# On what has been mentioned so far
Many partial answers have been given already. This is another partial answer.
Let us first carefully consider what the OP is actually asking.
* The answer to the question in the title is practically-certain to be negative. There does not exist anything worthy of the name 'database' for these 6-tuples. There do exist scattered results. *The whole topic seems frustratingly difficult*. There seems to be a controversy around whether it has been proved that a perfect parallelepiped *with all angles right angles* does not exist.
* The answer to the first question in the body of the OP
>Is there a collection of lists of six integer edge lengths that form a tetrahedron?
is positive. An answer of Igor Rivin has identified such a 'collection'. An answer of M. Winter has identified another way to obtain some such 6-tuples.
## So-far unmentioned, very relevant aspect: perfect parallelepipeds
No one so far has mentioned (though M. Winter did make a step in this direction when they mentioned 'Euler bricks')
* the widely known fact that any parallelepiped in Euclidean three-space can be 'partitioned'<sup>(terminology)</sup> into six tetrahedra. A pictorial representation of such a 'partition' is
[![enter image description here][1]][1]
(<sub>constructed with [Geogebra](https://de.wikipedia.org/wiki/GeoGebra) and [GIMP](https://de.wikipedia.org/wiki/GIMP) </sub>)
* the fact that this can be combined with recent discoveries of what is called 'perfect parallelepipeds'<sup>(terminological suggestion)</sub>='non-cube parallelepipeds such that all three edge-lengths, and all three minor-face-diagonal-lengths, and all three major-face-diagonal-lengths, and all four [body diagonal](https://en.wikipedia.org/wiki/Space_diagonal)-lengths are positive integers.
The first published specification of such a perfect parallelepiped appears to be:
> [Jorge F. Sawyer, Clifford A. Reiter: *Perfect parallelepipeds exist*. Mathematics of Computation Vol. 80, Number 274, 1037-1040](http://www.ams.org/journals/mcom/2011-80-274/S0025-5718-2010-02400-7/)
wherein one finds
[![enter image description here][2]][2]
and I would like to point out to the OP that
> because of the above-mentioned 'partition' of perfect parallelepipeds into integer-edge-length tetrahedra, **each perfect parallelepiped is a source of 6-tuples of the kind you are requesting**.
To make the central argument of the present suggestion explicit:
> **The property that all edges and diagonals of a perfect parallelepiped have integer lengths implies that the above decomposition into tetrahedra does not incur any tetrahedron with a non-integer edge-length.**
For example, the theorem of Sawyer and Reiter reproduced and color-coded above seems to correspond to a perfect parallelepiped of which the following picture is a not-to-scale representation
[![enter image description here][3]][3]
(<sub> The colors in the above picture correspond to the color-coding in the excerpt further above. </sub>)
Moreover, it seems that you will be able to obtain *infinitely-more* than the "thirty" 6-tuples you request by
* carefully studying
> [Benjamin D. Sokolowsky, Amy G. VanHooft, Rachel M. Volkert and Clifford A. Reiter: *An infinite family of perfect parallelepipeds*.Math. Comp. 83 (2014), 2441-2454](http://www.ams.org/journals/mcom/2014-83-289/S0025-5718-2013-02791-3/)
* using the decomposition I mentioned above
* making it very clear in whatever you are writing that the notion of 'dissimilar' used in op. cit. suitably translates into a notion of 'dissimilarity' among the tetrahedra my proposal gives you (I did not look into *that*)
Please note that opening sentence
> A **rational** parallelepiped is determined by three edge vectors $\vec{u}$, $\vec{v}$, $\vec{W}$. [emphasis added]
of the second section of op. cit. would be flat-out wrong (to see this: flatten the skeleton) if the word 'rational' were erased, since the skeleton of a cube is not even a locally-[rigid](https://en.wikipedia.org/wiki/Structural_rigidity) framework, let alone a globally-rigid one<sup>(further reading on frameworks)</sup>
Finally, facile<sup>(facile)</sup> skepticist warnings
* the verification of whether the parallelepipeds specified by Sokolowsky--VanHooft--Volkert--Reiter actually do have all the properties claimed neither seems to have been done by any human being, nor seems to be *doable* in reasonable time by a human being,
* said verification seems to have been delegated to Mathematica,
* only few people seem to be allowed to look under the hood of the marvellous machine that it Mathematica,
* even if one is allowed to look under hoods of large machines, it is very very difficult to be sure whether they work as expected
* so in the unlikely case that you are thinking about using these 6-tuples for some critical application, you should think twice.
To summarize,
* this answer suggests to the OP to make use of the fact that currently more thought and energy is going into finding someting *stronger* than what they are asking for (namely: stepparallelepipeds),
* this answer suggests that the OP could usefully try to contact the authors of op. cit. about their problem,
* I do not know 'how complete' the method is that I am proposing here (i.e.: to procure the integer-edge-length-tetrahedra from steppable parallelepipeds), i.e., whether any integer-edge-lenght tetrahedron can be so obtained, but **I strongly doubt it and expect it to be trivial to show that not all can be so obtained, but will have to leave the subject now**.
* While personally I would prefer not to even know what you need these 6-tuples for, I would very much like to understand what 'stance' you are planning to take towards the 6-tuples you get from various sources. How, if at all, do you plan to verify whether a given integer 6-tuple is indeed realization by a tetrahedron in Euclidean 3-space? Also: what is known about the complexity-theoretic status of this problem in general, i.e., how complex is it to decide, given $a\in\omega^6$, whether there exists a perfect parallelepiped with these edge-lengths?
<sup>(facile)</sup> <sub> It is facile because it is so easy to simply doubt anything, and since I know what an extremely well-discussed topic this is, for decades already. I am mentioning this to remind the OP that the references I give seem (0) not to contain humanly comprehensible proofs (more technically: a relevant *satisfaction-relation* seems not be checkable by human intuition), (1) seem to have only be checked with a software which is amazingly good, yet not formally verified. To be fair to both op. cit. and Mathematics, one should point out that, in a sense, they don't need each other: the kind of computations in op. cit. could easily be implemented in many other systems; the power of the mentioned package resides elsewhere.
Moreover, I now the 'but-there-may-equally-well-be-a-bug-in-one's-brain' objection to insistence on intuitive surveyability, and I don't have an answer to that. Moreover, I do not mean to imply that op. cit. is not a proof, but I think it is not a traditional proof relative to any of the respected proof systems; more technically speaking, these proofs contain *atomic sentences involving a new-fangled equality symbol* $=_{\tiny\text{CAS}}$, in addition to the usual $=_{\tiny{\text{HumanIntuition}}}$, with 'CAS' standing for 'computer algebra system', and for which the satisfaction-relation $\vDash$ becomes interestingly problematic in the sense that a claim of the form ' $\mathrm{World}$ $\vDash (A=_{\tiny\mathrm{CAS}}B) $ ' cannot be checked by human intuition. </sub>
<sup>(further reading on frameworks)</sup> <sub>The most recent publication on the topic of rigid frameworks, at least spiritually relevant to the OP, is: [Jessica Sidman, Audrey St. John: *The Rigidity of Frameworks: Theory and Applications* Notices of the AMS, October 2017](http://www.ams.org/journals/notices/201709/rnoti-p973.pdf) </sub>
<sup>(terminology)</sup> <sub> Needless to say, saying 'partitioned' here is not quite rigorous, since there are issues about sets of Lebesque-measure zero, which are of course irrelevant to this context. Of course, the OP did not even mention whether their 'tetrahedra' are open or closed point-sets. I will gloss over this. </sub>
<sup>(terminological suggestion)</sup> <sub>Personally I find the term 'perfect parallelepiped' very unfortunate and unimaginative. In particular, it seems to have nothing to do with 'perfect numbers'. My suggestion would be to rename this class of parallelepipeds '[steppable](https://en.wiktionary.org/wiki/steppable) parallelepipeds', maybe even [portmanteau](https://en.wikipedia.org/wiki/Portmanteau_(disambiguation))ed to '**stepparellipiped**'. These would make for more neutral and more informative art terms, for obvious reasons: all relevant lengths (i.e. the blue, the yellow, the green, and the red ones in the above picture) of stepparellepipeds can be constructed by a machine which is only allowed (0) arbitrary rotations in space (while staying put), and (1) unit-length steps. </sub>
[1]: https://i.sstatic.net/ki5r3.png
[2]: https://i.sstatic.net/8g6Og.png
[3]: https://i.sstatic.net/g2hLA.png
[4]: https://i.sstatic.net/lnH02.png