Collatz Map Basics

4.2 Rozier–Terracol Theorem 2.2

Definition 125
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The lower bound sequence \(L_j(q)\) is defined for \(j, q \in \mathbb {N}\) by:

\begin{equation} L_j(q) = \frac{3^q - 2^q}{2^j}. \end{equation}
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Definition 126
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The upper bound \(R(q)\) is defined for \(q \in \mathbb {N}\) by:

\begin{equation} R(q) = \frac{3^q - 2^q}{2^q} = \left(\frac{3}{2}\right)^q - 1. \end{equation}
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Theorem 127 Rozier–Terracol Theorem 2.2
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For every positive integer \(j\) and any \(n \in \mathbb {N}\), let \(q\) be the number of odd steps in the first \(j\) iterations of the compact Collatz map on \(n\). Then the correction ratio \(E_j(n)\) satisfies:

\begin{equation} L_j(q) \le E_j(n) \le R(q). \end{equation}
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Proof

We proceed by induction on \(j\). The base case \(j=1\) is trivial. For the inductive step \(j+1\), let \(q\) be the number of odd steps in the first \(j\) iterations and \(x \in \{ 0, 1\} \) be the \((j+1)\)-th parity bit. Then the number of odd steps in \(j+1\) iterations is \(q+x\). By the recursive definition, \(E_{j+1}(n) = \frac{3^x}{2} E_j(n) + \frac{x}{2}\). If \(x=0\), \(E_{j+1}(n) = E_j(n)/2\). Since \(L_{j+1}(q) = L_j(q)/2\) and \(R(q)/2 \le R(q)\), the bounds hold. If \(x=1\), \(E_{j+1}(n) = (3E_j(n) + 1)/2\). Since \(L_{j+1}(q+1) = (3L_j(q) + 2^q/2^j)/2\) and \(2^q/2^j \le 1\) (as \(q \le j\)), and \(R(q+1) = (3R(q) + 1)/2\), the bounds follow from the induction hypothesis.

Theorem 128
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For \(j {\gt} 0\), the upper bound \(E_j(n) = R(q)\) is reached if and only if all \(q\) odd steps occur at the end of the first \(j\) iterations. That is, the parity vector \(V_j(n)\) consists of \(j-q\) zeros followed by \(q\) ones.

Proof

By induction on \(j\). The condition \(E_{j+1}(n) = R(q+x)\) requires \(E_j(n) = R(q)\) and \(x=1\) (if \(q{\gt}0\)), or \(E_j(n)=0\) and \(q=0\) and \(x=0\). This forces the bits to align at the end.

Theorem 129
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Lean declarations:

For \(j {\gt} 0\), the lower bound \(E_j(n) = L_j(q)\) is reached if and only if all \(q\) odd steps occur at the beginning of the first \(j\) iterations. That is, the parity vector \(V_j(n)\) consists of \(q\) ones followed by \(j-q\) zeros.

Proof

Similar to the upper bound, by induction on \(j\). Equality \(E_{j+1}(n) = L_{j+1}(q+x)\) forces the bits to be "front-loaded."