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
36 James P. Jones and Péter Kiss suppose TV is a double and N < F nF n+Let r = R(N). We will that r < n. Suppose not. Suppose n < r. Let N = H r(a,b) where a are as in (2.16). By Lemma 3.1 (ii), since TV is a double, F r +1 < a. 1 N = H r(a, b) = aFr-i + bF r > F r+ lF r + F r > F n+ 2F n > F n • contradicting N < F nF n+ Therefore r < n. Theorem 4.7. For n > 1, the number of double numbers N < F r is equal to Fn — l F n-2 + F n — 1 2 Proof. By Lemma 4.6 and Theorem 4.5, the number of double nu] N < F nF n + 1 is £ F~ ilF r 2-' +1 ) = \i&-,+F rt) r= 3 r=3 /n-2 n-2 \ = 2 £ ^ + £ Fi = 2 + F n - V • What proportion of integers iV are double numbers? We shall shov on average approximately 7.3% of numbers are doubles. We shall shoi by proving that for n sufficiently large, approximately /2 of the nui N up to F nF n +1 are doubles. Here ß = (1 - Vh)/2 = -61803... sc ß*/2 = .072949016.... Theorem 4.8. The probabihty that N is a double number is at totic to ß*/2. Proof. Let a = (1 + Vb)/2 and ß = (1 - >/5)/2. Then a/5 = It is known that F n is asymptotic to a n j\J 5, i.e. that limi^/a 7 1 as By Lemma 4.6 and Theorem 4.7, the number of double numbers N F nF n +1, divided by the number of N up to F nF n +1 is equal to \F n—iF n—2 + F n — l)/2F nF n +i « F n-iF n-2/^F nF n +i « ((a"" 1 /V5)(a n~ 2/V5)) / (^2(a n /Vb)(a n+ l /Vb)) = a nla n~ 2/ 2a na n+ 1 = 1/2 a 4 = /3 4/2.
