Let (gi)i≥1 be a sequence of Chebyshev polynomials, each with degree at least two, and define (fi)i≥1 by the following recursion: f1=g1, fn=gn鈭榝n−1, for n≥2. Choose 伪∈Q such that is an infinite set. The main result is as follows: If is written in lowest terms, then for all but finitely many n>0, the numerator, An, has a primitive divisor; that is, there is a prime p which divides An but does not divide Ai for any i<n.
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