As noted in David Coffron’s answer to another question, the moonblade (Dungeon Master’s Guide, pg. 217) has no fixed limit on how much damage it can do: there is no stated limit to the number of runes on the blade. It states:
A moonblade has one rune on its blade for each master it has served (typically 1d6 + 1). The first rune always grants a +1 bonus to attack and damage rolls made with this magic weapon. Each rune beyond the first grants the moonblade an additional property. The DM chooses each property or determines it randomly on the Moonblade Properties table.
Some of the items on the table include:
91–92: When you hit with an attack using the moonblade, the attack deals an extra 1d6 slashing damage.
93–94: When you hit a creature of a specific type (such as dragon, fiend, or undead) with the moonblade, the target takes an extra 1d6 damage of one of these types: acid, cold, fire, lightning, or thunder.
But as David notes in a comment, some runes have limits on how many times they may be gained. That means, once the moonblade has gained these runes, it cannot gain them again, and the odds of gaining the remaining runes—including the quoted damage-improving runes—improves in the future.
So then, assuming that each rune is determined randomly per the table, rerolling any rolls that result in a rune that would not stack with the runes already on the moonblade, and that the creature-type-specific runes are distributed evenly among the creature types (feel free to assume fractional runes for that purpose), what is the expected value of the damage roll of a moonblade with 𝑛 runes? Assume that the hit has already taken place, so changes to the weapon’s accuracy, such as making it finessed, or that make it easier to target enemies in the first place, such as making it thrown, are irrelevant. Also, absent any information about the target, ignore the damage benefit that vorpal might have. The best answer would include the chance of a critical hit, including the possibility that the critical threat range has been doubled by the relevant rune.
Do not assume any GM intervention limiting the moonblade or otherwise making any particular ruling here: instead use a rules as written interpretation of the text.
Best Answer
I wrote a python script to calculate answers accurate to several decimal places. I created a Moonblade object, and added a specified number of runes just as the rules suggest (with the one alteration you made that you always re-roll non-stackable properties). It then calculates the expected damage from this moonblade taking into account all bonuses, bonus dice, critical hits, creatures, etc.... I made a couple of assumptions beyond what you have explicitly stated.
With these assumptions, I get that with 99% confidence, the true expected damage values given \$n\$ runes are these, \$\pm 0.05\$.
\begin{array}{ll} 0\text{ runes}&9.737\\ 1\text{ runes}&10.244\\ 2\text{ runes}&10.752\\ 3\text{ runes}&11.252\\ 4\text{ runes}&11.727\\ 5\text{ runes}&12.166\\ 6\text{ runes}&12.557\\ 7\text{ runes}&12.899\\ 8\text{ runes}&13.196\\ 9\text{ runes}&13.459\\ \end{array}
This plot demonstrates the almost linear relationship.
For the first 10 or so runes, the expected damage is about \$9.912 + 0.421 n\$. However, the linearity drops off at about \$n=10\$. This is because we expect to have reached a bonus of +3 by this point.
If we extend to a truly absurd number of runes, we see a transition to another linear behavior. I think this is because the dominant factor in the damage is the steadily growing number of d6's added to the roll. Because all other runes eventually stop stacking, in the end, all we can do is add another d6 for each rune, meaning we either add 1d6 to the damage, or add 1d6 to the damage for a particular type of creature. After adjusting to account for critical hits, this averages out to about 2.07 damage per rune, which is what we expect the slope to eventually be. Therefore, for large numbers of runes, the expected damage is calculated as \$\overline{d}\approx-93.24 + 2.07 n\$