Fractionality of a given sequence.

A recursive formula for $a_n$ is $a_n = a_{n-2} - a_{n-1},$ with $a_0 = 1, a_1 = \frac{\sqrt{5}-1}{2}$ .

Proof . Let $k \in \mathbb{Z}_{\geq 0}$ . If $k\in \{0, 1\}$ , the definition of $a_n$ already proofs the claim. Therefore, assume $k \geq 2$ . It follows that $a_{k} = \left(\frac{\sqrt{5}-1}{2}\right)^k = \left(\frac{\sqrt{5}-1}{2}\right)^{k-1} \cdot \frac{\sqrt{5}-1}{2} = \left(\frac{\sqrt{5}-1}{2}\right)^{k-1} \cdot \frac{\sqrt{5}+1-2}{2} = \frac{\sqrt{5}+1}{2} \left(\frac{\sqrt{5}-1}{2}\right)^{k-1} - \left(\frac{\sqrt{5}-1}{2}\right)^{k-1} \Leftrightarrow$ $a_k = \frac{(\sqrt{5}+1)(\sqrt{5}-1)}{4} \left(\frac{\sqrt{5}-1}{2}\right)^{k-2} - a_{k-1} = \frac{5-1}{4} a_{k-2} - a_{k-1} = a_{k-2} - a_{k-1},$ as expected.

Now note that $a_{k}$ can always be written in the form $\frac{a+b\sqrt{5}}{2}$ , where $a,b \in \mathbb{Z}$ , as $a_0$ and $a_1$ are in this form and $a_k$ is just the (positive) difference between $a_{k-2}$ and $a_{k-1}$ . In order to get rid of the fraction, both $a$ and $b$ must be even numbers.

Furthermore, it can be proven that the difference between an odd number and an even number is odd and that the difference between two odd numbers is always even (see lemma below). With this information, we can see that if $a_k = \frac{a+b\sqrt{5}}{2}$ with $a, b$ even numbers and $a_{k+1} = \frac{c+d\sqrt{5}}{2}$ with $c,d$ odd numbers, then $a_{k+2} = \frac{a-c+(b-d)\sqrt{5}}{2} = \frac{e+f\sqrt{5}}{2}$ , where $e,f$ are both odd. We can, applying the same method, also see that in this case $a_{k+3} = \frac{c-e+(d-f)\sqrt{5}}{2} = \frac{g+h\sqrt{5}}{2}$ , where $g,h$ are both even and that $a_{k+4} = \frac{g-e+(h-f)\sqrt{5}}{2} = \frac{i+j\sqrt{5}}{2}$ , where $i,j$ are both odd. Notice that the pattern will now repeat itself, as it has returned to the situation it was in the beginning. The pattern should therefore be like even, odd, odd, even, odd, odd, ... Also, notice that $a_0$ and $a_1$ have the properties which $a_k$ and $a_{k+1}$ have, respectively.

As stated earlier, to get rid of the fraction, we need to have for arbitrarily chosen $l \in \mathbb{Z}_{\geq 0}$ that $a_l = \frac{l_1 + l_2\sqrt{5}}{2}$ , where $l_1, l_2$ are both even. With the pattern even, odd, odd, even, odd, odd, ..., it should be clear that the probability of $a_k$ being fractionless after simplifying is $\boxed{\frac{1}{3}}$ .

Lemma: (1) The difference between an odd number and an even number is odd and (2) the difference between two odd numbers is even.

Proof . We first proof (1). An even number can be written as $2n$ , where $n \in \mathbb{Z}$ and an odd number can be written as $2m+1$ , where $m \in \mathbb{Z}$ . Therefore their difference can be written as $2m+1-2n = 2(m-n)+1 = 2k+1$ , where $k = m-n$ . The difference between two integers is an integer, so $k \in \mathbb{Z}$ and we conclude that the difference between an odd number and an even number is odd.

Now we proof (2). We can write the two odd numbers as $2n+1$ and $2m+1$ respectively, where $n,m \in \mathbb{Z}$ . Their difference therefore is $2m+1-(2n+1) = 2m - 2n = 2(m-n) = 2k$ , where $k = m-n$ . Again, the difference between two integers is an integer, so $k \in \mathbb{Z}$ and we conclude that the difference between two odd numbers is even.