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

34 James P. Jones and Péter Kiss Lemma 4.1. For all n > 2, F

nF

n

+\ is a double number. Proof. Suppose 2 < n. Recall that by Corollary 1.10, R(F

nF

n+1) = n. We have F

nF

n+1 = F

n (F

n_i + F

n) ~ F

nF

n-i + F

nF

n = H

n(F

n,F

n). On the other hand, F

nF

n+1 = (

F

n + F

n)F

n1 + (F

n — F

n-i)F

n = H

n{2F

n,F

n - F

n_0 = H

n(2F

n, F

n-2)0 < F

n-2 since n > 2. The two representations of F

nF

n+i are distinct since F

n F

n—2 • Lemma 4.2. For n > 4, if N = F

n(F

n+1 - 1), then R(N) = n and N is a double number. Proof. By an argument similar to that in the proof of Lemma 4.1 it is easy to see that (4.2) N = H

n(F

n,F

n - 1) = H

n(2F

n, F

n_

2 - 1). To prove that R(N) = n we will use the IVL. Obviously n < R(N). Suppose that n + 1 < R(n). Then by the IVL there exist a > 1 and b > 1 such that N = H

n

+i(a,b). Hence F

n(F

n+

1 - 1) = aF

n + bF

n+Then F

n | 6, since (F

n,F

n

+1) = 1. Let b — eF

n

, where 1 < e. Then we have a+(e —l)F

n+

1 < 0, a contradiction. Thus R(N) = n. We give next a formula for the number of double numbers N with a fixed R value r. For this it is necessary first to characterise double numbers. From section 2 we have the following result. Lemma 4.3. Suppose R(N) = r. Then N is a double number iff (~l)

rF

r2N + l F

r1 + 1 (-1 )

rF

rXN - 1 F

r Proof. See the remark following Theorem 2.22 that N is a double iff +1= [h

r(N)\. Theorem 4.4. N is a double number and R(N) = r iff there exist unique positive integers x and y such that (4.4) N = H

T{x,y

) and i < y < x < F

r. Proof. For the proof of one part of the theorem, suppose N is a double number and R{N) = r. By Lemma 3.1, there exist positive integers c and