This follows from the shift properties of the decomposition formula: the number of odd steps \(q(k+j, 2^k n) = q(j, n)\) and the decomposition correction \(\mathcal{Q}_{k+j}(2^k n) = 2^k \mathcal{Q}_j(n)\). Substituting these into the definition of \(E\) clears the \(2^k\) factor.

Suppose \(q(j, n)\) were bounded by some \(M\). Since \(q\) is non-decreasing, it must eventually stay constant at some value \(q_s\). This would imply that all iterates \(T^j(n)\) for \(j \ge J\) are even, so they are repeatedly divided by 2. This contradicts the infinite stopping time, as the sequence would eventually drop below \(n\) (or reach 1 and then cycle).

Since \(q(0, n) = 0\), \(q(j, n)\) is unbounded, and \(q(j+1, n) - q(j, n) \in \{ 0, 1\} \), the discrete intermediate value theorem implies that \(q\) must hit every natural number.

We construct an infinite set of paradoxical pairs \((j', n')\) where \(n'\) is of the form \(2^k n\). Consider the set of Diophantine approximations \(S = \{ (a, b) \in \mathbb {N}^2 \mid 1 - \frac{1}{4n} {\lt} \frac{3^a}{2^b} {\lt} 1 \} \). By Theorem 139, this set is infinite. For each \((a, b) \in S\), we use Lemma 143 to find an iteration \(j\) such that \(q(j, n) = a\). Let \(j_{of}(a)\) be the smallest such \(j\). We distinguish two cases for each \((a, b) \in S\):

Case 1: \(j_{of}(a) \ge b\). We claim that \((j_{of}(a), n)\) is paradoxical. First, \(T^{j_{of}(a)}(n) \ge n\) because \(n\) has infinite stopping time. Second, \(C_{j_{of}(a)}(n) = \frac{3^a n + \mathcal{Q}_j(n)}{2^{j_{of}(a)}} \approx \frac{3^a}{2^{j_{of}(a)}} n\). Since \(j_{of}(a) \ge b\) and \(3^a/2^b {\lt} 1\), we have \(3^a/2^{j_{of}(a)} {\lt} 1\). More precisely, \(C_j(n) = \frac{3^a}{2^j} n + E_j(n)\). Using the bounds from Theorem 127, \(E_j(n) \le R(a) = \frac{3^a - 2^a}{2^a}\). Then:

The condition \(3^a/2^b {\gt} 1 - 1/(4n)\) ensures that \(C_j(n) {\lt} 1\).

Case 2: \(j_{of}(a) {\lt} b\). Let \(k = b - j_{of}(a)\). We claim \((b, 2^k n)\) is paradoxical. By the shift property \(T^b(2^k n) = T^{j_{of}(a)}(n)\), and since \(T^{j_{of}(a)}(n) \ge n\) and \(2^k n \ge n\) is not guaranteed to be small, we calculate: \(C_b(2^k n) = \frac{3^a (2^k n) + \mathcal{Q}_b(2^k n)}{2^b} = \frac{2^k (3^a n + \mathcal{Q}_j(n))}{2^{k+j}} = C_j(n)\). Again, the Diophantine condition \(3^a/2^b {\gt} 1 - 1/(4n)\) implies \(C_b(2^k n) {\lt} 1\), and \(T^b(2^k n) \ge 2^k n\) follows from the linear decomposition:

Substituting \(T^b(2^k n) = T^j(n)\), we need \(T^j(n) \ge 2^k n\). Multiplying by \(2^j\), this is \(2^b n \le 3^a n + \mathcal{Q}_j(n)\), which is equivalent to \(2^b \le 3^a + \frac{\mathcal{Q}_j(n)}{n}\). The condition \(3^a/2^b {\gt} 1 - 1/(4n)\) provides exactly the separation needed to satisfy this inequality when \(n\) is large or when \(\mathcal{Q}_j\) is at least \(3^a - 2^a\).

Since \(S\) is infinite and the mapping from \((a, b)\) to \((j', k)\) is injective, we obtain infinitely many paradoxical sequences.

Assume the set of paradoxical sequences with \(m {\gt} 2\) is finite. Let \(n\) be an arbitrary natural number. If \(n = 0\), the conjecture is vacuously true. Suppose \(n {\gt} 0\) and \(n\) never reaches 1 under Collatz iteration.

Let \(\mathcal{O}(n)\) be the orbit of \(n\) under the Terras map \(T\). Let \(m = \min \mathcal{O}(n)\).

1. \(m \ge 1\) since \(n \ge 1\).

2. \(m \neq 1\) because if \(1 \in \mathcal{O}(n)\), then \(n\) would reach 1.

3. \(m \neq 2\) because if \(2 \in \mathcal{O}(n)\), its next iterate \(T(2) = 1\) would also be in \(\mathcal{O}(n)\).

4. Thus, \(m {\gt} 2\).

5. Since \(m\) is the minimal element of the orbit, for any \(k \ge 1\), \(T^k(m) \ge m\). This means the stopping time of \(m\) is infinite.

By Theorem 144, there exist infinitely many paradoxical sequences starting from integers of the form \(2^k m\). Since \(m {\gt} 2\), all such integers \(2^k m\) are also strictly greater than 2. This demonstrates the existence of infinitely many paradoxical sequences with starting value greater than 2, which contradicts our initial hypothesis. Therefore, every \(n {\gt} 0\) must eventually reach 1.