Words equal to the concatenation of their odd and even position - characters
It would be more convenient to index word positions starting with zero.
Let us take a look at $L \cap \Sigma^{2^n}$. Any word $w \in L \cap \Sigma^{2^n}$ satisfies
$$
w = w_0 w_2 \ldots w_{2^n-2} w_1 w_3 \ldots w_{2^n-1}.
$$
You can check that for $0 \leq m &lt; 2^{n-1}$ and $b \in \{0,1\}$, this is equivalent to
$$
w_{2^{n-1}b + m} = w_{2m+b}.
$$
In other words, if we think of the index as an $n$-bit string, then this is the same as $w_i = w_{\mathrm{rot}(i)}$, where $\mathrm{rot}(i)$ is obtained by rotating $i$ to the left. 

A necklace is an equivalence class of words under rotation. What we have shown above is that $w \in L \cap \Sigma^{2^n}$ iff $w_i$, interpreted as a function $\{0,1\}^n \to \Sigma$, is constant on necklaces. Denoting the number of necklaces of length $n$ by $N_2(n)$, we deduce that
$$
|L \cap \Sigma^{2^n}| = 2^{N_2(n)}.
$$
The explicit formula for $N_2(n)$ shows that $N_2(n) \sim 2^n/n$.



It is known that if $L$ is a regular language then there exists a modulus $m \geq 1$ such that for every $0 \leq i &lt; m$ there exist constants $C,d,\lambda \geq 0$ such that for $N \equiv i \pmod{m}$,
$$
|L \cap \Sigma^N| \sim C N^d \lambda^N.
$$
We can calculate $\lambda$ as $\lim_{N\to\infty} |L \cap \Sigma^N|^{1/N}$ (the limit taken over $N \equiv i \pmod{m}$).

Now suppose that our $L$ were regular. Choose $i$ so that infinitely many values $2^n$ satisfy $2^n \equiv i \pmod{m}$. We have
$$
|L \cap \Sigma^{2^n}|^{1/2^n} = 2^{N_2(n)/2^n} = 2^{O(1/n)} \longrightarrow 1,
$$
and so $\lambda = 1$. However, that implies that $|L \cap \Sigma^{2^n}|$ should have polynomial growth (in $2^n$), whereas it clearly has superpolynomial growth.