The Lucas V-sequence has V0(P,Q)=2, V1(P,Q)=P and Vn+2=PVn+1−QVn. Thus its first terms are 2,P,P2−2Q,P3−3PQ. For (P,Q)=(1,−1) these are the Lucas numbers 2,1,3,4,7,….
For roots α,β of t2−Pt+Q, one has Vn=αn+βn, including the repeated-root case. This gives V2n=Vn2−2Qn immediately. The identity Vn=2Un+1−PUn relates it to the Lucas U-sequence without requiring division.