Let $T \subset \Bbb{Z}$ be the first $N$ primes, and $a_n = n, n \geq 2$. Define $\widehat{T}$ to be the sequence operator that deletes $a_m$ from $a$ if $p_k | m$ for some $k \leq N$. Thus $\widehat{T}a$ starts on the prime $p_{N+1}$. I've observed that $(\widehat{T}a)_n - (\widehat{T}a)_{n-1}$ is periodic for $N = 1, 2, 3, 4$, but have no idea how to compute the period or prove that it's always periodic, for all $N\geq 1$.
Any ideas?
Let $N = 2$. Then $\widehat{T}a = 5, 7, 11, 13, 17, 19, 23, 25, \dots$ with differences:
$2, 4, 2, 4, 2, 4, 2, \dots$ so the period is $2$.
Let $\displaystyle P = \prod_{k=1}^N p_k$. Then clearly any $a_m$ is removed from the sequence if and only if $a_{m+P}$ is removed from the sequence because for every $1 \leqslant k \leqslant N$: $p_k \mid m \iff p_k \mid m+P$.
Let $S = \# \{ n \in \mathbb{N} : 1 \leqslant (\widehat{T} a)_n \leqslant P \}$.
Then we prove $S = \# \{ n \in \mathbb{N} : A+1 \leqslant (\widehat{T} a)_n \leqslant A+P \}$ by induction on $A$.
Now fix $n$ and let $A+1 = (\widehat{T} a)_n$. Then $(\widehat{T} a)_n + P$ is the least element of the sequence $\widehat{T} a$ lying above the set $\{ n \in \mathbb{N} : A+1 \leqslant (\widehat{T} a)_n \leqslant A+P \}$ and to reach it we skip $S$ terms, so $(\widehat{T} a)_{n+S} = (\widehat{T} a)_n + P$.
Hence $(\widehat{T} a)_{n+S} - (\widehat{T} a)_{n+S-1} = (\widehat{T} a)_n - (\widehat{T} a)_{n-1}$, i.e. $(\widehat{T} a)_n - (\widehat{T} a)_{n-1}$ is periodic with period $S$.