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Transcendental continued fractions. (English) Zbl 0531.10035

Let \(A=a_ 1+\frac{1}{a_ 2+}\frac{1}{a_ 3+}...\), \(B=b_ 1+\frac{1}{b_ 2+}\frac{1}{b_ 3+}... \). The author proves Theorem 1: Suppose that there exists a real number \(r>1\) such that \(r^{-1}a_ n\geq b_ n\geq a^{n-1}_{n-1}\) holds for \(n=1,2,3,... \). Then A and B are algebraically independent. Next he proves a theorem from which he deduces the following Corollary: Let \(a_ n\geq a^ 2_{n-1}\), \(b_ n\geq b^ 2_{n-1}\) hold for all \(n\geq 3\) and \(\log a_ n\quad \log b_ n=o(\log \min(a_{n+1},b_{n+1}))\) and \(\log b_{n+1}=o(b_ n\quad \log b_ n)\) as \(n\to \infty\). Then \(A^ B\) is transcendental. A striking example is this: Let C and \(a_ 1\) be positive integers. Put \(a_{n+1}=C^{a_ n}\) for \(n=1,2,... \). Then \(A^ A\) is transcendental.
Reviewer: K.Ramachandra

MSC:

11J81 Transcendence (general theory)
11J70 Continued fractions and generalizations
11J85 Algebraic independence; Gel’fond’s method
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