4.6 More Paradoxical Sequences
Let \(j, n \in \mathbb {N}\) with \(n {\gt} 0\). If the pair \((j, n)\) is paradoxical, then:
From the paradoxical condition \(C(j, n) {\lt} 1\), we get \(0 {\lt} 1 - C(j, n)\). By Lemma 109, \(T^j(n) = C(j, n) \cdot n + E(j, n)\). The paradoxical condition \(T^j(n) \ge n\) therefore gives \(C(j,n) \cdot n + E(j,n) \ge n\), hence \(E(j,n) \ge (1 - C(j,n)) \cdot n\), and dividing by \(n {\gt} 0\) yields the result.
Let \(j, n, m \in \mathbb {N}\) with \(m {\gt} 0\). Suppose that every odd iterate \(T^k(n)\) with \(k {\lt} j\) satisfies \(T^k(n) \ge m\). Let \(q = \mathit{q}(j, n)\) be the number of odd steps in the first \(j\) iterations. Then:
By induction on \(j\). The base case \(j = 0\) is trivial.
For the inductive step, assume the bound holds for \(j\). We consider two cases depending on the parity of \(T^j(n)\):
Odd case (\(X(T^j(n)) = 1\)): By definition,
Combining with the inductive hypothesis \(T^j(n) \le n(3 + 1/m)^q/2^j\) and using \(q_{\text{new}} = q + 1\):
Even case (\(X(T^j(n)) = 0\)): By definition, \(T^{j+1}(n) = T^j(n)/2\). Since \(q\) stays the same,
Under the same hypotheses as Lemma 147, we have:
By Lemma 147:
Substituting the factored form \(T^j(n) = C(j,n) \cdot n + E(j,n)\) from Lemma 109 and dividing through by \(n {\gt} 0\):
Since \(C(j, n) = 3^q / 2^j\), rearranging gives the result.
Since the harmonic mean \(h\) is not less than the minimum \(m\) of the odd iterates of \(T\), Lemma 148 is a weaker version of Theorem \(4.1\) in Rozier–Terracol.
Let \(j, q, m \in \mathbb {N}\) with \(j {\gt} 0\) and \(m {\gt} 0\). Suppose the following inequalities hold for the rationals \(3^q/2^j\) and \((3+1/m)^q/2^j\):
Then the ratio \(q/j\) satisfies:
For the upper bound, \(1 - 3^q/2^j {\gt} 0\) implies \(3^q {\lt} 2^j\). Taking logs gives \(q \log 3 {\lt} j \log 2\), so \(q/j {\lt} \log 2 / \log 3\). For the lower bound, the second inequality implies \(2^j - 3^q \le (3+1/m)^q - 3^q\), which simplifies to \(2^j \le (3+1/m)^q\). Taking logs gives \(j \log 2 \le q \log (3+1/m)\), yielding the final result.
Let \((j, n)\) be a paradoxical sequence starting at \(n {\gt} 0\). Let \(m {\gt} 0\) be a lower bound for all odd iterates \(T^i(n)\) with \(i {\lt} j\). Let \(q\) be the number of odd steps. Then:
Since the harmonic mean \(h\) is not less than the minimum \(m\) of the odd iterates of \(T\), Lemma 150 is a weaker version of Corollary \(4.2\) in Rozier–Terracol.
For any \(j \ge 2\) and any \(k \in \mathbb {N}\), the value \(j \cdot \log 2 / \log 3\) is not an integer:
If \(j \log 2 / \log 3 = k\), then \(\xi = \log 2 / \log 3 = k/j \in \mathbb {Q}\), contradicting the irrationality of \(\xi \) (Lemma 135).
For any \(j \ge 2\), the floor of \(j \log 2 / \log 3\) is strictly less than \(j \log 2 / \log 3\):
By definition, \(\lfloor x \rfloor \le x\). The floor must be strictly less because \(j \log 2 / \log 3\) is not an integer by Lemma 151.
Let \(j, q, m \in \mathbb {N}\) with \(j, q, m {\gt} 0\). Suppose the ratio bounds
hold. Then:
\(2^{j/q} - 3 {\gt} 0\), i.e., \(2^{j/q} {\gt} 3\);
\(m \le \dfrac {1}{2^{j/q} - 3}\).
The upper bound \(q/j {\lt} \log 2 / \log 3\) implies \(q \log 3 {\lt} j \log 2\), hence \(\log 3 {\lt} (j/q) \log 2 = \log (2^{j/q})\), so \(2^{j/q} {\gt} 3\).
The lower bound \(\log 2 / \log (3+1/m) \le q/j\) implies \(j \log 2 \le q \log (3+1/m)\), i.e., \((j/q)\log 2 \le \log (3+1/m)\), so \(2^{j/q} \le 3 + 1/m\). This gives \(2^{j/q} - 3 \le 1/m\), and since \(2^{j/q} - 3 {\gt} 0\), we can take reciprocals (reversing the inequality) to get \(m \le 1/(2^{j/q} - 3)\).
Let \(j, q, m \in \mathbb {N}\) with \(j \ge 2\) and \(m {\gt} 0\). Suppose the ratio bounds of Lemma 149 hold for \(q/j\). Define:
Then \(2^r - 3 {\gt} 0\) and \(m \le 1/(2^r - 3)\).
From \(q \le \lfloor j \log 2/\log 3 \rfloor \) (by the upper bound \(q {\lt} j \log 2/\log 3\)), we have \(r = j/\lfloor j \log 2/\log 3 \rfloor \le j/q\). This means \(2^r \le 2^{j/q}\), so \(1/(2^r - 3) \ge 1/(2^{j/q} - 3)\). By Lemma 153, \(m \le 1/(2^{j/q}-3) \le 1/(2^r - 3)\).
To show \(2^r {\gt} 3\): since \(\lfloor j\log 2/\log 3 \rfloor {\lt} j \log 2/\log 3\) (strictly, by Lemma 152), \(r = j/\lfloor j\log 2/\log 3 \rfloor {\gt} j/(j\log 2/\log 3) = \log 3/\log 2\). Therefore \(r \log 2 {\gt} \log 3\), i.e., \(\log (2^r) {\gt} \log 3\), so \(2^r {\gt} 3\).
Let \((j, n)\) be a paradoxical pair with \(j \ge 2\) and \(n {\gt} 0\). Let \(m {\gt} 0\) be a lower bound for all odd iterates \(T^k(n)\) with \(k {\lt} j\). Define \(r = j / \lfloor j \log 2 / \log 3 \rfloor \). Then:
Since the harmonic mean \(h\) is not less than the minimum \(m\) of the odd iterates of \(T\), Lemma 155 is weaker than Corollary \(4.3\) in Rozier–Terracol.