Thanks for giving the fractional tinas for the commas. They led to the thought that the dot up and dot down don't have to be ±¹⁄₂ t. They could be ±¹⁄₃ t. Is there an accurate EDO that's approximately 3 × 8539edo? That would also mean that the maximum error allowed for the various commas would be ±¹⁄₆ t.

And if we impose a requirement that n>7 tina commas must be 5-schisma complements of n<7 tina commas, then the fact that there are 13.90 tinas in the 5-schisma, not 14, implies even tighter constraints on the errors.

Those below 7 tinas can have an error of -0.167 t but only +0.067 t. And vice versa for those above 7 tinas.

It seems your list is still not complete, because when I subtract your 9.21 tina comma from the 5-schisma, I obtain the following 4.69 tina comma that isn't in your list.

tinas comma monzo interval cents limit pop 13.90 5s [-15 8 1 0 0 0 0 0 ⟩ 32805/32768 1.953721 5 5 9.21 41503n [ -9 -4 0 3 2 0 0 0 ⟩ 41503/41472 1.293601 11 43 4.69 ?????n [ -6 12 1 -3 -2 0 0 0 ⟩ ?????/????? ?.?????? 11 48

But these do not make an acceptable pair for 5 and 9 tinas, because they are more than a sixth of a tina away from 5 and 9.

The obvious 9 tina comma is:

8.93 539n [ 17 -5 0 -2 -1 0 0 0 ⟩ 131072/130977 1.255240 11 25

It's annoying that its 5-schisma complement (a candidate for 5 tinas) has a 3-exponent of 13:

4.97 ???n [-32 13 1 2 1 0 0 0 ⟩ ??????/?????? ?.?????? 11 30

There is a clear winner for 8 tinas. It has the best popularity ranking by far, and it is the 5-schisma complement of the 6 tina (2 mina) comma.

7.98 13:77n [-20 11 0 1 1 -1 0 0 ⟩ 13640319/13631488 1.121197 13 31

The idea occurs, to obtain the 2 tina comma as the difference between the 8 and 6 tina commas, but no, it's complete rubbish, with a 3-exponent of 14 and a popularity rank of 67.

So what about the 5 tina as the difference between the 8 and 3 tinas? Perhaps unsurprisingly, it gives us the same comma we found for 5 tinas as 14 - 9 tinas. i.e. the one with a 3-exponent of 13 (popularity 30). Maybe we should use it anyway.

It seems, at first, that there are two potential complementary 4 tina/10 tina pairs:

3.84 17:245n [-12 10 -1 -2 0 0 1 0 ⟩ 1003833/1003520 0.539891 17 36 10.06 17:1225n [ -3 -2 2 2 0 0 -1 0 ⟩ 1225/1224 1.413829 17 41 and 4.07 7:3025n [ -4 -3 2 -1 2 0 0 0 ⟩ 3025/3024 0.572403 11 39 9.83 7:605n [-11 11 -1 1 -2 0 0 0 ⟩ 1240029/1239040 1.381317 11 34

But that 9.83 tinas is actually 9.830 or 9.829 and so is more than 1/6th of a tina away from 10 tinas. So it has to be the 17-limit pair.

Unfortunately the 5 tina comma obtained as 9 tinas minus 4 tinas has a 3-exponent of -15 (popularity 33).