Stacking identical sequences with minimal overlap I stumbled upon the following problem. As computer scientist, I don't know all the correct terminology, which makes it hard to search for existing discussion surrounding this or related problems. If those exist, I guess a pointer to them would already bring me closer to understanding my problem.
The Problem
Given a sequence of elements ( using [ 1, 2, 3, 4, 5, … ] as example in the following), I want to stack multiple copies of this sequence in such a way, that each element overlaps at most once with another element. Pursuing a greedy strategy (always placing the next sequence as soon as possible), leads to the following construction:
Seq. 1:  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
Seq. 2:    1 2 3 4 5 6 7 8  9 10 11 12 13 14 ...
Seq. 3:        1 2 3 4 5 6  7  8  9 10 11 12 ...
Seq. 4:                1 2  3  4  5  6  7  8 ...
Seq. 5:                              1  2  3 ...

Offsets: |1|<2>|<--4-->|<-----5----->|<------...

I say that two elements overlap if they appear in the same column. Thus, since the 1 of the $2^{nd}$ sequence overlaps with the 2 of the $1^{st}$ sequence, any following sequences' 1 cannot overlap with a 2 of any other sequence, or vica versa. I do think, however, that, if this is solved for one element, it holds for all elements.
I define the offset of the $n^{th}$ sequence as how much further it is indented compared to the $(n-1)^{th}$ sequence. Thus, the offsets for the $2^{nd}$ to $5^{th}$ sequence, as displayed here are [1, 2, 4, 5]. The list for more subsequent offsets looks like this: [1, 2, 4, 5, 8, 10, 14, 21, 15, 16, 26, 25, 34, 22, 48, 38, 71, 40, 74, …]. Here, I honestly expected exponential growth. Instead, the $n^{th}$ offset may be smaller than the $(n-1)^{th}$ offset, which is counterintuitive to me.
My questions

*

*Is this, or a problem this can be reduced to, known under a common name that I could look up?

*Is my greedy strategy, to always place the sequence as soon as possible, optimal? Or, does there exist some other strategy, where the sum of all offsets is smaller for the same number of stacked sequences?

*Does there exist a formula that computes the list of offsets, without having to exhaustively try all possibilities?

 A: Instead of looking at the offsets between adjacent sequences, I want to look at the offset from the start: in your example, the five sequences have total offsets $0, 1, 3, 7, 12$.
If the first $n$ sequences start at total offsets $a_1, a_2, \dots, a_n$, then the overlap condition essentially says that all the pairwise differences $a_i - a_j$ must be distinct. (That's because the $1$ of the $i^{\text{th}}$ sequence overlaps with $1 - a_i + a_j$ of the $j^{\text{th}}$ sequence.) A finite choice of offsets of this type is called a Golomb ruler.
The greedy strategy is already not optimal for five sequences: total offsets of $0, 1, 4, 9, 11$ also work. (This corresponds to offsets of $1, 3, 4, 2$ in your notation.) In fact, for four sequences, $0, 1, 4, 6$ is better than the greedy solution $0, 1, 3, 7$.
As a lower bound, $a_n \ge \binom n2$, because the $\binom n2$ differences $a_j - a_i$ with $1 \le i < j \le n$ must all be in the range $\{1, 2, \dots, a_n\}$. If we just want to solve the problem for a finite $n$, then the Wikipedia article mentions the Erdős–Turán construction $$a_{k+1} = 2pk+ (k^2 \bmod p) \qquad k=0, \dots, n-1$$ where $p$ is the smallest (odd) prime $\ge n$, and this satisfies $a_n = O(n^2)$.

A repeated difference $a_i - a_j = a_k - a_\ell$ is equivalent to a repeated sum $a_i + a_\ell = a_j + a_k$. (Here, we allow repetition between the indices $i,j,k,\ell$, except for the trivial case $a_i - a_j = a_i - a_j$.) Sequences without such a repeated sum are called Sidon sequences. I am mentioning them because the Wikipedia page quotes some result about infinite Sidon sequences as well as finite one, if that's what you're interested in.
