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. . . 25 with k = r + 1, by Lemma 1.8 we get (8/5) r_ 1 < F r+ l < N. Taking logs of both sides we have (r - l)ln(8/5) < ln(jV). Hence we have r — 1 < ln(/V)/ln(8/5) < ln(iV)/(47/100) < ln(JV) • 2.128, proving the lemma. Lemma 1.16. If 1 < N, then [1.5 + .893 -]n(N)] < R(N). Proof. Let r = R(N). Lemma 1.11 implies N < F rF r+ l. If N < 6, the inequality can be checked by cases. Suppose 7 < N . Then 4 < r. We will use the following elementary inequality which is easy to prove using the fact that x 2 > x + 1 for x = 7/4. (1.17) F k<{ 7/4)*" 2 (3<ifc). Using the inequahty twice, with k — r and k — T + 1, we get (1.18) N < F rF r+ l < (7/4) r~ 2(7/4) r_ 1 = (7/4) 2 r~ 3. Taking logs of both sides, ln(iV) < (2r - 3)ln(7/4). Hence ln(A r)/ln(7/4) < 2(r - 1.5). Therefore 2~ x • ln(A r)/ln(7/4) < r - 1.5. Consequently 1.5 + 2" 1 • ln(iV)/ ln(7/4) < r. Hence [1.5 + 2" 1 • ln(A r)/ln(7/4)l < r. Therefore [1.5 + .893 • ln(7V)] < r. Corollary 1.19. For N > 1, [1.5 + .893 • ln(A r)] < R(N) < [1 + 2.128 • ln(7V)J. Proof. By Lemma 1.14 and Lemma 1.15. Corollary 1.20. If R(N) = r, then F r+ l < N < F rF r+ l. Proof. Suppose i?(A r) = r. By Lemma 1.8, F r +i < N . By Lemma 1.11, N < F rF r +i . The equation N = H r(a,b ) sometimes has two solutions (a, b) in positive integers with r = R(N). E.g. if N = 6, then R(6) = 3, 6 = // 3(2.2) and 6 = H 3(4,1). Definition 1.21. N is called a double number if there exist a,b,c,d > 1 such that N = H r(a,b) = H r(c,d),a ^ c or b f- d, (equivalently if a ^ c and b / d ), where r — R(N). If N is not a double number, then N is called a single number. Examples 1.22. Some representations of N in the form N = H r(a,b) with R = R(N): N = 1, R= 1, a= 1, 6 = 1, N = 10, R = 4, a = 2, 6 = 2,
