Four-Number Door Puzzle

Question

So I had an idea for a number-based door puzzle for a TTRPG campaign that could readjust itself every time a wrong guess is made. Here's the basic premise:
Given two numbers, find two more numbers in such a way that:

no number differs from any other number in the same way,
and none of these differences are part of the series themselves,
and no number is bigger than it has to be, but not zero.

It seems fully deterministic to me, so let me give you an example:
Given are two numbers (1, 4). The difference is 3.
2 is out because its difference from 1 is part of the series, and in itself would be the difference from 4. 3 is out because of its difference from 2 and 4 being the same, and that difference would be 1, which is additionally part of the series. 5 is not valid either, but 6 is. 7 is not valid due to its difference to 6 being part of the series, and 8 is invalid because of the difference between (6, 8) and (4, 6). 9-6=4-1, and 10-6=4. 11-6=5, 11-1=10, 11-4=7, but 11-6=6-1. 12-6=6, but 13 works: the correct sequence is (1, 4, 6, 13).
It seems simple enough for a door puzzle, and given the above set of rules, it seems absolutely deterministic. I was considering to give the following (more vague) clue to its solution:
No siblings differ alike, nor match they their differences, and none grow taller than they must.
I need it to sound like a mystical/fantasy riddle, but still be concise and accurate and its description of the puzzle. If anyone has any suggestions, I would greatly appreciate the help.
I suppose I have four questions for this community:

Is there a unique solution for any two starting numbers below 10?
Can you think of a more interesting variant, like perhaps by minimising the sum of the numbers (but not having any predetermined numbers given - (1, 4, 6, 13) being the best solution I found so far)?
It seems to me there should be some way to determine the two missing numbers in a way that minimizes the sum of all four, which is not the same as choosing the lowest possible numbers one after the other. Is there such a way?
Is there a better way to phrase the problem in a single sentence to my players, without being too obvious or sounding too scientific?

And some more example solutions for your convenience:

(3, 4) → (9, 11)
(2, 5) → (6, 13)
(1, 9) → (3, 13)
(1, 3) → (7, 12)
(1, 5) → (7, 15)
(4, 6) → (1, 13)
(8, 1) → (3, 12)

WhatsUp · Accepted Answer

Here are some experimental results.

The "heuristic algorithm", namely choosing smallest numbers one after the other, does not always produce the smallest possible sum.
Examples:
$[1, 9]$: heuristic: $[1, 9, 3, 13]$; optimal: $[1, 9, 4, 11]$;
$[3, 5]$: heuristic: $[3, 5, 9, 16]$; optimal: $[3, 5, 11, 12]$;
and many more.

The example of $[3, 5, 11, 12]$ probably has the largest sum of two added numbers, being $11 + 12 = 23$.
The second largest is $[1, 5, 7, 15]$, being $7 + 15 = 22$.
I checked these for all starting values $leq 50$. One can probably prove that above $50$ there must exist small numbers that fulfill the task (in many cases, the smallest pair $[1, 3]$ works).

Four-Number Door Puzzle

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