The Lucas numbers $L(n)$ are defined by $L(0)=2$, $L(1)=1$, and $L(n)=L(n-1)+L(n-2)$, for $n\ge2$. Looking at the sequence $\{L(n)\}$ modulo various numbers, we are lead to conjecture that $\{L(n) \mod m\}$ contains a complete residue system modulo $m$ if and only if $m$ is one of the following: $2, 4, 6, 7, 14, 3^k$, $k\ge1$. Example: Modulo 5 we have the sequence $2,1,3,4,2,1,\dots$, and since the sequence repeats we never obtain $0 \mod 5$. Example: Modulo 6 we have $2,1,3,4,1,5,0,\dots$ and we obtain a complete residue system $\mod 6$.

The corresponding problem for the Fibonacci sequence was solved by S. A. Burr in "On Moduli for Which the Fibonacci Sequence Contains a Complete System of Residues," Fibonacci Quarterly, December 1971, pp. 497-504. The sequence $\{F(n) \mod m\}$ contains a complete residue systems modulo $m$ if and only if $m$ is one of $5^k, 2\cdot5^k, 4\cdot5^k, 3^j\cdot5^k, 6\cdot5^k, 7\cdot5^k, 14\cdot5^k$, $k\ge0$, $j\ge1$.

Although 40 years later, I don't find that anyone has looked at the Lucas version. If someone supplies a reference or a proof, I would appreciate it. Thank you.