I’m going to try and fill in the fudge factors. There is a unique way of doing this, given known results and assuming the only allowed constants are the Thurston-Bennequin number and the number of components.  First let me assemble the known results.

We take out of [HR]:

Definition.  For a ruling $\rho$, we write $-\chi (\rho )$ for the number of switches minus the number of right cusps, or equivalently, minus the Euler characteristic of the canonical surface of the ruling.  By comparison, the Thurston-Bennequin number is the number of crossings minus the number of right cusps, or minus the would-be Euler characteristic of a total ruling.

The ruling polynomial is by definition

$$R(z)=\sum _{\rho }{z}^{-\chi (\rho )}$$

Example.  The standard unknot has a single ruling, and the corresponding disk has Euler characteristic $-1$.  So its ruling polynomial is $z^{-1}$.

Aside.  This sort of genus expansion is entirely standard in string theory; see e.g. any equation in the paper where Ooguri and Vafa made the Ooguri-Vafa conjecture.  They all have exactly this exponent, although the curves being counted there are holomorphic whereas the curves here are Lagrangian.  Given that in this case as well we are computing the HOMFLY polynomial (even with very similar correction factors, see the introduction to [STZ]) it seems like there should be in some cases some other version of the topological string which explains this coincidence, or perhaps some way to “just make a hyperkahler rotation” although we have never succeeded in making actual sense of such a rotation.

The main result of [HR] is:

Theorem. [HR].   Let $\mathrm{\Lambda }$ be a Legendrian knot, and $\mathrm{\#}Aug({\mathbb{F}}_q)$ the (naive) number of augmentations over ${\mathbb{F}}_q$. Then

$$R(q^{1/2}–q^{-1/2})=(q^{1/2}–q^{-1/2}{)}^{-c}{q}^{?}\mathrm{\#}Aug({\mathbb{F}}_q)$$

Moreover, if we’re counting ungraded augmentations, introduce the broken Euler characteristic

$${\chi }^{\ast }=\sum _{i\ge 0}(-1{)}^i{a}_i+\sum _{i<0}(-1{)}^{i+1}{a}_i$$

where $a_i$ is the number of Reeb chords of DGA degree $i$.  Then the question mark can be filled in:

$$R(q^{1/2}–q^{-1/2})=q^{(-c-{\chi }^{\ast })/2}\cdot (q^{1/2}–q^{-1/2}{)}^{-c}\cdot \mathrm{\#}Aug({\mathbb{F}}_q)$$

Example.  The DGA of the unknot has a single generator, which has degree 1.  So ${\chi }^{\ast }=-1$.  It has one component, so $c=1$.  It has one augmentation, since the degree one generator must be sent to zero.  The equation asserts

$$(q^{1/2}–q^{-1/2}{)}^{-1}=q^{(-1+1)/2}(q^{1/2}–q^{-1/2}{)}^{-1}$$

Remark.  In [HR], they leave off the $c$ in the exponent of $q$.  But, then you get the wrong thing for the unknot, so that can’t be right.

 

On the other hand, in [STZ], we showed that for rainbow braid closures,

$$R(q^{1/2}–q^{-1/2})=q^{-tb/2}\mathrm{\#}\underset{―}{Aug}({\mathbb{F}}_q)$$

Here, $\mathrm{\#}\underset{―}{Aug}({\mathbb{F}}_q)$ was in [STZ] naive groupoid cardinality of the category, but since all objects in these cases are (not complexes of) sheaves, there are no negative homs, so it’s the same as the cardinality we want presently.

Thus the desired conjecture, normalized, is:

Conjecture.

$$\mathrm{\#}\underset{―}{Aug}({\mathbb{F}}_q)=q^{(tb-{\chi }^{\ast })/2}\cdot (q–1{)}^{-c}\cdot \mathrm{\#}Aug({\mathbb{F}}_q)$$