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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.

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    $\begingroup$ I feel that 'cluster algebras' tag may be appropriate. $\endgroup$
    – Fedor Petrov
    Commented Aug 30, 2016 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.
    Commented Aug 30, 2016 at 13:11
  • $\begingroup$ So it is unlikely 'cluster algebras'. $\endgroup$
    – Alexey Ustinov
    Commented Aug 31, 2016 at 7:29
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    $\begingroup$ @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). $\endgroup$
    – Carlo Beenakker
    Commented Oct 24, 2020 at 20:30
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    $\begingroup$ 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. $\endgroup$
    – Martin Rubey
    Commented Oct 25, 2020 at 20:33

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