Primary/Secondary On Call Rotations
Given $n$ persons, this question is about a sequence of unordered pair of persons, called a two-person schedule(TPS). A TPS is good if any of its initial contiguous subsequences satisfies the following two conditions.

(fairness requirement) Any person cannot appear two times more than any other person.
(non-repeating requirement) Any unordered pair of persons can appear at most once.

This question asks the best way to generate a good TPS that is as long as possible.
Let us define a perfect TPS as a TPS that satisfies the following requirement.

(inclusive requirement) Each unordered pair of persons is included.

It turns out that a good TPS of the longest length must be a perfect TPS.
In fact, if each unordered pair of person compete with each other, this question is none other than the round-robin tournament as explained by Wikipedia, which has complete solutions. Because of the durability and reliability of wikipedia as well as the size of the solutions, readers are encouraged to check the wikipedia item as well as the references listed thereof. Also there are plenty of resource on the web if you search for &quot;round robin tournament scheduling&quot;, such as a nice explanation of Round Robin Tournament Schedule and Javascript code to schedule round robin tournament.
A program that produces Perfect TPS
You can play with a scheduling algorithm in this Javascript playground. Modify the number of persons as you want to on the last line of the code, document.writeln(JSON.stringify(robin(8))). Hit Run code button. You will see output like the following.
[[[1,8],[2,7],[3,6],[4,5]],[[1,7],[8,6],[2,5],[3,4]],[[1,6],[7,5],[8,4],[2,3]],[[1,5],[6,4],[7,3],[8,2]],[[1,4],[5,3],[6,2],[7,8]],[[1,3],[4,2],[5,8],[6,7]],[[1,2],[3,8],[4,7],[5,6]]]