Does anyone know where I can find examples of quartic K3 surfaces for which the Picard group is known? I'm really interested in examples where there are explicit constructions of the divisors generating the Picard group (rather than examples where we only know the Picard rank). I appreciate that this is not easy to compute in general, so any useful references you can suggest will be most welcome.
It is known that the Picard number of the Fermat quartic $$S=\{x_0^4+x_1^4+x_2^4+x_3^4=0\}$$ is $20$ (i.e., $S$ has maximal Picard rank) and that the NéronSeveri group of $S$ is generated by lines. Furthermore, a basis of $\textrm{NS}(S)$ given by lines can be written explicitly. See Section 6 of
Matthias Schütt, Tetsuji Shioda, and Ronald van Luijk, Lines on Fermat surfaces, J. Number Theory 130 (2010), no. 9, 19391963.

$\begingroup$ 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. $\endgroup$ – Kenny Wong May 4 '16 at 13:24

1$\begingroup$ 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 sixtangent 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: wwwfourier.ujfgrenoble.fr/sites/ifmaquette.ujfgrenoble.fr/… $\endgroup$ – Francesco Polizzi May 5 '16 at 6:04

$\begingroup$ @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. $\endgroup$ – potentially dense May 5 '16 at 8:33
The first example of a smooth quartic surface in $\mathbb{P}^3$ with (geometric) Picard number 1 was found by Ronald van Luijk (who, incidentally, was my PhD advisor). The following is cited from one of his sets of slides:
The quartic surface in $\mathbb{P}^3(x,y,z,w)$ given by $$ wf= 3pq−2zg $$ with $$ f = x^3−x^2y−x^2z+x^2 w−xy^2−xyz + 2xyw+ xz ^2 + 2 xzw + y^3 + y^2z − y^2w + yz^2 + yzw − yw^2 + z^2w + zw^2 + 2w^3, $$ and $$g = xy^2 + xyz − xz^2 − yz^2 + z^3 ,$$ and $$ p = z^2 + xy + yz,$$ and $$ q = z^2 + xy$$ has geometric Picard number 1 (and infinitely many rational points).
Since the rank is minimal in this case, writing down explicit divisors whose classes generate the Picard group is completely trivial.
See this link for the slides, and also Van Luijk's paper "K3 surfaces with Picard number one and infinitely many rational points", Algebra and Number Theory, Vol. 1, No. 1 (2007), 115.
The following paper gives a large family of K3 surfaces for which the authors prove that the geometric Picard number is 3. The generators are quite explicit, since the surfaces are defined by $(2,2,2)$ forms on $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$.
Baragar, Arthur; van Luijk, Ronald; K3 surfaces with Picard number three and canonical vector heights. Math. Comp. 76 (2007), no. 259, 1493–1498. MR2299785
EDIT: Sorry, just realized that you want quartic K3's sitting in $\mathbb P^3$, not K3's in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. But maybe this paper will still be of interest, so I won't delete this answer. In any case, based on the first three answers, it seems that the correct answer is "ask van Luijk", since he's the leading expert in proving this sort of thing.
There are some simple such examples if you want to know the generators but not explicit equations for the surfaces:
 A very general quartic in $\mathbb P^3$ has Picard number $1$ and hence its Picard group is generated by any minimal degree curve.
 A quartic, very general among those containing a fixed line has Picard number $2$ and its Picard group is generated by that line and the hyperplane class (or the complementary cubic curve).
 A quartic, very general among those containing a fixed conic has Picard number $2$ and its Picard group is generated by that conic and the hyperplane class (or the complementary conic).
 A quartic, very general among those containing two fixed skew lines has Picard number $3$ and its Picard group is generated by those lines and the hyperplane class.
If you are only interested in generating $\mathrm{Pic}\otimes\mathbb Q$, then you have that
 $\mathrm{Pic}\otimes\mathbb Q$ is generated by the smooth rational and elliptic curves on it for any $K3$ with Picard number at least $5$.
 $\mathrm{Pic}\otimes\mathbb Q$ is generated by the smooth rational curves on it for any $K3$ with Picard number at least $12$.

$\begingroup$ In what way are you using the word "general" here? Does it refer to the complement of a Zariski closed (proper) subset of the space $\mathbb{P}^{34}$ of quartics, or do you e.g. need to exclude a countably infinite number of closed subsets? $\endgroup$ – RP_ May 5 '16 at 16:56

1$\begingroup$ Alas one does need to exclude a countable infinity of codimension1 subspaces. $\endgroup$ – Noam D. Elkies May 5 '16 at 16:59

$\begingroup$ Sorry, should have said "very general". $\endgroup$ – Sándor Kovács May 5 '16 at 17:44