Genus of non-reduced curves Let $X$ be a smooth projective variety of dimension $3$, and $L$ an ample line bundle with $h^0(X,L)\geq 2$. Let $s$ and $t$ be two generic linearly independent sections of $L$, and $C$ the curve that they define.
Assume we know all intersection numbers $L^3$, $L^2K_X$, $LK_X^2$ and $K_X^3$.
Of course we can compute the genus of $C$ by adjunction. Assume now that $C$ is irreducible but non-reduced, and of multiplicity $2$. In this case, we can also compute the genus of $C_{red}$ by adjunction: restricting $L$ to $\{s=0\}$ we get $2H$ for some line bundle $H$, and $C_{red}$ is defined by a section of $H$.
Now, take different sections and assume that the curve $C'$ you get has two irreducible components, say $A$ and $B$, with $A$ reduced and $B$ non-redued of multiplicity two.
Can I say something about the genus of $B_{red}$? For instance, is it smaller than the genus of $C_{red}$ computed in the previous paragraph?
 A: This does not answer the question, since the question concerns both genera of residual and of non reduced curves, but addresses only the issue of residual genus formulas, and in the case of $\mathbb{P}^3$. The formula is taken from Semple and Roth, page 91. But it may help.
Ok, as in Semple and Roth, pages 90-91, consider a space curve $C$ of degree $n$ and genus $p$, lying on a surface $S$ of degree $d$. Then pass another surface $S’$ of degree $e$ through $C$, so that it cuts further on $S$, another residual curve $C’$ of degree $n’$ and genus $p'$. ($C’$ is not necessarily connected, so $p’$ could be negative.)
Then we claim:  $2(p-p’) = (n-n’)(d+e-4)$. 
This follows from the two adjunction formulas for $C$ and for $C’$, both taken say on $S$, and subtracted.
By the adjunction formula for the curve $C$ on the surface $S$, the degree of the canonical bundle $K$ of $C$, equals $C\cdot C + (d-4)n$, where $C\cdot C$ is the intersection number on the surface $S$. Similarly the degree of the canonical bundle of $C’$ computed as a curve on the same surface $S$, equals $C’\cdot C’ + (d-4)n’$.
Subtracting, the difference of the degrees of the two canonical bundles equals  $\deg(K) - \deg(K’) = 2(p-p’) = (d-4)(n-n’) + C\cdot C - C’\cdot C’$.
Now since $S$ intersects $S’$ in the union $C+C’$, I claim $(C+C’)\cdot C = en$, and $(C+C’)\cdot C’ = en’$. Subtracting, and canceling, gives $C\cdot C - C’\cdot C’ = e(n-n’)$. Substituting above, gives $2(p-p’) = (d-4)(n-n’) + e(n-n’) = (d+e-4)(n-n’)$. Of course also $n+n’ = de$.
For example, if $C$ is two disjoint lines lying on a cubic surface $S$, and $S’$ is a quadric surface passing through both lines, then $C’$ is a quartic curve.  Then $n=2$, $n’=4$, $p = -1$, $d=3$, $e=2$, so $2(-1 -p’) = -2$, so $p’ = 0$, and $C’$ is rational. This is example (2) at the top of p. 93, of Semple and Roth.
