We are given the relation $R = \{(a, b) : a = 2b + 1\}$ on the set $A = \{1, 2, 3, ..., 100\}$. We need to find the largest integer $k$ such that there exists a sequence $(a_1, a_2), (a_2, a_3), (a_3, a_4), ..., (a_k, a_{k+1})$ of $k$ elements of $R$.

Since $(a_i, a_{i+1}) \in R$ for all $i = 1, 2, ..., k$, we have $a_i = 2a_{i+1} + 1$. We want to find the largest possible $k$.

We can express the terms of the sequence as follows:

• $a_1 = 2a_2 + 1$
• $a_2 = 2a_3 + 1$
• $a_3 = 2a_4 + 1$
• ...
• $a_k = 2a_{k+1} + 1$

Since $a_i \in A$ for all $i$, we have $1 \le a_i \le 100$. We want to find the smallest possible value for $a_{k+1}$ to maximize $k$.

Let's start with $a_{k+1} = 1$. Then:

• $a_k = 2(1) + 1 = 3$
• $a_{k-1} = 2(3) + 1 = 7$
• $a_{k-2} = 2(7) + 1 = 15$
• $a_{k-3} = 2(15) + 1 = 31$
• $a_{k-4} = 2(31) + 1 = 63$
• $a_{k-5} = 2(63) + 1 = 127$

Since $a_i \le 100$, we stop at $a_{k-4} = 63$. The sequence is $63, 31, 15, 7, 3, 1$. The length of this sequence is 6. So, $k = 6$.

Now, let's try to find a longer sequence. We need to find the smallest possible $a_{k+1}$ such that all $a_i$ are less than or equal to 100.

If we start with $a_{k+1} = 0$, then $a_k = 1, a_{k-1} = 3, a_{k-2} = 7, a_{k-3} = 15, a_{k-4} = 31, a_{k-5} = 63, a_{k-6} = 127$. But $a_{k+1}$ must be in A, so it must be at least 1.

Let's consider the sequence in reverse. We want to find the largest $a_1$ such that $a_1 \le 100$. We have $a_i = 2a_{i+1} + 1$. We want to find the largest $k$ such that $1 \le a_i \le 100$ for all $i$.

We can rewrite the relation as $a_{i+1} = \frac{a_i - 1}{2}$. We want to find the largest $k$ such that $a_{k+1} \ge 1$.

Let's start with $a_1 = 100$. Then $a_2 = \frac{100 - 1}{2} = 49.5$, which is not an integer. So, we need to choose $a_1$ such that $a_i$ are integers.

We need $a_i$ to be odd for all $i$. Let $a_{k+1} = 1$. Then $a_k = 2(1) + 1 = 3$. $a_{k-1} = 2(3) + 1 = 7$. $a_{k-2} = 2(7) + 1 = 15$. $a_{k-3} = 2(15) + 1 = 31$. $a_{k-4} = 2(31) + 1 = 63$. $a_{k-5} = 2(63) + 1 = 127$. Since $a_i \le 100$, we have the sequence $63, 31, 15, 7, 3, 1$. The length of the sequence is 6.

Let's try $a_{k+1} = 2$. Then $a_k = 2(2) + 1 = 5$. $a_{k-1} = 2(5) + 1 = 11$. $a_{k-2} = 2(11) + 1 = 23$. $a_{k-3} = 2(23) + 1 = 47$. $a_{k-4} = 2(47) + 1 = 95$. $a_{k-5} = 2(95) + 1 = 191$. The sequence is $95, 47, 23, 11, 5, 2$. The length of the sequence is 6.

Let's try $a_{k+1} = 3$. Then $a_k = 2(3) + 1 = 7$. $a_{k-1} = 2(7) + 1 = 15$. $a_{k-2} = 2(15) + 1 = 31$. $a_{k-3} = 2(31) + 1 = 63$. $a_{k-4} = 2(63) + 1 = 127$. The sequence is $63, 31, 15, 7, 3$. If we start with 99, we have (99-1)/2 = 49, (49-1)/2 = 24, (24-1)/2 = 11.5.

Consider the sequence $a_1, a_2, ..., a_{k+1}$. We have $a_i = 2a_{i+1} + 1$. We want to maximize $k$. Let $a_{k+1} = x$. Then $a_k = 2x + 1$, $a_{k-1} = 2(2x+1) + 1 = 4x + 3$, $a_{k-2} = 2(4x+3) + 1 = 8x + 7$, $a_{k-3} = 16x + 15$, $a_{k-4} = 32x + 31$, $a_{k-5} = 64x + 63$. We want $64x + 63 \le 100$, so $64x \le 37$, which means $x \le \frac{37}{64}$. Since $x \ge 1$, this is not possible.

Let $a_{k-5} = 63$. Then $a_{k-4} = 31$, $a_{k-3} = 15$, $a_{k-2} = 7$, $a_{k-1} = 3$, $a_k = 1$. This gives a sequence of length 6. If we start with 95, we have 95, 47, 23, 11, 5, 2. This also gives a sequence of length 6. If we start with 99, we have 99, 49, 24, 11, 5, 2.

Consider the sequence 99, 49, 24, 11, 5, 2. This doesn't work because 24 is not 2(11) + 1. Consider the sequence 95, 47, 23, 11, 5, 2. This works. Consider the sequence 63, 31, 15, 7, 3, 1. This works.

Let's try to find a sequence of length 7. We need $a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8$. $a_i = 2a_{i+1} + 1$. $a_7 = 1$. $a_6 = 3$. $a_5 = 7$. $a_4 = 15$. $a_3 = 31$. $a_2 = 63$. $a_1 = 127$. This doesn't work. $a_7 = 2$. $a_6 = 5$. $a_5 = 11$. $a_4 = 23$. $a_3 = 47$. $a_2 = 95$. $a_1 = 191$. This doesn't work.

The largest integer $k$ is 6.