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Oliver Chambers delivers this months puzzle challenge - A Rational Sequence.
A sequence $$\{\alpha_0, \alpha_1, \alpha_2 \dots\}$$ is defined such that $$\alpha_{n+1} = \alpha_n + \frac{1}{\lfloor \alpha_n \rfloor}$$, for all $$n \ge 1$$, where $$\lfloor x \rfloor$$ denotes the greatest integer less than or equal to $$x$$.
a) If $$\alpha_0 = \frac{43}{11}$$, what is the smallest $$k$$ such that $$\alpha_k$$ is an integer? (computers allowed)
b) Show that if $$\alpha_0$$ is rational and $$\alpha_0 > 1$$ then the sequence must contain an integer.
The winner for this edition is Neil Jain, well done Neil! The solution is below:
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