Consider the following languages. $$L_1 = \{a^p \mid p \text{ is a prime number}\}$$ $$L_2 = \{a^n b^m c^{2m} \mid n \ge 0, m \ge 0\}$$ $$L_3 = \{a^n b^n c^{2n} \mid n \ge 0\}$$ $$L_4 = \{a^n b^n \mid n \ge 1\}$$ Which of the following are CORRECT? I. $L_1$ is context-free but not regular. II. $L_2$ is not context-free. III. $L_3$ is not context-free but recursive. IV. $L_4$ is deterministic context-free.

Let us evaluate each statement:

• Statement I: $L_1 = \{a^p \mid p \text{ is prime}\}$ is defined over a unary alphabet $\Sigma = \{a\}$. By Parikh's Theorem, any context-free language over a unary alphabet is regular. Since $L_1$ is not regular (primes cannot be recognized by a finite automaton), it cannot be context-free either. Thus, statement I is incorrect.

• Statement II: $L_2 = \{a^n b^m c^{2m} \mid n \ge 0, m \ge 0\}$ is context-free (and deterministic context-free) because a pushdown automaton can read any number of $a$'s without using the stack, push two symbols onto the stack for each $b$, and pop one symbol for each $c$. Thus, statement II is incorrect.

• Statement III: $L_3 = \{a^n b^n c^{2n} \mid n \ge 0\}$ requires simultaneously comparing the number of $a$'s, $b$'s, and $c$'s, which cannot be done with a single stack. Hence, $L_3$ is not context-free (it is context-sensitive), and it is clearly recursive. Thus, statement III is correct.

• Statement IV: $L_4 = \{a^n b^n \mid n \ge 1\}$ is a standard deterministic context-free language (DCFL) recognizable by a deterministic pushdown automaton (DPDA). Thus, statement IV is correct.

Statements III and IV are correct, corresponding to option (D).