To find a closed-form formula for the sequence $c_n$, we must carefully analyze the definitions and relationships of the sequences $a_n$, $b_n$, and $c_n$.

Step 1: Understanding $a_n$

We start with: $a_n = \left\lfloor \frac{n^2}{3} \right\rfloor.$

This means $a_n$ is the greatest integer less than or equal to $\frac{n^2}{3}$. Observe that $a_n$ grows approximately quadratically as $n$ increases.

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Step 2: Understanding $b_n$

The sequence $b_n$ is defined as: $b_n = \min \{ k : a_k \geq n \}.$

This means $b_n$ is the smallest index $k$ such that $\left\lfloor \frac{k^2}{3} \right\rfloor \geq n$. Rearranging the inequality: $\frac{k^2}{3} \geq n \implies k^2 \geq 3n \implies k \geq \sqrt{3n}.$

Since $k$ must be an integer, we round $\sqrt{3n}$ up to the nearest integer: $b_n = \left\lceil \sqrt{3n} \right\rceil.$

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Step 3: Understanding $c_n$

The sequence $c_n$ is defined as: $c_n = \min \{ k : b_k \geq n \}.$

Substituting $b_k = \left\lceil \sqrt{3k} \right\rceil$, we need: $\left\lceil \sqrt{3k} \right\rceil \geq n.$

This implies: $\sqrt{3k} \geq n \implies 3k \geq n^2 \implies k \geq \frac{n^2}{3}.$

Since $k$ must be an integer, we round $\frac{n^2}{3}$ up to the nearest integer: $c_n = \left\lceil \frac{n^2}{3} \right\rceil.$

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Final Answer:

The closed-form formula for $c_n$ is: $c_n = \left\lceil \frac{n^2}{3} \right\rceil.$

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Would you like a detailed explanation of any step or verification for specific values? Here are some additional questions for deeper exploration:

1. What are the first few terms of $c_n$, $b_n$, and $a_n$ to verify the relationships?
2. Can you graph $a_n$, $b_n$, and $c_n$ to visualize their growth?
3. How does $c_n$ behave asymptotically as $n \to \infty$?
4. Can you find a simple approximation for $c_n$ instead of the exact formula?
5. What happens if the floor/ceiling functions are removed from $a_n$, $b_n$, or $c_n$?

Tip: Carefully distinguish between the use of floor and ceiling functions when analyzing such sequences to avoid misinterpretation of inequalities.