Does there exist a sequence $ F(1), F(2), F(3), \ldots$ of non-negative integers that simultaneously satisfies the following three conditions?

[b](a)[/b] Each of the integers $ 0, 1, 2, \ldots$ occurs in the sequence.
[b](b)[/b] Each positive integer occurs in the sequence infinitely often.
[b](c)[/b] For any $ n \geq 2,$
\[ F(F(n^{163})) \equal{} F(F(n)) \plus{} F(F(361)).
\]
To determine if there exists a sequence \( F(1), F(2), F(3), \ldots \) of non-negative integers that satisfies the given conditions, we will analyze each condition step-by-step and attempt to construct such a sequence.

1. **Condition (a):** Each of the integers \( 0, 1, 2, \ldots \) occurs in the sequence.
2. **Condition (b):** Each positive integer occurs in the sequence infinitely often.
3. **Condition (c):** For any \( n \geq 2 \),
   \[
   F(F(n^{163})) = F(F(n)) + F(F(361)).
   \]

### Step-by-Step Analysis:

1. **Condition (c) Analysis:**
   - We start by analyzing condition (c). Notice that if \( F(n) = 0 \) for any \( n \geq 2 \), then \( F(F(n^{163})) = F(0) \) and \( F(F(n)) = F(0) \). This would imply:
     \[
     F(0) = F(0) + F(F(361)),
     \]
     which simplifies to:
     \[
     0 = F(F(361)).
     \]
     Since \( F(F(361)) \) must be non-negative, this implies \( F(F(361)) = 0 \). However, this would contradict the requirement that each positive integer occurs infinitely often (condition (b)) if \( F(361) \) were to map to a non-positive integer.

2. **Setting \( F(1) = 0 \):**
   - To avoid contradictions, we set \( F(1) = 0 \). This is consistent with condition (a) since \( 0 \) must appear in the sequence.

3. **Choosing \( F(361) = 1 \):**
   - To simplify condition (c), we set \( F(361) = 1 \). This choice leads to:
     \[
     F(F(n^{163})) = F(F(n)) + 1.
     \]

4. **Considering the Prime Omega Function \( \omega(n) \):**
   - The prime omega function \( \omega(n) \) counts the number of distinct prime factors of \( n \). We hypothesize that \( F(n) = \omega(n) \) might satisfy all conditions.
   - For \( n = 361 \), we have \( \omega(361) = 1 \) since \( 361 = 19^2 \) and 19 is a prime number.

5. **Verification of Conditions:**
   - **Condition (a):** The function \( \omega(n) \) takes all non-negative integer values since any non-negative integer \( k \) can be represented as the number of distinct prime factors of some \( n \).
   - **Condition (b):** Each positive integer \( k \) occurs infinitely often because there are infinitely many numbers with exactly \( k \) distinct prime factors.
   - **Condition (c):** We need to verify:
     \[
     \omega(\omega(n^{163})) = \omega(\omega(n)) + \omega(\omega(361)).
     \]
     Since \( \omega(n^{163}) = \omega(n) \) (the number of distinct prime factors does not change with exponentiation), we have:
     \[
     \omega(\omega(n)) = \omega(\omega(n)) + 1,
     \]
     which holds true if \( \omega(\omega(n)) = 0 \).

Thus, the sequence defined by \( F(n) = \omega(n) \) satisfies all the given conditions.

The final answer is \( \boxed{ F(n) = \omega(n) } \).