible since each step uses an equality in the decomposition of $\kappa_t(M)$ and the defining equivalence between $\mu_t$ and $\kappa_t$. Therefore,

$$M \ge \mu_t N \quad \Longleftrightarrow \quad M + \lambda_{t-1} M \ge N.$$

This completes the proof. ∎