Let $R$ be a reduced curve singularity over an algebraically closed field $k$ and $\tilde{R}$ its integral closure in its total ring of fractions.  

The $k$-dimension of $\tilde{R}/R$ is finite.  If we assume $R$ is non-planar and Gorenstein, then **how small can this number be?**

The ring $R = k[[x,y,z]]/(xy = z^2, z x = y^2)$ is a complete intersection, hence Gorenstein, and the dimension of $\tilde{R}/R$ is $4$.  The question is thus "is $2$ or $3$ possible?"

For the sake of concreteness, let's say that a curve singularity is a $1$-dimensional quotient of $k[[x_1, \dots, x_n]]$ for some $n$.

**Edit:** I had thought that the $k$-dimension of $\tilde{R}/R$ was widely known as the $\delta$-invariant; I think this the notation Serre uses in _Algebraic Groups and Class Fields_.  From the comments, it seems this is non-standard and I have edited accordingly.

As Graham points out, the number $\operatorname{dim}(\tilde{R}/R)$ is also the colength of the conductor ideal.  The number also comes up in computing the (arithmetic) genus of a singular curve.