TITLE: Vassilliev invariants of knots and their cables 
QUESTION [7 upvotes]: 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.)

REPLY [9 votes]: 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
$$v_2(K_{p,q})=pv_2(K)+\frac{1}{24}(p^2-1)(q^2-1).$$