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2

Here is yet another proof. It is known that the number of integers at most $n$ which are the product of two primes is $o(n)$, see for example this answer, which gives the asymptotic $\frac{n\log\log n}{\log n}$. This means that your language is infinite yet has vanishing asymptotic density. This is impossible for a regular unary language.

3

Just pump up $(M+1)$ $y$'s. Now you get $xy^{M+1}z=a^{(M+1)j+M-j}=a^{M(j+1)}$. Since $M$ is a product of two primes, $M(j+1)$ is a product of at least 3 primes, so $a^{M(j+1)}\notin L_1$, which proves $L_1$ is not regular by the pumping lemma.

1

There is an alternative to the “pumping” lemma which I find easier: After each possible input, determine the set of continuations that would complete a string of the language. You can use each of those sets as a state in the finite state machine for the language, so if there is a finite number of those sets then the language is regular- if there are ...

2

Huffman's algorithm is known to be optimal, that is, produce a code which minimizes the average codeword length (with respect to the input distribution). Let us notice now that there is a code in which each codeword has length $\lceil \log n \rceil$, and in particular the average codeword length is $\lceil \log n \rceil$ (with respect to any distribution). ...

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