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

Representation of integers as terms of a. . . 23 Earlier we saw that n = 1 is realizable as a value of n such that N — Il

n(a,b) for a > 1, b > 1. In the next lemma we shall show that all values of n < R(N

) are realizable as values of n such that N = H

n(a

, b). We shall call this the Intermediate Value Lemma (IVL). Lemma 1.5. (I.

V.L.) If n < R(N), then there exist a,b such that N = fí

n(a, b), 1 < a and 1 <6. Proof. Suppose r = R(N) and n < r. There exist a > 1, b > 1 such that N = H

r(a,b). Let k = r — n. Then 0 < k. By Lemma 1.3 (ii), N = H

r(a, b) = H

r.

k(H

k(a,b),H

k+

l(a,b)) = II

n{H

k(a, b), H

k+

l (a, b)) where 1 < H

k(a,b

) and 1 < II

k+i(a,b), since 0 < k and a, b > 1. Lemma 1.6. If n > 1 then R(F

n

+1

) = n. Proof. Let r = R(F

n

+i). Since F

n+

X = F

n_i + F

n = H

n(l,l),n < r. Conversely, F

n+

1 = H

r(a.b

) = aF

r_i + bF

r > F

r-i + F

r = iv+i • Hence n > r. Therefore n — r. Lemma 1.7. If n > 2, then R(L

n

+1) = n + 1. Proof. Here we need the inequality L

N

+I +1 < -F

n+3- Let r = R(L

N

+1). Since L

N may be defined by LQ = 2, L\ =1 and L

N+2 = LN + L

N+\ > we have L

n — II

n{

2,1) and so = H

n+i(2,l). Hence n-f-1 < r. Conversely, L

n+i — H

r(a,b

) — aF

r_i + bF

r > F

r-\ + F

r = F

r

+\. Hence F

r+1 < L

n+iTherefore F

r

+

1 + 1 < X

n+i + 1 < ^n+3 and s

o ^V+i < ^

n+3

• Therefore r + 1 < n + 3. Hence r < n -f 2. Therefore r < n + 1. So r = n + 1 and Ä(i„+i) = 1. Lemma 1.8. If N < F

n+1, then R(N) < n. Proof. Let R(N) = r. Then there exist a, b > 1 such that N — H

r(a

, b). Hence we have F

n

+i > N — H

r(a,b

) = aF

r-\ + bF

r > F

r_i + F

r = F

r+

1. Thus F

n

+i > F

r

+

1. Hence we have n-fl > r + 1 so that n > r. In otherwords n > R(N). Corollary 1.9. If 1 < n and N < F

n+

lF

n+

2, then R(N) < 2n. Proof. If 1 < n, then F

n+

2 < L

n

+1- Hence F

n

+iF

n+

2 < F

n

+iL

n+i = F

2n+

2. Therefore N < F

2n+

2. Hence by Lemma 1.8, R(N) <2n + l. Therefore R(N) < 2n. Lemma 1.10. Let A be an arbitrary positive integer and suppose 0 < n. Then (0 (ii) n = R(AF

n

+i

) if A<F

n

, n<ft(AF

n

+i) if F

n < A.