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. . . 33 t < II k(a,b ) - H k+ l(a,b). Since x < F n, we also have the inequality iifc(a, b)—tF n < which implies If 0 < k, then by Lemma 3.5 we have t < 0 and 0 < t + l so that —1 < / < 0. This is a contradiction since t is an integer. Hence k = 0. Thus r — n and hence R{N ) — n. Remark. Condition (3.6) cannot be replaced by the weaker condition N = H n(x , y) and 1 < y < x, This condition is not strong enough to imply n = R{N). For example if N = 96, then R(N) = 6 but N = H 5( 17,9) and 9 < 17. Also N = H 5(12,12) and 12 < 12. Theorem 3.7. Let r be fixed nonnegative integer. Then the number of N such that R(N ) = r is exactly Proof. Let r be fixed nonnegative integer. We will use Theorem 3.6 to count the number of N such that R(N) = r. We will count pairs (x, y) such that 1 < y < x < F r. For each such pair, we put N = H r(x,y). For each N there is only one pair (x,y) satisfying N = H r(x,y) and 1 < y < x < F r, by Theorem 2.6. How many pairs (x,y ) are there such that 1 < x < F rl For each such x, there are x choices of y such that 1 < y < x. Hence the number of N such that R(N) — r is given by the sum Example 3.7. The number of N such that R(N) = 5 is F 5(F 5 +l)/2 = 5 • 6/2 — 15. By Corollary 1.20, these 15 N all he in the interval 8 = F 6 < N < F S F Q = 40. They are the 15 values N = 8,11,14,16,17,19,20,22,24, 25,27,30,32,35 and 40. In this section we first prove that there are infinitely many double numbers. Then we give a combinatorial formula for the number of double numbers N having a fixed value of R. Last we give an asymptotic estimate for the number of double numbers up to F nF n +i . H k(a,b)/F n <t + l. F r {F r + 1) 2 x=l 4. Double numbers
