In his $G\_2$ paper, Kuperberg gives the following numbers of acyclic freeways
for n=0...6: 1, 0, 1, 1, 4, 10, 35. (Which is identical to $dim Inv(V^{⊗n}\_{1,0})$, the spanning size of the tangle vector space. (??). But that is NOT
identical with the number of crossingless trivalent tangle graphs:
for n=6, the "hexagon" can't be resolved into the 34 acycling tangle graphs
and must be added to the linear independent set.)
The $G\_2$ numbers (or so I think) are identical for the whole $E\_7$ family
(except $B\_3(\Lambda\_3)$, $A\_1(3\Lambda\_1)$ and $A\_1\*A\_1(\Lambda\_1\*\Lambda\_1)$ where the last number should be 30,34 and 25 if I computed correctly). (But since you now
must use 2 irrep colors for any other than $G\_2$, 5 of the 34 above graphs are forbidden.)  

With some jump of faith, I suppose also for all members of the $E\_8$ family the spanning
space numbers are the same: 1, 0, 1, 1, 5, 15, 70 (???), except for some
"special" groups of that family.


Can you verify my $dim Inv(V^{⊗n}\_{1,0})$ values for $E\_8$ ? (Maybe with a list of exceptional members ? Directly I can only compute for $A\_1(4\Lambda\_1)$ and this surely gives less than 70.)



Span of tangle vector space for different Lie groups
Here are the dimensions of the space of invariant tensors in the tensor powers of the adjoint representation:


F4: 1 0 1 1 5 16 80 436 2891 ...


E6: 1 0 1 1 5 17 90 542 3962 ...


E7: 1 0 1 1 5 16 80 436 2877 ...


E8: 1 0 1 1 5 16 79 421 2674 ...


The reason the numbers for E6 are too high is that there is a diagram automorphism. The correct category of representations is the representations of the semidirect product of the Lie group and a group of order two. Then these numbers suggest that the sequence starts


1 0 1 1 5 16 80 436 ...


and that for E8 there is an extra relation for n=6. However it is not clear what should happen for n=8 or n>8.


