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May 5, 2016 at 8:33 comment added Lazzaro Campeotti @FrancescoPolizzi: a bit more simply, I guess one could take $F_6$ to be a sextic with $d$ tritangent lines; the double cover will then have Picard rank $1+d$. For small values of $d$ this should be easy enough to do by hand.
May 5, 2016 at 6:04 comment added Francesco Polizzi It is possible to construct a K3 surface $S$ with Picard rank $2$ by taking a double cover of the plane of the form $w^2=F_6(x, \, y, \, z)$, where $F_6$ is a smooth plane sextic curve which admits a six-tangent conic. The surface $S$ does not sit naturally in $\mathbb{P}^4$, but in the weighted projective space $\mathbb{P}(1, \, 1, \, 1, \, 3)$. See Example 20 here: www-fourier.ujf-grenoble.fr/sites/ifmaquette.ujf-grenoble.fr/…
May 4, 2016 at 13:24 comment added user80337 This is very helpful indeed! Thank you! Do you also know of any examples with Picard rank between 2 and 19 (inclusive)? I ask this because I'm trying to test some claims, so the more examples I can find, the easier my life will be.
May 4, 2016 at 12:15 history answered Francesco Polizzi CC BY-SA 3.0