The count of odd steps over the first \(a+b\) iterates splits into the count over the first \(a\) iterates starting from \(n\), plus the count over the next \(b\) iterates starting from \(T^a(n)\). This follows directly from the additivity of finite sums and the identity \(T^{a+j}(n) = T^j(T^a(n))\).

By induction on \(j\). The base case \(j = 0\) is trivial. For the inductive step, one uses \(T(1) = 2\) (since \(1\) is odd: \(T(1) = (3 \cdot 1 + 1)/2 = 2\)) and \(T(2) = 1\) (since \(2\) is even: \(T(2) = 2/2 = 1\)), so \(T^{2(j+1)}(1) = T^2(T^{2j}(1)) = T^2(1) = T(2) = 1\).

We have \(T^{2j+1}(1) = T(T^{2j}(1)) = T(1) = 2\) by Lemma 157 and the fact that \(T(1) = 2\).

Write \(i = 2j\) or \(i = 2j + 1\). In the even case, \(T^{2j}(1) = 1\) by Lemma 157, and \(1\) is odd so \(X(1) = 1\). In the odd case, \(T^{2j+1}(1) = 2\) by Lemma 158, and \(2\) is even so \(X(2) = 0\).

By induction on \(j\). The base case is trivial. For the inductive step, use Lemma 156 to split the \(2(j+1)\) steps as \(2j\) steps followed by \(2\) steps starting from \(T^{2j}(1) = 1\) (by Lemma 157). The count for the last \(2\) steps starting from \(1\) is \(1\) (since \(T^0(1) = 1\) is odd and \(T^1(1) = 2\) is even), so the total is \(j + 1\).

Apply Lemma 156 to split the \(2j+1\) steps as \(2j\) steps followed by \(1\) step starting from \(T^{2j}(1) = 1\) (by Lemma 157). The count for the first \(2j\) steps is \(j\) by Lemma 160, and the single additional step starts at \(1\) which is odd, contributing \(1\) more. Total: \(j + 1\).

Consider the two cases according to the parity of \(m\).

• If \(m = 2j\), then \(s = j\) by Lemma 160, so \(2s = m\) and both bounds hold with equality (up to \(\pm 1\)).

• If \(m = 2j + 1\), then \(s = j + 1\) by Lemma 161, so \(2s = 2j + 2 = m + 1\), and again both bounds hold.

Let \(s_k = Q(k, n)\). For \(k \ge k_0\), write \(k = k_0 + (k - k_0)\) and apply Lemma 156:

using that \(T^{k_0}(n) = 1\). Let \(c = Q(k_0, n)\), which is a fixed constant with \(c \le k_0\). By Lemma 162 applied to \(m = k - k_0\), we have \(2Q(k-k_0, 1) \in \{ k-k_0,\, k-k_0+1\} \), so:

Using the order-topology characterisation of convergence, we show the limit equals \(1/2\):

• Lower bound: For \(a {\lt} 1/2\), set \(\varepsilon = 1 - 2a {\gt} 0\). Choose \(N\) large enough that \((k_0 + 1)/\varepsilon {\lt} N\) and \(k \ge \max (N, k_0+1)\). Then from \(k \le 2s_k + (k_0 + 1)\) we get \((k - s_k)/k {\gt} a\).

• Upper bound: For \(b {\gt} 1/2\), set \(\varepsilon = 2b - 1 {\gt} 0\). Choose \(N\) large enough that \((k_0 + 1)/\varepsilon {\lt} N\) and \(k \ge \max (N, k_0+1)\). Then from \(2s_k \le k + (k_0 + 1)\) we get \((k - s_k)/k {\lt} b\).

In both cases, the bound follows by a simple linear arithmetic argument once \(k\) is large enough relative to \((k_0 + 1)/\varepsilon \).

Verified by explicit computation for each value \(n \in \{ 1, 2, \ldots , 9\} \): \(T^2(1)=1\), \(T^1(2)=1\), \(T^5(3)=1\), \(T^2(4)=1\), \(T^4(5)=1\), \(T^6(6)=1\), \(T^{11}(7)=1\), \(T^3(8)=1\), \(T^{13}(9)=1\).

This is equivalent to \(30n + 10 \le 31n\), i.e. \(10 \le n\), which holds by assumption.

By induction on \(k\).

Base case (\(k = 0\)): both sides equal \(n\).

Inductive step: Assume the bound holds for \(k\), with \(s_k = Q(k,n)\), and that \(T^j(n) \ge 10\) for all \(j \le k+1\). Let \(n_k = T^k(n)\) and \(s_{k+1} = s_k + X(n_k)\), where \(X(n_k) \in \{ 0,1\} \) is the parity indicator.

Case \(n_k\) even (\(X(n_k) = 0\)): Then \(T(n_k) = n_k/2\), so \(2T(n_k) = n_k\), and \(s_{k+1} = s_k\).

where the last step uses the induction hypothesis.

Case \(n_k\) odd (\(X(n_k) = 1\)): Then \(T(n_k) = (3n_k+1)/2\), so \(2T(n_k) = 3n_k + 1\), and \(s_{k+1} = s_k + 1\). Since \(n_k \ge 10\) and \(n_k\) is odd, Lemma 165 gives \(10(3n_k+1) \le 31 n_k\).

where the second-to-last step uses the induction hypothesis.

By strong induction on \(s\), with step size \(3\).

Base cases:

• \(s = 0\): Need \(1 {\lt} 2^k\). From the hypothesis \(5 \cdot 0 + 1 \le 3k\) we get \(k \ge 1\), so \(2^k \ge 2 {\gt} 1\).

• \(s = 1\): Need \(31 {\lt} 2^k \cdot 10\). From \(5 + 1 \le 3k\) we get \(k \ge 2\), so \(2^k \cdot 10 \ge 40 {\gt} 31\).

• \(s = 2\): Need \(961 {\lt} 2^k \cdot 100\). From \(11 \le 3k\) we get \(k \ge 4\), so \(2^k \cdot 100 \ge 1600 {\gt} 961\).

Inductive step (\(s \ge 3\), write \(s = s' + 3\)): Assume \(31^{s'} {\lt} 2^{k'} \cdot 10^{s'}\) for all valid \(k'\), where \(5s' + 1 \le 3k'\). Apply the induction hypothesis with \(k' = k - 5\) (valid since \(5(s'+3)+1 \le 3k\) implies \(5s'+1 \le 3(k-5)\), and \(k \ge 6\)):

Using the key arithmetic fact \(31^3 = 29791 {\lt} 32000 = 2^5 \cdot 10^3\):

Suppose for contradiction that \(T^j(m) {\lt} m\) for some \(j\). The case \(j = 0\) gives \(m {\lt} m\), a contradiction. For \(j = j' + 1\): let \(n' = T^{j'+1}(m)\). Since \(n' {\lt} m\) and \(n' \ge 1\) (because iterates of positive integers stay positive), the minimality of \(m\) gives some \(k \ge 1\) with \(T^k(n') = 1\). But then \(T^{k + (j'+1)}(m) = T^k(T^{j'+1}(m)) = T^k(n') = 1\), contradicting the assumption that \(m\) never reaches \(1\).

Since the limit is \(1/2\), the sequence eventually lies in the open interval \((2/5, 3/5)\). Choose \(N\) such that for all \(k \ge N\) the ratio \((k - s_k)/k {\gt} 2/5\), where \(s_k = Q(k,n)\). Set \(k_0 = \max (N, 1)\). Then \((k_0 - s_{k_0})/k_0 {\gt} 2/5\), i.e.

so \(\tfrac {3}{5} k_0 {\gt} s_{k_0}\), which gives \(5 s_{k_0} {\lt} 3 k_0\), i.e. \(5 s_{k_0} + 1 \le 3 k_0\).

(2.4 \(\Rightarrow \) 2.5). Let \(n \ge 1\). By Conjecture 2.4, there exists \(k_0 \ge 1\) with \(T^{k_0}(n) = 1\). Lemma 163 then directly gives the desired Tendsto conclusion.

(2.5 \(\Rightarrow \) 2.4). Suppose for contradiction that Conjecture 2.4 fails. Then there exists some \(n_0 \ge 1\) with \(T^k(n_0) \ne 1\) for all \(k \ge 1\). Among all such counterexamples, let \(m\) be the minimal one (using Nat.find); formally, let

and set \(m = \min \{ n : P(n)\} \). The properties of \(m\) are:

1. \(m \ge 1\) and \(T^k(m) \ne 1\) for all \(k \ge 1\).

2. For every \(n {\lt} m\) with \(n \ge 1\), there exists \(k \ge 1\) with \(T^k(n) = 1\).

Step 1: \(m \ge 10\). By Lemma 164, every \(n\) with \(1 \le n \le 9\) reaches \(1\). So if \(m \le 9\), property (1) would be contradicted. Hence \(m \ge 10\).

Step 2: \(T^j(m) \ge m\) for all \(j\). This is Lemma 168 applied to \(m\) with its minimality property. In particular, all iterates satisfy \(T^j(m) \ge m \ge 10\).

Step 3: Extract \(k\) with a good odd-step ratio. Since Conjecture 2.5 holds (by assumption), the ratio \((k - s_k)/k \to 1/2\) for \(n = m\). By Lemma 169, there exists \(k \ge 1\) with

Step 4: Derive a contradiction. Let \(s = Q(k, m)\).

• From Step 2, all iterates \(T^j(m) \ge 10\) for \(j \le k\), so Lemma 166 gives:

  \[ 10^s \cdot 2^k \cdot T^k(m) \le 31^s \cdot m. \]

• Lemma 167 with \(5s + 1 \le 3k\) gives:

  \[ 31^s {\lt} 2^k \cdot 10^s. \]

Combining these two inequalities with \(T^k(m) \ge m\) (Step 2):

Dividing by \(10^s \cdot m {\gt} 0\):

i.e. \(31^s \ge 2^k \cdot 10^s\), contradicting Lemma 167.