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Let $f:S\rightarrow \mathbb{P}^1$ be an elliptic K3 surface. Assume that $\mathrm{Pic}(S)\cong U$, where $U$ stands for the hyperbolic lattice. I think that the elliptic fibration has only singular fiber of Kodaira Type $I_{1}$ and $II$, and $m+2n=24$ where $n (m)$ is the number of singular fiber of Kodaira Type $I_{1}(II)$.

Can we say something similar when $\mathrm{Pic}(S)\cong U(k)$ for some $k>1$?

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    $\begingroup$ If $S$ is elliptic, then the fiber and the section generate a sublattice in $Pic(S)$ which is isomorphic to $U$. So if $S$ is elliptic and $Pic(S) \cong U(k)$ you will have $k = 1$. Did you mean to ask for $S$ to be only genus one fibered and $Pic(S) \cong U(k)$ ? $\endgroup$
    – Tony Pantev
    Commented Sep 3, 2012 at 1:50
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    $\begingroup$ Yes: a genus-1 fibration gives a rank-2 sugroup of Pic, and if there were any reducible fiber the rank would exceed 2. (I guess you're using the definition of an elliptic fibration that does not require a section, else the section together with the fiber already generate a copy of $U$ in Pic.) Likewise for K3 surfaces of higher degree: if Pic has rank 2 then there are no reducible fibers (and $m+2n=12\chi$). $\endgroup$
    – Noam D. Elkies
    Commented Sep 3, 2012 at 1:54

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