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 (a1, a2), (a2, a3), ..., (ak, ak+1) of k elements of R.

Since a = 2b + 1, we can write b = (a - 1) / 2. We need to find a sequence of pairs such that the second element of one pair is the first element of the next pair.

Let's start with the largest possible value for a1, which is 100. Then a1 = 2a2 + 1, so a2 = (100 - 1) / 2 = 99 / 2 = 49.5, which is not an integer. So, 100 cannot be the first element.

Let's try a1 = 99. Then a2 = (99 - 1) / 2 = 98 / 2 = 49. Now, a3 = (49 - 1) / 2 = 48 / 2 = 24. Next, a4 = (24 - 1) / 2 = 23 / 2 = 11.5, which is not an integer.

We need to find an a1 such that all subsequent terms are integers. This means that all terms must be odd to produce integer values for the next term.

Let's start with a1 = 99. Then a2 = 49, a3 = 24, a4 = 11, a5 = 5, a6 = 2, a7 = 0.5 (not an integer). So, this sequence stops at a6 = 2.

To maximize the length of the sequence, we should start with an odd number. Let's try to find the smallest odd number that generates a long sequence.

Consider the sequence: a(i+1) = (a(i) - 1) / 2. We want to find the largest k such that a(i) is an integer and 1 <= a(i) <= 100 for all i = 1, 2, ..., k+1.

Let's start with a small value for a(k+1), say 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, which is greater than 100.

So, the sequence is 63, 31, 15, 7, 3, 1. The length of this sequence is 6. Thus, k = 6.

Let's start with 65. 65, 32, 15.5. Not good.

Consider the sequence starting from 99: 99, 49, 24, 11, 5, 2. Here k = 5.

Consider the sequence starting from 63: 63, 31, 15, 7, 3, 1. Here k = 5.

Consider the sequence starting from 65: 65, 32. Not good.

Let's try starting with 67: 67, 33, 16, 7.5. Not good.

The sequence 63, 31, 15, 7, 3, 1 has length 6. So k = 5.

Let's try to find a sequence of length 7. We need a7 = 1. Then a6 = 3, a5 = 7, a4 = 15, a3 = 31, a2 = 63, a1 = 127. This is not possible since a1 > 100.

However, if we start with a2 = 63, then a1 = 2*63 + 1 = 127 > 100. So, we can't have a sequence of length 6 starting from 127.

Let's try starting with 99. 99, 49, 24, 11, 5, 2. This has length 5. So k = 5.

Let's try starting with 97. 97, 48, 23.5. Not good.

Let's try starting with 95. 95, 47, 23, 11, 5, 2. This has length 5. So k = 5.

Let's try starting with 7. 7, 3, 1. Then a1 = 7, a2 = 3, a3 = 1. So k = 2.

Let's start with 15. 15, 7, 3, 1. So k = 3.

Let's start with 31. 31, 15, 7, 3, 1. So k = 4.

Let's start with 63. 63, 31, 15, 7, 3, 1. So k = 5.

Let's start with 127. Not possible.

If we start with 2, we have no sequence.

If we start with 3, we have 3, 1. So k = 1.

If we start with 5, we have 5, 2. So k = 1.

If we start with 7, we have 7, 3, 1. So k = 2.

If we start with 15, we have 15, 7, 3, 1. So k = 3.

If we start with 31, we have 31, 15, 7, 3, 1. So k = 4.

If we start with 63, we have 63, 31, 15, 7, 3, 1. So k = 5.

If we start with 99, we have 99, 49, 24, 11, 5, 2. So k = 5.

If we start with 100, we have no sequence.

Consider the sequence 95, 47, 23, 11, 5, 2. k = 5.

Consider the sequence 91, 45, 22, 10, 4, 1. k = 5.

Consider the sequence 87, 43, 21, 10, 4, 1. k = 5.

Consider the sequence 83, 41, 20, 9, 4, 1. k = 5.

Consider the sequence 79, 39, 19, 9, 4, 1. k = 5.

Consider the sequence 75, 37, 18, 8, 3, 1. k = 5.

Consider the sequence 71, 35, 17, 8, 3, 1. k = 5.

Consider the sequence 67, 33, 16, 7, 3, 1. k = 5.

Consider the sequence 63, 31, 15, 7, 3, 1. k = 5.

Consider the sequence 99, 49, 24, 11, 5, 2. k = 5.

Consider the sequence 7. 7, 3, 1. k = 2.

Consider the sequence 15. 15, 7, 3, 1. k = 3.

Consider the sequence 31. 31, 15, 7, 3, 1. k = 4.

Consider the sequence 63. 63, 31, 15, 7, 3, 1. k = 5.

Consider the sequence 127. Not possible.

If we start with 2, we have no sequence.

If we start with 3, we have 3, 1. So k = 1.

If we start with 5, we have 5, 2. So k = 1.

If we start with 7, we have 7, 3, 1. So k = 2.

If we start with 15, we have 15, 7, 3, 1. So k = 3.

If we start with 31, we have 31, 15, 7, 3, 1. So k = 4.

If we start with 63, we have 63, 31, 15, 7, 3, 1. So k = 5.

If we start with 99, we have 99, 49, 24, 11, 5, 2. So k = 5.

If we start with 100, we have no sequence.

The longest sequence is 63, 31, 15, 7, 3, 1. But the question asks for the number of elements in R, which is 5.

The sequence is (63, 31), (31, 15), (15, 7), (7, 3), (3, 1). So k = 5.