Adapted from a past Putnam problem

Let ( a n ) n = 1 (a_n)_{n=1}^{\infty} be a sequence of 1 1 s and 2 2 s such that a 1 = 1 a_1 = 1 and a n a_n is the number of 2 s 2s between the n t h n^{th} and ( n + 1 ) t h (n+1)^{th} 1 s 1s in the sequence. So the sequence starts out like this: 1 , 2 , 1 , 2 , 2 , 1 , 2 , 1 , 2 , 2 , 1 , 2 , 2 , 1 , 2 , 1 , . . . 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, ...

Prove (if you wish) that there exists a real number r r such that a n = 1 a_n = 1 if and only if n = 1 + r k n = 1 + \lfloor rk \rfloor for some k N k \in \mathbb{N} .

If r r can be written in simplest form as r = α + β γ r = \frac{\alpha+\sqrt{\beta}}{\gamma} where α , β , γ \alpha, \beta, \gamma are coprime natural numbers, find α 2 + β 2 + γ 2 \alpha^2+\beta^2+\gamma^2 .


The answer is 38.

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1 solution

Wesley Low
Feb 23, 2021

This sequence is a typical rabbit sequence, or equivalently a Fibonacci word. One way to construct this sequence is to draw a line y = k x y=kx on the square grid where k k is irrational and mark a 0 0 where the line crosses a horizontal and a 1 1 at a vertical. This method is known as a cutting sequence. For this sequence, k = ϕ = 1 + 5 2 k=\phi=\dfrac{1+\sqrt{5}}{2} .

Given that the nth 1 1 must be on the line x = n 1 x=n-1 , we count the number of "cuts" the line makes with the square grid.

( n + 1 ) + n ϕ \left(n+1\right)+\lfloor n\phi\rfloor = 1 + n ( ϕ + 1 ) =1+\lfloor n\left(\phi+1\right)\rfloor = 1 + 3 + 5 2 n =1+\lfloor\dfrac{3+\sqrt{5}}{2}n\rfloor

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