Basis of invariant tensors of rank n in three dimensions [This is a question motivated by theoretical physics, so apologies if the language is rough...]
In three dimensions the spaces of invariant (or isotropic) tensors of rank $n$ have dimensions 1, 0, 1, 1, 3, 6, 15, 36, 91, 232, ... (see MathWorld). Equivalently, this is the number of trivial representations in the $n$-fold tensor product of the fundamental representation of $SO(3)$.
Any invariant tensor can be expressed as a linear combination of products of Kronecker deltas and (for odd rank) one Levi-Civita symbol (see this note, which contains references to the original literture), but in general there will be linear independences.
The dimensions are given by the Riordan numbers, of which there are many combinatoric interpretations. My question is: is there an explicit bijection between one of these interpretations and a linearly independent basis of invariant tensors?
For example, in the case of $SU(2)$ there is such a bijection, expressed in terms of noncrossing  pairings on a circle, with each pairing corresponding to $\epsilon_{ab}$ (in quantum mechanics we say a two-spin singlet state). This is originally due to Rumer, Teller, and Weyl, and generalized by Temperley and Lieb. See this paper for a more recent discussion
 A: Planar partitions with no singletons works.  You need to pick for each $n>1$ some map with certain properties.  One way to do this is to just fix a preferred trivalent tree of each size and interpret each vertex as a cross product.  For example, one arbitrary choice gives the map $V^{\otimes n} \rightarrow \mathbb{C}$ given by $$v_1 \otimes \ldots \otimes v_n \mapsto (((v_1 \times v_2) \times v_3) \ldots \times v_{n-1})\ldots) \cdot v_n.$$
It’s a little awkward that the above formula isn’t invariant under cyclic permutation of the variables, so you have to arbitrarily pick a "starting point" for each connected component.  A better way is to write the non-crossing partition of 2n strands where each end connects to its neighbor and then attach a bunch of “2nd Jones-Wenzl projectors” (i.e. the projection from the square of the 2d to the 3d) along the boundary in an interleaving way (ie connecting each strand to its other neighbor).
As discussed in comments, the JW2 part can be interpreted as rewriting the vector in terms of the corresponding sum of Pauli matrices.  If we let $\sigma_v$ denote the sum of Pauli matrices corresponding to $v$, then the rotationally invariant map is just the trace of the product of Pauli matrices:
$$v_1 \otimes \ldots v_n \mapsto \mathrm{Tr}(\sigma_{v_1} \sigma_{v_2} \cdots \sigma_{v_n}).$$
This was the formula for a single connected component, so the general formula is a product over connected components of the trace of the product of Pauli matrices for each component.
Another approach, using a different set, is to take n pairs of boundary points and look at non-crossing partitions of 2n points such that no pair connects to itself.  Then just interpret this in SU(2) by embedding each 3-dimensional irrep inside the square of the 2-dimensional irrep.
