$L=\{a^ib^ja^k \mid j \text{ is odd, then } k=i^2+j ;\ j \text{ is even, then } k =i+j\}$
I tried writing $L$ as the union of the language created with $j$ odd and the one with $j$ even.
When $j$ is odd, I can prove using the pumping lemma that it is not a cfl.
When $j$ is even, I can prove that it is a cfl by writing a context-free grammar for it.
But then the union of a cfl and a non-cfl doesn't really help me prove $L$ is a cfl or not.
How do I proceed?
We can prove $L$ is not a CFL using the pumping lemma by choosing the string $a^pb^{1}a^{p^2+1}$, where $p$ is the constant in the pumping lemma.
Since $|vwx|\le{p}$, we can have:
If $v$ or $x$ contains more than one type of letter, so when we pump the order breaks.
If both $v$ and $x$ contain only $a$'s then when we pump the sum breaks because we either don't add enough letters or we add too much(the case with $i^2$ when pumping the first $a$'s).
The only other case is when either $v=b$ or $x=b$ that we solve by pumping $v$ down to 0, so the new string will match the $j$ even case with $j=0$, but $i + j = p^2$ while $k = p^2 + 1$.
Note that the context-free languages are closed under left-quotient as well as intersection with regular languages, as John L. noted.
Assume, for contradiction, that $L$ was context-free. Then $$L_1 := L \cap a^*ba^* = \{a^iba^{i^2 + 1} : i \geq 0\}$$ and $$L_2 := a^*ba \setminus L_1 = \{a^{i^2} : i \geq 0\}$$ must be context-free as well. If $L_2$ is context-free then it must also be regular, since it only contains one letter.
But using the Pumping lemma for regular languages it can be shown that $L_2$ isn't regular, contradicting our assumption that $L$ is context-free ↯
