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. . . 31 be finite. By Lemma 1.11, the number of such N must be less than or equal to F rF r+i . We shall give an exact formula for this number. First we need some lemmas. Lemma 3.1. Suppose R(N) = r and a,b,c,d are as defined in (2.16) and (2.17). Then N = H r(a,b) = H r(c,d). If N is a single, then c = a,d=b, (i) 1 <b<a< F t and 1 < b < F R If N is a double, then we have c = a — F r, d = b -f F r-\ , b < a, (ii) F r-i < d < c < F r , F r +1 < a < 2 F r and 1 < b < F r2Proof. Suppose a,6,c and d are as above and N > 1. Let r = R(N). Then N = H r(a,b ) = H r(c,d). Suppose first N is a single. By (2.16) and (2.17), c = a,d = 6, 1 < a and 1 < b. By Lemma 1.2, N = H r(a,b) = H r(a -f F r,b — F r1). Hence b < F r_i , else N would be a double. By Lemma 1.4, 6 < a. By Lemma 1.2 we know N — II r(a,b) = II r(a — F r,b+ iv-i)Hence a < F r, else N would be a double. Therefore (i) holds. Next suppose N is a double. Then by (2.15), (2.16) and (2.17), c = a — F t , d — 6-f F r-i , 1 < a,b,c,d. By Lemma 2.20, a < 2F R . Since c = a — F R, this implies c < F R. By Lemma 1.4, since N = H r(c,d), d < c. Hence d < F R. Since d < F R and d = b-\- F r_i , b-\- F R_I < F R, so that b < F R — F R_I = F r_ 2 5 i.e. b < F R2- Since 0 < b and d = b -f iv_i, F R-\ < d. Since F r_i < d and d < c, F R~\ < c. Since a = c + F R , this implies that F R +1 < a. Hence statement (ii) holds. Lemma 3.2. If R(N ) = r, then there exist unique positive integers x and y satisfying (3.2) N = H t(X, y) and 1 < y < x < F r. Proof. By Lemma 3.1. If /V is a single, then we can let x = a and y — b. If N is a double, then we can let x — c and y — d and we will have ^ V £ £ ^ F r. x and y are unique by Theorem 2.22, to the effect that N = H r(x,y) has at most two solutions. Every N is either a single or a double. Note that if TV is a double, then x = a and y = b won't satisfy 1 £ V < x < F r since F r +1 < a. Lemma 3.3. Suppose R(N) = r. Then all solutions (x.y) of N = H r(x,y ) in positive integers satisfy either (3.3.1) 1 <y <x < F T
