24
$\begingroup$

The terms of the sequence A276123, defined by $a_0=a_1=a_2=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-3}}\;,$$ are all integers (it's easy to prove that for all $n\geq2$, $a_n=\frac{9-3(-1)^n}{2}a_{n-1}-a_{n-2}-1$).

But is it also true for the sequence A276175 defined by $a_0=a_1=a_2=a_3=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)(a_{n-3}+1)}{a_{n-4}} \;\;?$$

Remark : This question has been asked previously on math.SE ; one participant gave an interesting answer, but partial.

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ I feel that 'cluster algebras' tag may be appropriate. $\endgroup$ Commented Aug 30, 2016 at 10:07
  • 2
    $\begingroup$ When using generic $a_0,a_1,a_2,a_3$, one does not get Laurent polynomials (starting with $a_8$). $\endgroup$ Commented Aug 30, 2016 at 13:11
  • $\begingroup$ So it is unlikely 'cluster algebras'. $\endgroup$ Commented Aug 31, 2016 at 7:29
  • 2
    $\begingroup$ @darij grinberg : perhaps in view of today's edit it would be helpful to indicate why the MSE proof is not complete (I note that also OEIS lists this as a proven statement). $\endgroup$ Commented Oct 24, 2020 at 20:30
  • 1
    $\begingroup$ Initially I thought all of the recurrences $a^{(k)}_n = \prod_{i=1}^{k-1} (a^{(k)}_{n-i}+1)/a^{(k)}_{n-k}$ (with initial values all $1$) might have all terms integral - where $a^{(3)}$ is oeis.org/A276123 and $a^{(4)}$ is oeis.org/A276175. But it seems that $a^{(6)}_{12}$ is non-integral. $\endgroup$ Commented Oct 25, 2020 at 20:33

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.