Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1998. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 25)
JONES, J. P. and Kiss, P., Representation of integers as terms of a linear recurrence with maximal index
28 (2.8) James P. Jones and Péter Kiss h r(N) = (~ 1) r f'-' A f - 1 Then g r(N) and h r(N ) are reals. For a and 6 as in (2.6), we have 1 < a iff t < h r(N ) and 1 < b iff g r{N) < t. Hence (a, b) is a positive solution of aF r_ 1 + bF r = N iff (2.9) g r(N) < t < h r(N). Since t is integer valued, condition (2.9) is equivalent to (2.10) g r(N) < \g T(N]\ < t < [/i. r(7V)J < h r(N). Condition (2.9) is in turn equivalent to [g r(7V)] < h T(N) and also to g r(N) < [h r(N)\. From (2.3), (2.7) and (2.8), it is easy to see that (2.11) h r(N)-g r(N) = N - F, r+l fi'r — 1 F r The functions g r(N) and h r(N) give us a new algorithm to compute R(N). We have Theorem 2.12. Suppose N >1. Then R(N ) is the largest integer r > 1 such that (2.12) ( —l) riV-2ÍV -f- 1 F r1 < (-iyFr-iN-1 Fr Furthermore, the set of r > 1 satisfying (2.12) is the set {2, 3,..., R(N)}. Hence (2.12) can be used as an algorithm to calculate R(N). Proof. By Lemma 1.8, if r < R(N ), then F r+\ < N and hence by (2.11), g r(N) < h r(N). Thus (2.13) r < R(N) => g r(N) < h r(N). By (2.9) and the IVL, for all r < R(N), there exist t(g r(N) < t < h r(N)), and this imphes [^ r(7V)] < j_/i r(7V)J . On the other hand, by (2.9), when R(N) < r, there is no integer t such that g r(N) < t < h r(N) and so we have not [^(jV)] < [h r(N)J. This shows that the set of r > 1 satisfying (2.12) is an interval.
