This conjecture is false. See this post
Time is running out Suggest notation for Steps 1 and 2. Earn the bonus.
For each $k\in\mathbb{Z^{+}}$.
Step 1: Create a list $(1,1,1,1...,1)$ of length $k^2+2k$.
Step 2: For all n: $1< n \leq \pi(k)$, at index multiples of $p_n$, insert $p_n$ into the list. Start with the greatest $p_n$ and work down. This identifies composites with least prime factors less than $k$.
Within this list, when $1< n < (k+1)^2$, all $1$s represent primes, because if any are composite they would have prime factors $\leq p_{\pi(k)}$, but we have already accounted for all of those in Step 2. $1$s above that limit may represent primes or composites.
Example For $k=5$,
Steps 1 and 2 identify all composites having least prime factors $\leq5$. These lpf are persistent.
$(1, 2, 3, 2, 5, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2$,
$3, 2, 1, 2, 5, 2, 3, 2, 1, 2, 1, 2, 3, 2, 5, 2, 1, 2, 3, 2$,
$1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 5, 2, 3, 2, 1, 2)$
$\begin{array}{l} (1,2,3,2,5) \\ (2,1,2,3,2) \\ (1,2,1,2,3) \\ (2,1,2,1,2) \\ (3,2,1,2,5) \\ (2,3,2,1,2) \\ (1,2,3,2,5) \\ (2,1,2,3,2) \\ (1,2,1,2,3) \\ (2,1,2,1,2) \\ (3,2,1,2,5) \\ (2,3,2,1,2) \\ \end{array}$
Above, we see that all unique Step 2 $k$-tuples are repeated. The tuple $(2, 1, 2, 1, 2)$ has the first encounter with a composite $(2,47,2,49,2)$. But looking ahead, we see that it identifies all primes at $(2,107,2,109,2)$, which led to the conjecture.
The conjecture: There are infinitely many $k$-tuple patterns when tuple index $\geq (k+3)$, where all $1$s represent primes.
Q: Are there better notations that I can use to improve this post?
My answer is limited by my inability to understand the question beyond the sieve described initially in an extended chat session. Since the OP now is asking about steps 1 and 2 perhaps the bar is low enough for me to answer.
Step 1 is clear as currently written. Form a list of $k^2+2k$ ones.
Step 2. Modify the list from Step 1 as follows (by way of example, let $k = 5$). For the nth element of the list in Step 1, if n is a composite containing at least one factor less than k, replace the 1 by the lowest prime factor of n. If n is prime or a composite of primes $p> 5$ retain 1 as the nth element. This is done for each of the $k^2+2k$ elements of the list in Step 1. So in the case of the interval $[1,35]$ the list of 35 ones becomes:
$$\{1,2,3,2,5,2,1,2,3,2,1,...,2,5\}.$$
This is a complete step. What follows should be given a new heading in case there are questions about it.
This could have been done in one step. Many of the difficulties of this idea arise because it was originally expressed as Mathematica code.
Step 3. Divide the list into sublists of length 5. (The OP has done this above.)
OP observes that some of the sublists formed in this way are duplicates. Also by design, ones in the list in Step 2 correspond to primes or composites of primes greater than $k.$ That is, because a one occurred in the nth position in the original list, n a prime or composite of primes g.t. $k,$ it is no surprise that the ones appear where they do.
This is where I lose the thread of the idea. The conjecture as currently worded is difficult to understand
There is ample disussion of sieve techniques online$^1$ including some that have borne fruit. This particular approach does not seem likely to lead to new theorems about primes in short intervals but as I do not understand the idea beyond this, I can't speak to it.
$^1$ And in texts such as An introduction to sieve methods and their applications. London Mathematical Society Student Texts 66. Cambridge University Press. Cojocaru, Murty, (2005).