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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.

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  • 3
    $\begingroup$ The obvious Mathematica code (and an hour on my laptop) confirmed integrality up to $n=41$. $\endgroup$ – Kevin O'Bryant Feb 24 '19 at 4:37
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    $\begingroup$ These numbers get big fast! I noticed that the last digits seem to be always be 1 or 6. Moreover, after some initial "noise," the repeating pattern is 6, 6, 6, 6, 1, 1, 1 (starting at $n=11$). Looking modulo 30, from the same point the repeating pattern is 16, 16, 16, 16, 1, 1, 1. $\endgroup$ – Brian Hopkins Feb 24 '19 at 19:03
  • $\begingroup$ I can't find a proof that for $k\ge5$, we have $(a^{(k)}_{2k-1} + 1) \equiv 2^4 \pmod{2^5}$. Would you give more hints? thanks! $\endgroup$ – René Gy Feb 25 '19 at 19:37
  • $\begingroup$ Kevin, thanks for the computations. It's always nice to have additional verification. Brian, thanks for the interesting observation. @RenéGy for k=5 one can computer mod 32 and find 1, 1, 1, 1, 1, 16, 8, 4, 26, 15. For k>5 it will be an initial segment of 1s, followed by some 0s, followed by 16, 8, 4, 26, 15. Note the 0s don't really change things since we multiply by 0+1=1 in the recurrence. $\endgroup$ – John Machacek Feb 26 '19 at 15:54
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    $\begingroup$ I'm not 100% convinced the $k=4$ case is really resolved. $\endgroup$ – Sam Hopkins Feb 27 '19 at 16:24

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