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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.

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.

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.

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.