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. . . 29 This approach to R(N), thru g r(N) and h r(N), also gives a new algorithm to decide whether TV is a single or a double. From (2.9) and (2.10) we have (2.14) TV is a single iff [^(TV)] = [M^OJAlso (2.15) TV is a double iff |"^ r(iV)] < [h r(N)\. From (2.5), (2.6), (2.7) and (2.8) we can obtain explicit formulas for a,b,c and d: (2.16) a = A r(N)- \g r(N)]F r and b = B r{N) + r<7r(A01^-i , (2.17) c = A r(N) - [h r(N)\F r and d = B r{N) + [h r{N)\F r^ . If TV is a single, c = a and d = b. If TV is a double, c = a — F r and d — 6+F r_i . Thus when r = Ä(TV), formulas (2.16) and (2.17) can be used as definitions of a.b.c and d. The ratio on the right side of (2.11) is not always less than 2 however, even when r — R(N). In this case, when r = R(N), we have only the weak inequality (2.18) R(N) < r => ~ < a + 1. r r — 1 * T Here a = (1 -f \fb)/2 = 1.61803 ... so that a + 1 = 2.61803 .... The idea of the proof is the following: From Lemma 1.11 we see that R(N) < r implies TV < F rF r+1. Then (F rF r +\ — F r-i)F r-\F r < a + 1 can be shown using Fr < aFiF r_i + F r+i . Next we shall prove that there are no triples. The following lemmas will be used. Lemma 2.19. If 1 < r, then F rF r+ l < (1 + 2F r+ l )F r_i + F r. Proof. If 1 < r, then F r < 1 + 2F r_i . Hence F rF r+i < (1 + 2F r_! )F r +\ = F r +i + 2F r-i • F r+1 = F r-i + 2 • F r+ l + Fr = (1 + 2 F r+ 1 )F r_i + F r. Lemma 2.20. Suppose 1 < TV, TV = H r(a,b), R(N) = r and 1 < b. Then a < 2F r. Proof. Let r = R(N) and TV = H r(a,b). Since 1 < TV and r = Ä(TV), we have 1 < r. We claim (2.21) a <b + 2F r+ l.
