Let $$A = \{1, 2, 3, \ldots, 100\}$$ and R be a relation on A such that $$R = \{(a, b) : a = 2b + 1\}$$. Let $$(a_1, a_2), (a_2, a_3), (a_3, a_4), \ldots, (a_k, a_{k+1})$$ be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k, for which such a sequence exists, is equal to :

We are given $$A = \{1, 2, 3, \ldots, 100\}$$ and the relation $$R = \{(a, b) : a = 2b + 1\}$$ on $$A$$.

Step 1: Identify all ordered pairs in R

For $$(a, b) \in R$$, we need $$a = 2b + 1$$ where both $$a$$ and $$b$$ are in $$A$$.

Since $$a = 2b + 1$$, $$a$$ must be odd. The valid pairs are:

$$(3, 1), (5, 2), (7, 3), (9, 4), (11, 5), (13, 6), \ldots, (99, 49)$$

That is, for each $$b$$ from $$1$$ to $$49$$, we get the pair $$(2b+1, b)$$.

Step 2: Understand the chain condition

We need a sequence $$(a_1, a_2), (a_2, a_3), (a_3, a_4), \ldots, (a_k, a_{k+1})$$ of $$k$$ elements of $$R$$ such that the second entry of each pair equals the first entry of the next pair.

This means: $$a_1 = 2a_2 + 1$$, $$a_2 = 2a_3 + 1$$, $$a_3 = 2a_4 + 1$$, and so on.

Step 3: Express $$a_1$$ in terms of $$a_{k+1}$$

From the recurrence $$a_i = 2a_{i+1} + 1$$, we can write:

$$a_1 = 2a_2 + 1 = 2(2a_3 + 1) + 1 = 4a_3 + 3$$

$$a_1 = 4(2a_4 + 1) + 3 = 8a_4 + 7$$

In general: $$a_1 = 2^k \cdot a_{k+1} + (2^k - 1)$$

Step 4: Find the maximum $$k$$

We need $$a_1 \leq 100$$ and $$a_{k+1} \geq 1$$.

Setting $$a_{k+1} = 1$$ (the smallest possible value):

$$a_1 = 2^k \cdot 1 + (2^k - 1) = 2^{k+1} - 1$$

We need $$2^{k+1} - 1 \leq 100$$, so $$2^{k+1} \leq 101$$.

Checking: $$2^6 = 64 \leq 101$$ ✓ and $$2^7 = 128 \gt 101$$ ✗

So $$k + 1 \leq 6$$, which gives $$k \leq 5$$.

Step 5: Verify with $$k = 5$$

With $$k = 5$$ and $$a_6 = 1$$:

$$a_6 = 1, \quad a_5 = 2(1) + 1 = 3, \quad a_4 = 2(3) + 1 = 7$$

$$a_3 = 2(7) + 1 = 15, \quad a_2 = 2(15) + 1 = 31, \quad a_1 = 2(31) + 1 = 63$$

The chain is: $$(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)$$

All values are in $$A = \{1, 2, \ldots, 100\}$$ ✓

Therefore, the largest integer $$k$$ is $$\mathbf{5}$$.

Hence, the correct answer is Option C.