Understanding the Relation and Sequence

The problem defines a set $A = \{1, 2, 3,...., 100\}$ and a relation $R = \{(a, b) : a=2b+1\}$. We are looking for the longest sequence of ordered pairs in $R$, denoted as $(a_1, a_2), (a_2, a_3), ..., (a_k, a_{k+1})$, where the second element of each pair matches the first element of the next pair. This implies a chain relationship: $a_i = 2a_{i+1}+1$ for $i = 1, 2, ..., k$. All elements $a_1, a_2, ..., a_{k+1}$ must belong to the set $A$. The relation $a = 2b+1$ means that $a$ must be an odd integer and $a > b$ for any pair $(a, b)$ in $R$. Since $a_i = 2a_{i+1}+1$, the sequence of elements $a_1, a_2, ..., a_{k+1}$ must be strictly decreasing.

Finding the Maximum Chain Length

To find the largest possible value of $k$ (the number of pairs), we need to construct the longest possible chain of elements $a_1, a_2, ..., a_{k+1}$ that are all within the set $A$. Since the sequence must decrease, we should start building the chain from the smallest possible valid value for $a_{k+1}$ and work backwards.

The smallest element in set $A$ is 1. Let's assume $a_{k+1} = 1$. We can then find the preceding elements using the relation $a_i = 2a_{i+1}+1$:

• If $a_{k+1} = 1$, then $a_k = 2(1) + 1 = 3$. ($3 \in A$)
• If $a_k = 3$, then $a_{k-1} = 2(3) + 1 = 7$. ($7 \in A$)
• If $a_{k-1} = 7$, then $a_{k-2} = 2(7) + 1 = 15$. ($15 \in A$)
• If $a_{k-2} = 15$, then $a_{k-3} = 2(15) + 1 = 31$. ($31 \in A$)
• If $a_{k-3} = 31$, then $a_{k-4} = 2(31) + 1 = 63$. ($63 \in A$)
• If $a_{k-4} = 63$, then $a_{k-5} = 2(63) + 1 = 127$.

Determining the Value of k

The value $a_{k-5} = 127$ is greater than 100, which means it is not in the set $A$. Therefore, the chain must terminate with the element 63. The longest sequence of elements within $A$ is $63, 31, 15, 7, 3, 1$. This sequence corresponds to the following ordered pairs in $R$:

• $(63, 31)$
• $(31, 15)$
• $(15, 7)$
• $(7, 3)$
• $(3, 1)$

This sequence contains 5 ordered pairs. The question defines $k$ as the number of these ordered pairs in the sequence $(a_1, a_2), (a_2, a_3), ..., (a_k, a_{k+1})$. Thus, the largest integer $k$ for which such a sequence exists is 5.