The following is perhaps a standard question, but i could not find a plain enough answer 
by simply searching online.


Q: Given a knot $K$ and its $(p,q)$-cable $K\_{p,q}$ what is a relation
between the Vassiliev invariants of $K$ and $K\_{p,q}$?


In particular, I would be happy with a formula for the 2nd coefficient
of the Conway polynomial. (One may attempt to
solve this via the A-polynomial since it has a simple relationship for
satellites, however relating the resulting A-polynomial back
to the Conway p. seems nontrivial, ... at least for me. Also it is
seems likely that somebody could have already worked this out.)



Vassilliev invariants of knots and their cables 
As I mentioned in a comment, for the degree $2$ invariant $v\_2$ which is the coefficient of $z^2$ in the Conway Polynomial, we have that $v\_2(K\_{p,q})=av\_2(K)+b$. If $K$ is the unknot, this implies that $b=v\_2(T\_{p,q})$, where $T\_{p,q}$ is the $(p,q)$-torus knot (assuming here $p,q$ are relatively prime.) Alvarez and Labastida wrote down formulas for Vassiliev invariants of torus knots, and in particular they showed $$v\_2(T\_{p,q})=\frac{1}{24}(p^2-1)(q^2-1).$$ So that gives you your constant term. Using Ryan's formula for the Alexander polynomial, one should be able to show that $a=p$. This is because when you make the substitution $t\mapsto t^p$, in the conversion to the the Conway polynomial we have $z^2=t+t^{-1}-2$, and so $z^2\mapsto t^p+t^{-p}-2$. It's a lemma that $t^p+t^{-p}=2+pz^2+\cdots$, so the coefficient of $z^2$ will get multiplied by $p$. So the answer will be
$$v\_2(K\_{p,q})=pv\_2(K)+\frac{1}{24}(p^2-1)(q^2-1).$$ 


