Let $p(n) = 1 $ if $n$ is not prime or $1$ and $p(n) = 0$ if $n$ is prime or $1$. Define for prime $q$ the sequence $\hat{q}(n) = 1 $ if $n = 0 \pmod q$ and $\hat{q}(n) = 0 $ otherwise. Let $\oplus$ be pointwise boolean "or". Then $\hat{2}(n)\oplus \hat{3}(n)\oplus p(n)$ up to $n = 5^2 -1$ looks like (as a concatenated bitstring):
$$ (1\cdot 011101 \cdot 011101 \cdot 011101 \cdot 011101) $$
Check this by counting $0,1,2,3, \dots$ and verifying that there's a $1$ in each position of a composite or $2,3$. When we say prime we mean unital primes: the primes as well as $1$.
Define $\hat{0} = 10000\dots \ $ and $x(n) = 0011101000\dots \ $ where we have used bitstring notation since we're only dealing with two sequence values.
Then $\hat{2} \oplus \hat{3} \oplus p(n)$ up to $n = 5^2 - 1$ also can be written:
$$ \hat{0}\oplus x(n) \oplus x(n -6) \oplus x(n-12) \oplus x(n-18) = y(n) $$
where of course negative indices get assigned a sequence value of $0$. Of course $p(n)$ up to $7^2 - 1$ is given by:
$y(n) \oplus y(n - 25) \oplus \hat{5}$
My question is, can we keep on constructing like this (up to the next prime squared minus 1) or do weird things start happening on the boundaries?
You are close to the Sieve of Erastosthanes. If you change the definition of $\hat q(n)$ to be zero for the prime you can just take the or over the primes and get zero for all the primes and one for all the composites. Your version using $6,12,18$ is trying to cover for the fact that you didn't define it properly for the prime, so if you fix $\hat q(n)$ you don't need that.