A sequence potentially consisting of only integers

Question

I will first ask the question which can be stated very simply. Afterwards I will explain some motivation and give references to related sequences.

Consider the sequence defined by $$b_n = \frac{(b_{n-1} + 1)(b_{n-2} + 1)(b_{n-3} + 1)(b_{n-4}+1)}{b_{n-5}}$$ where $b_0 = b_1 = b_2 = b_3 = b_4 = 1$.

Question: Is $b_n \in \mathbb{Z}$ for all $n \geq 0$?

Provided my program is correct, I have found that $b_n \in \mathbb{Z}$ for $0 \leq n \leq 36$.

I came to this thinking about MO323867 which asks for sequences that have all integer values but do not satisfy the Laurent phenomenon. More generally we can consider the sequence defined by

$$a^{(k)}_n = \frac{\prod_{i=1}^{k-1} (a^{(k)}_{n-i} + 1)}{a^{(k)}_{n-k}}$$

with $a_0 = a_1 = \cdots = a_{k-1} = 1$.

• When $k=2$ we obtain $a^{(2)}_n a^{(2)}_{n-2} = a^{(2)}_{n-1} + 1$ which can be seen coming from a type $A_2$ cluster algebra which implies the Laurent phenomenon and integrality (see A076839).

• When $k=3$ we have the Laurent phenomenon and hence the sequence is all integers (see A276123). The Laurent phenomenon follows for a more general result in Theorem 6.2.3 of the thesis of Matthew Russell.

• When $k = 4$ the sequence does not exhibit the Laurent phenomenon, but it is still a sequence of integers (see A276175, MO248604, MSE1905063).

• When $k = 5$ we have the sequence from the question asked above.

• When $k > 5$ the sequence does not consist of only integers with the first violation being $a^{(k)}_{2k} \not\in \mathbb{Z}$. This can be seen by looking at the 2-adic valuation. In particular, it can be shown $\nu_2(a^{(k)}_{2k}) = 5-k$ for $k \geq 5$. To see this we can work $\pmod{2^5}$ and find that $(a^{(k)}_{2k-1} + 1) \equiv 2^4 \pmod{2^5}$. It is clear that $a_k = 2^{k-1}$ and $a_{n}$ is even, equivalently $(a_{n}+1)$ is odd, for $k \leq n < 2k - 1$.

So, the sequence $b_n$ corresponding to $k=5$ is the only sequence in the family which I do not know if all terms are integers.

Answer 1

This is more of a comment on your sequences than an answer to your question; unfortunately I do not have enough points to post a comment yet.

The sequences $(a_n^{(k)})_{n=0}^\infty$ labelled by $k=2, 3, \ldots $ that you consider all exhibit a "hidden" positive Laurent Phenomenon, i.e. a positive Laurent Phenomenon after a suitable change of variables.

More precisely, set $a_0^{(k)} = y_0$ and $a_j^{(k)} = y_j \prod_{i=0}^{j-1}(1+a_i^{(k)})$ for $j=1, \ldots, k-1$.

Claim:([Lemma 2.13, dSG23]) For all $n$, the element $a_n^{(k)}$ is a Laurent polynomial, with positive integer coefficients, in the variables $y_0, y_1, \ldots, y_{k-1}$.

In particular, the sequence $(a_n^{k})$ evaluated at $y_0 = \ldots = y_{k-1}=1$, or equivalently at $a_0^{(k)}=1$ and $a_j^{(k)} = (a_{j-1}^{(k)})^2+a_{j-1}^{(k)}$ for $j = 1, \ldots, k-1$, consists entirely of positive integers. Such a sequence is called a $\textit{unitary Y-frieze pattern}$ in [dSG23].

Example: For $k=4$, we write $a_k^{(4)} = a_n$ for convenience. Set $a_0 = y_0$, $a_1 = y_1(1+y_0)$, $a_2 = y_2(1+y_0)(1+y_1(1+y_0))$ and $a_3 = y_3(1+y_0)(1+y_1(1+y_0))(1+y_2(1+y_0)(1+y_1(1+y_0)))$. The claim is that every $a_n$ is a Laurent polynomial in $y_0, y_1, y_2$ and $y_3$, with positive integer coefficients.

In particular, specialising $y_0 = \cdots= y_3 = 1$, or equivalently $a_0=1$, $a_1=2, a_2=6, a_3=42$, we obtain a sequence of positive integers.

References

[dSG23]: Y-frieze patterns, A. de Saint Germain, arxiv:2311.03073

Answer 2

I think I can at least show that the only denominators are powers of two.

We have the following recurrence: $$b_n=\frac{b_{n-1}(b_{n-1}+1)}{2}\frac{b_{n-6}b_{n-10}b_{n-14}\dots}{b_{n-4}b_{n-8}b_{n-12}\dots}$$ where $b_i$ is interpreted as $1$ if $i\leqslant 0$ so that the products on top and bottom have a finite number of terms that aren't equal to $1$.

Now suppose that $p$ is an odd prime, and let $n(p)$ be the first value of $n$ such that $\nu=\nu_p(b_n)\ne 0$. Then $p$ is a prime divisor of $b_{n(p)-1}+1$, and from $n(p)$ onwards the values of $\nu_p(n)$ go as follows: $\nu,\ \nu,\ \nu,\ \nu,\ 0,\ 0,\ 0,\ 0,\ 0,\ \dots$ until a further $\nu_p(b_{n-1}+1)$ is non-zero. But this can only happen if $\nu_p(b_{n-1})$ is zero, at which point another sequence of the same form is added in, and so on. The point is that this shows $\nu_p(b_n)$ always stays non-negative.

Finally, we need to deal with $\nu_2$. This seems more mysterious to me. The problem is to understand $\nu_2(b_n+1)$ when $b_n$ is odd.