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I am not an expert, but I would assume, from reading Walker's plane curves for the polynomial case, that the delta invariant is the sum of the numbers (1/2)(n)(n-1), $\frac{n(n-1)}{2}$, summed over all multiplicities n $n$ at the given point and all "infinitely near" points. Hence it would seem this number is minimal when there are no infinitely near singular points, e.g. when the point is "ordinary" of multiplicity n. $n$. Thus the minimal delta invariant at a point of multiplicity n $n$ would be (1/2)(n)(n-1).$\frac{n(n-1)}{2}$.

This ordinary case has $r = n n$ branches, the maximum number of branches, so from the formula above, this would seem also to minimize the milnor number, as $n(n-1)+1-n = (n-1)^2.n-1)^2$.

Does this seem plausible?

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I am not an expert, but I would assume, from reading Walker's plane curves for the polynomial case, that the delta invariant is the sum of the numbers (1/2)(n)(n-1), summed over all multiplicities n at the given point and all "infinitely near" points. Hence it would seem this number is minimal when there are no infinitely near singular points, e.g. when the point is "ordinary" of multiplicity n. Thus the minimal delta invariant at a point of multiplicity n would be (1/2)(n)(n-1).

This ordinary case has r = n branches, the maximum number of branches, so from the formula above, this would seem also to minimize the milnor number, as n(n-1)+1-n = (n-1)^2.

Does this seem plausible?