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. . . 35 d such that N = H r(c,d ) and F r _i < d < c < F r. Let x = c and y = d. Then (4.4) holds. Also since the condition F r_i < y < x < F r implies 1 < y < x < F r, x and y are unique by Lemma 3.1. For the proof of the second part, suppose (4.4) for some positive integers x and y. Then since 1 < r, 1 < y < x < F r. Hence R(N) = r by Theorem 3.6. N cannot be a single since in that case, by Lemma 3.1, we would have x — a,y = b and b < F r Hence N is a double. Note that if (x, y) satisfies F r-\ < y < x < F r, then (x + F r, y — F r_1) satisfies F r+\ < x < 2F r and 1 < y < F r_Also if (x,y ) satisfies F r+ 1 < x < 2F r and 1 < y < F r_ 2, then (x — F r iy + F r_ 1) satisfies F r_! < y < x < F r. So one could also prove a version of Theorem 4.4, with condition (4.4) replaced by N = H r(x,y ), F r+1 < x < 2 F r and 1 < y < F r_ 2. Theorem 4.5. Let r > 3. The number of N such that N is a double number and R(N) = r is exactly Proof. Suppose r is a fixed positive integer. To count the number of double numbers N such that R(N) = r we will use representation (4.4) of Theorem 4.4. We can determine the number of double numbers N such that R(N) — r by counting pairs of integers (2, y) such that F r_ 1 < y < x < F r. For each such pair (x, y) we can let N = H r{ xi y) since N depends uniquely on (x,y). How many pairs of integers (x,y) are there such that F r_ 1 < y < X < F r1 Since F r — F r_ 1 = F r_ 2, there are F r_ 2, there Eire F r_ 2 choices for x such that F r_ 1 < x < F r. For each choice of x, there are x choices for y such that F r-i < y < x. Therefore the numbers N such that R(N) — r is given by the sum Example. The number of N such that N is a double and R(N) = 6 is F 4(F 4 + l)/2 = 3-4/2 = 6. By Corollary 1.20 and Theorem 2.23 with n = 5 these N He in the interval 18 = 5 • 8 + 5 2 + 13 = F 5F 6 + F 5 2 + F x < N < F 6F 7 = 8 • 13 = 104. They are N = 78,83,88,91,96 and 104. Lemma 4.6. For all double numbers TV, N < F nF n + 1 iff R(N) < n. Proof. The first part of the lemma is the contrapositive of Lemma 1.11, if R(N) < n then N < F nF n+ 1. For the proof of the second part F r. 2 (/V-2 + 1) 2 F -
