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
30 James P. Jones and Péter Kiss If not, then b -f 2F r +i < a. Since 1 < b and N = H r(a , b), by Lemma 2.19 and Lemma 1.11 we have N = aF r-i+bF r > (b + 2F r+ 1)F r. l+bF r > {l + 2F r+ l)F r_ l+F r > F rF r+ l. But this contradicts Lemma 1.11 which says that N < F rF r+1, since r = R(N). Hence (2.21) holds. Now it is easy to see that N = aF r-.\ -f bF r = (6 + 2 F r+ l - a)F r + (a - 2F r)F r+ 1 . Supposing 2F r < a and using (2.21), we get the contradiction R(N ) > r-f 1. So a < 2F r. Theorem 2.22. If R(N) = r, then the equation N = H r(a, b) has at most two solutions in positive integers a, b. There are no triples. Proof. Suppose the equation N = H r(a, b) has three solutions in positive integers, say (a,6),(c, d) and a third solution (x,y ). Then c = a — F r , d = 6 -f F t_ 1, x = a — 2F r and y = b + 2F r_i . But by Lemma 2.20, a < 2F r. Hence x < 0, a contradiction. From Theorem 2.22, if r = R(N), then LM-AOJ < fPr(^)] + 1. And so in (2.15), when ^(iV)] < [h T(N)J, we have r<7 r(A r)l + 1 = [hr(N)\. Following F nF n +i there is a very long interval consisting entirely of singlels . Suppose R(N) = r. Recall from Corollary 1.20 that if R(N) = r , then N must lie in the interval F r +i < N < F rF r +i . We can show that most N in this interval are singles. Theorem 2.23. If F nF n +1 < N < F nF n+ l + Fl + F n+ 2, then N is a single. We won't prove this result, (Theorem 2.23.). However it will be clear how to do so after we have proved Lemma 3.1 in the next section. Taking a limit as n —» oc, one finds that the interval [F nF n +i , F nF n+1 + l'n tn+ 2] occupies some 38% of the interval [I n t n+i , i n+i F n+ 2]. (, Q 2 = ((1 - \/5)/2) 2 = ( —.61803) 2 = .381966 . ..) Thus on average more than 28% of N are singles. Actually, in the next section, we shall prove that 92.7% of N are singles. 3. The number of N such that R(N) = r In this section we consider the problem of the number of N such that R(N) — r. Here r is a fixed positive integer. The number of such N must
