부산대학교 도서관

Given a set $A$ of relatively prime positive integers, the authorsconsider the set $\Gamma(A)$ of nonnegative linear combinations ofelements of $A$. The complement of $\Gamma(A)$ is known to be finite;the numbers $g(A)=$ the largest integer in $A$ and $n(A)=$ the numberof integers {\it not} in $A$ are of interest here.\parIf $|A| = 2$, the problem was solved by Sylvester:$$g(\{a,b\}) = (a-1)(b-1)-1\quad \text{and}\quad n(\{a,b\}) =\frac{1}{2}(a-1)(b-1).$$When the elements of $A$ are drawn from the Fibonacci and Lucassequences, some results for $g$ and $n$ have been determined. In thecurrent work, if $a$ and $b$ are relatively prime integers, the sets$$F = \{a, a+b, 2a+3b, \ldots , F_{2k-1}a + F_{2k}b\},$$where $\{F_n\}$ denotes the Fibonacci sequence, and$$L = \{a, a+3b, 4a+7b, \ldots , L_{2k-1}a + l_{2k}b\},$$where $\{L_n\}$ denotes the Lucas sequence, are considered as inputs tothe functions $g$ and~$n$.\parFor the Fibonacci sequence, it is shown that$$g(F) = a \left(\left\lceil \frac{F_{2k-1}(a-1)}{F_{2k}}\right\rceil -1\right) + (a-1)b$$and$$n(F) = \sum_{y=1}^{a-1} \left\lceil\frac{F_{2k-1}y}{F_{2k}}\right\rceil + \frac{1}{2}(a-1)(b-1),$$though the authors note that a simplification of the sum for $n$``appears difficult, but would be desirable''.\parFor the Lucas sequence, expressions for $g$ and $n$ are somewhat morecomplicated and depend on the value of $a\pmod 3$; the case $a < 12$requires special handling.