The conductivity sum rule applied to LaSn3, CeSn3, LaB6, and CeB6 between 1.5 and 4.35 eV shows that there is approximately one more electron contributing to the optical conductivity of the Ce compounds than to that of the analogous La compounds. From the data and band calculations we conclude that about 0.5 Ce 4f electron per atom contributes as initial states for CeSn3 in the above energy range, and about half that number of 4f electrons serve as initial states in LaSn3. This shift of oscillator strength from higher energies for atoms to lower energies for solids is attributed to hybridization. Much of the remaining Ce-La difference is from transitions to p-like final states, which are reduced in the La compounds due to hybridization with unoccupied 4f states.