부산대학교 도서관

Given a set $A$ of relatively prime positive integers, the authors consider the set $\Gamma(A)$ of nonnegative linear combinations of elements 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 number of integers {\it not} in $A$ are of interest here. \par If $|A| = 2$, 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 Lucas sequences, some results for $g$ and $n$ have been determined. In the current 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 to the functions $g$ and~$n$. \par For 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''. \par For the Lucas sequence, expressions for $g$ and $n$ are somewhat more complicated and depend on the value of $a\pmod 3$; the case $a < 12$ requires special handling.