Is a^n , n = 3j+4k , n>=0, a context-free language?

Question

I have no idea how to approach this question... How would I go about proving or disproving this? any explanation is appreciated.

Answer 1

As nothing is mentioned about $j,k$ I can think of 2 possible cases,

Case 1: when $j,k \in N$

$L = $ {$a^{n}$ | $n=3j+4k>=0$}

We have to find possible solutions of $3j+4k>=0$, for $j,k >=0$ this equation holds true, we can rewrite $a^{n}$ as follow

$a^{n} = a^{3j+4k}$ $=a^{3j}.a^{4k}$ $=(a^{3})^{j}.(a^{4})^{k}$

from here the language can be written as $L=(a^{3})^{*}(a^{4})^{*}$, as we have obtained a regular expression our language $L$ is regular as well as context-free

Case 2: when $j,k\in R$

Any whole number $n$ can be written as $3j+4k$ where $j=n/6, k=n/8$ hence our language $L$ becomes $a^{*}$ which is regular and content-free also.