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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.

Update : It now appears that the problem received a complete solution on math.SE.

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    $\begingroup$ I feel that 'cluster algebras' tag may be appropriate. $\endgroup$ – Fedor Petrov Aug 30 '16 at 10:07
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    $\begingroup$ When using generic $a_0,a_1,a_2,a_3$, one does not get Laurent polynomials (starting with $a_8$). $\endgroup$ – F. C. Aug 30 '16 at 13:11
  • $\begingroup$ So it is unlikely 'cluster algebras'. $\endgroup$ – Alexey Ustinov Aug 31 '16 at 7:29
  • $\begingroup$ At least a priori it could be Laurent polynomial not literally in $a_0,\dots,a_3$, but in something different, like, say, also $(a_1a_2+1)/(a_2+a_3)$ or whatever. $\endgroup$ – Fedor Petrov Sep 11 '16 at 21:58
  • $\begingroup$ @FedorPetrov they are actual polynomials in the first eight $a_{n-1}a_{n+1}/(a_n(a_n+1))$ but unfortunately for this particular sequence, those quantities are not all integers. $\endgroup$ – mercio Sep 15 '16 at 15:26

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