The sequence a_{n} is defined by a_{1} = 2, a_{n+1} = a_{n}^{2} - a_{n} + 1. Show that any pair of values in the sequence are relatively prime and that ∑ 1/a_{n} = 1.

**Solution**

We show by induction on k that a_{n+k} = 1 (mod a_{n}). Obviously true for k = 1. Suppose it is true for k. Then for some m, a_{n+k} = m a_{n} + 1. Hence a_{n+k+1} = a_{n+k}(m a_{n} + 1 - 1) + 1 = a_{n+k}m a_{n} + 1 = 1 (mod a_{n}). So the result is true for all k. Hence any pair of distinct a_{n} are relatively prime.

We show by induction that ∑_{1}^{n} 1/a_{r} = 1 - 1/(a_{n+1} - 1). For n = 2, this reduces to 1/2 = 1 - 1/(3 - 1), which is true. Suppose it is true for n. Then ∑_{1}^{n+1} 1/a_{r} = 1 - k, where k = 1/(a_{n+1} - 1) - 1/a_{n+1} = 1/(a_{n+1}^{2} - a_{n+1}) = 1/(a_{n+2} - 1), so it is true for n + 1. But a_{n} → ∞, so ∑ 1/a_{n} = 1.

© John Scholes

jscholes@kalva.demon.co.uk

4 Dec 1999