归纳定义集合序列求解：给定n时Tₙ中的(a,b)取值

Let's break down the recursive definition of (T_n) and derive clear conditions for when a pair ((a,b)) belongs to (T_n).

Recap of the Sequence Definition #

First, let's restate the sequence in plain terms to avoid confusion:

• (T_0 = {(0,0)})
• For each (n \geq 0), (T_{n+1}) is made of two distinct groups of pairs:
  1. Combination group: Pairs formed by taking one element from (T_n) and one element from any earlier set (T_k) (where (k \leq n)), then computing ((1+x_1+x_2, y_1+y_2))
  2. Increment group: Pairs formed by taking every element from (T_n) and adding 1 to its y-coordinate: ((x_1, 1+y_1))

Key Conditions for ((a,b) \in T_n) #

We can split this into cases based on the values of (a) and (b):

Base Case: (n=0)

Only the pair ((0,0)) belongs to (T_0).

Case 1: (b=0) (y-coordinate is 0)

For (n \geq 1), ((a,0) \in T_n) if and only if (1 \leq a \leq 2^n - 1).

• Why this works: The combination group lets us build larger x-values by merging existing pairs. For example:
  • (T_1) gets ((1,0)) by combining two elements from (T_0)
  • (T_2) adds ((2,0)) (merge (T_1) and (T_0)) and ((3,0)) (merge two (T_1) pairs)
  • Each step (n) lets us reach up to (2^n -1) (the largest n-bit binary number), and every integer between 1 and this maximum is achievable.

Case 2: (a=0) (x-coordinate is 0)

For (n \geq 1), ((0,b) \in T_n) if and only if (1 \leq b \leq n).

• Why this works: These pairs come only from the increment group. Starting from ((0,0)), we need exactly (b) increments to reach ((0,b)), so (n) must be at least (b) to include all those steps.

Case 3: (a \geq 1) and (b \geq 1)

A pair ((a,b)) is in (T_n) if either:

1. Increment from (T_{n-1}): The pair ((a, b-1)) is already in (T_{n-1}) (we just added 1 to the y-coordinate of an existing pair), OR
2. Combination of smaller pairs: There exist non-negative integers (x_1, x_2, y_1, y_2) such that:
   • (a = 1 + x_1 + x_2)
   • (b = y_1 + y_2)
   • ((x_1,y_1)) is in (T_{n-1})
   • ((x_2,y_2)) is in some (T_k) where (k \leq n-1)

Alternative View: Minimal (n) for ((a,b)) #

If you care about the smallest (n) where ((a,b)) first appears (let's call this (n(a,b))):

• (n(0,b) = b) (needs exactly (b) increment steps)
• (n(a,0) = \lceil \log_2(a+1) \rceil) (needs enough combination steps to build up (a))
• For (a,b \geq 1), (n(a,b)) is the smallest number such that either (n(a,b-1) = n-1), or you can split (a) and (b) into valid components from sets up to (T_{n-1})

内容的提问来源于stack exchange，提问作者OwlBandicoot