Answer
\[
T(2)=1,\quad
T(3)=1,\quad
T(4)=1,\quad
T(5)=1,\quad
T(6)=1,\quad
T(7)=1.
\]
All paths eventually reach \(T(1)=1\). This is, of course, the classic Collatz‐type behavior—though whether *all* integers ultimately reach 1 is the famous open “3n+1 problem.”
Work Step by Step
First, recall the given definition (which resembles the Collatz or “3n+1” iteration), for integers \(n \ge 1\):
\[
T(n) \;=\;
\begin{cases}
1, & \text{if }n=1,\\[6pt]
T\!\bigl(\tfrac{n}{2}\bigr), & \text{if }n\text{ is even},\\[6pt]
T\!\bigl(3n+1\bigr), & \text{if }n\text{ is odd and }n>1.
\end{cases}
\]
We want to compute \(T(2), T(3), T(4), T(5), T(6), T(7)\).
All of these eventually reduce to \(T(1)\), which is given as \(1\). Here are the step‐by‐step chains:
---
### 1. \(T(2)\)
- \(2\) is even, so \(T(2) = T(2/2) = T(1)\).
- \(T(1) = 1\) by definition.
Hence \(\boxed{T(2)=1}.\)
---
### 2. \(T(3)\)
\(3\) is odd (\(\neq 1\)), so \(T(3) = T(3\cdot3 + 1) = T(10)\).
\(10\) is even, so \(T(10) = T(10/2) = T(5)\).
\(5\) is odd (\(\neq 1\)), so \(T(5) = T(3\cdot5 + 1) = T(16)\).
\(16\) is even, so \(T(16) = T(16/2) = T(8)\).
\(8\) is even, so \(T(8) = T(8/2) = T(4)\).
\(4\) is even, so \(T(4) = T(4/2) = T(2)\).
We already found \(T(2)=1\).
So the chain \(3 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\) leads to \(T(3)=1\).
Hence \(\boxed{T(3)=1}.\)
---
### 3. \(T(4)\)
- \(4\) is even, so \(T(4) = T(4/2) = T(2)\).
- From above, \(T(2)=1\).
Thus \(\boxed{T(4)=1}.\)
---
### 4. \(T(5)\)
\(5\) is odd, so \(T(5) = T(3\cdot5 + 1) = T(16)\).
\(16\) is even \(\to T(16)=T(8)\).
\(8\) is even \(\to T(8)=T(4)\).
\(4\) is even \(\to T(4)=T(2)\).
\(T(2)=1\).
Hence \(\boxed{T(5)=1}.\)
*(We already saw most of this chain when doing \(T(3)\).)*
---
### 5. \(T(6)\)
\(6\) is even, so \(T(6) = T(6/2) = T(3)\).
From above, \(T(3)=1\).
So \(\boxed{T(6)=1}.\)
---
### 6. \(T(7)\)
\(7\) is odd, so \(T(7) = T(3\cdot7 + 1) = T(22)\).
\(22\) is even \(\to T(22)=T(11)\).
\(11\) is odd \(\to T(11)=T(34)\).
\(34\) is even \(\to T(34)=T(17)\).
\(17\) is odd \(\to T(17)=T(52)\).
\(52\) is even \(\to T(52)=T(26)\).
\(26\) is even \(\to T(26)=T(13)\).
\(13\) is odd \(\to T(13)=T(40)\).
\(40\) is even \(\to T(40)=T(20)\).
\(20\) is even \(\to T(20)=T(10)\).
\(10\) is even \(\to T(10)=T(5)\).
\(5\) is odd \(\to T(5)=T(16)\).
\(16\) is even \(\to T(16)=T(8)\).
\(8\) is even \(\to T(8)=T(4)\).
\(4\) is even \(\to T(4)=T(2)\).
\(T(2)=1\).
Thus \(\boxed{T(7)=1}.\)
