1) Note that we have \(f(1,0) = f(n,0) = 1\) by definition. Moreover, \(f\) is arithmetic and so the first equation in Lemma 1.2 is precisely the divisibility condition defining a flute.

2) By arguing recursively, one can construct a pattern \(f\) such that \( (f (1,0), \ldots , f (n,0)) = (a_1, \ldots , a_n)\). Moreover, such a frieze pattern is necessarily positive and \(\mathbb {Q}\)-valued. It remains to show that \(f\) is integer-valued. We begin by showing that \(f (2,m) \in \mathbb {Z}\) for all \(m \in \mathbb {Z}\). By the definition of a flute, \(f (2,0) \in \mathbb {Z}\), and for each \(i \in \{ 1,\ldots , n-2\} \), there exists a positive integer \(c_i\) such that

Using the first equation in Lemma 1.2, we deduce that \(f(2,i) \in \mathbb {Z}\) for \(i =0, \ldots , n-2\). Moreover, \( f (2,n-1) = f (n-1,0) \in \mathbb {Z}\) by assumption. Thus we have proved that \(f (2,m) \in \mathbb {Z}\) for \(m = 0, \ldots , n-1\). To see that \(f (2,n) \in \mathbb {Z}\), note from Lemma 1.2 that \(f (2,n) = f (n-1,1)\) can be expressed as a polynomial with integer coefficients in the variables \(f (2,1), f (2,2),\ldots , f (2,n-2)\). The claim for all \(m\) then follows from Proposition 1.3.

To see how this implies that \(f (i,m) \in \mathbb {Z}\) for all \(i \in \{ 2, \ldots , n\} \), it suffices to see, again from Lemma 1.2, that every \(f (i,m)\) can be expressed as a polynomial with integer coefficients in the variables \(f (2,m), f (2,m+1),\ldots , f (2,m+i-2)\).

The proof of Lemma 2.2 showed that \((1,1,\ldots , 1)\) is a flute. The claim then follows from 2) of Proposition 3.2.

By Proposition 3.2, every entry of an arithmetic frieze pattern of height \(n\) belongs to a flute of height \(n\). By Proposition 2.5, entries in a flute of height \(n\) are bounded above by \(F_n\). Thus \(u_n \leq F_n\) for all \(n\). On the other hand, Lemma 2.3 and 2) of Proposition 3.2 show that there exists an arithmetic frieze pattern of height \(n\) containing \(F_n\) as a value.