Let a_{0}, a_{1}, a_{2}, ... be a sequence such that: a_{0} = 2; each a_{n} = 2 or 3; a_{n} = the number of 3s between the nth and n+1th 2 in the sequence. So the sequence starts: 233233323332332 ... . Show that we can find α such that a_{n} = 2 iff n = [αm] for some integer m ≥ 0.

**Solution**

Let b_{n} be the index of the nth 2, so b_{0} = 0, b_{1} = 3, b_{2} = 7, b_{3} = 11, b_{4} = 14, ... . Then we are told that b_{n} = (a_{0} + 1) + (a_{1} + 1) + ... + (a_{n-1} + 1). Now each a_{i} = 2 or 3, so b_{n} = 4n - (number of b_{m}s in 0, 1, 2, ... , n-1).

Suppose that b_{n} = [nα] as claimed. Then the number of b_{m}s in 0, 1, ... , n-1 is the number of [mα] < n, which is the same as the number of (mα) < n (since α is irrational). And that is just 1 + [n/α]. So we have [nα] = 4n - 1 - [n/α]. But 4n - [n/α] = [4n - n/α] + 1 (since n/α is irrational), so [nα] = [4n - n/α] = [nβ], where β = 4 - 1/α. But this holds for all n, so we must have β = α. Solving the quadratic gives α = 2 ± √3. But clearly α > 3, so α = 2 + √3.

We are not done yet. All we have done is shown that *if* it is true that b_{n} = [nα] for some α, then α must be 2 + √3. But we can work the previous argument in reverse. Suppose we define d_{n} = [n(2 + √3)]. Then the argument shows that d_{n} = 4n - (number of d_{m}s in 0, 1, 2, ... , n-1) and that is sufficient to identify d_{n} and b_{n}.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001