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Are there any sequence given by a recurrence relation: $x_{n+t}=P(x_t,\cdots,x_{t+n-1})$, where $P$ is a positive Laurent Polynomial, satisfy:

  • if $x_0=\cdots=x_{n-1}=1$, then the sequence is only integer;

  • but does not exhibit Laurent Phenomenon ?

What if we allow $P$ to be a rational function?

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    $\begingroup$ Great question! I assume you want $P$ to have positive coefficients, since otherwise $x_n = \dfrac{x_{n-1} + x_{n-2} - x_{n-3}}{x_{n-4}}$ is a counterexample (indeed, the general formula for $x_8$ is not a Laurent polynomial, but if $x_0 = x_1 = x_2 = x_3 = 1$, then all values $x_i$ equal $1$). $\endgroup$
    – darij grinberg
    Commented Feb 22, 2019 at 20:07
  • $\begingroup$ @darij grinberg, thanks, I forgot the positivity $\endgroup$
    – Sylvester W. Zhang
    Commented Feb 22, 2019 at 20:26

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Consider $x_{n+3} = \frac{x_n+x_{n+1}}{x_{n+2}}$.

With $x_1=x_2=x_3 =1$ this gives the sequence $1,1,1,2,1,3,1,4,1,5,...$ but it certainly is not Laurent because for instance we have $x_5=\frac{x_2x_3+x_3^2}{x_1+x_2}$.

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  • $\begingroup$ Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right? $\endgroup$
    – darij grinberg
    Commented Feb 22, 2019 at 20:35
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    $\begingroup$ Nice counterexample!! $\endgroup$
    – darij grinberg
    Commented Feb 22, 2019 at 20:37
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This reminds me of a question I had seen on both MO and MSE. Sequence A276175 in the OEIS is defined by $$a_n = \frac{(a_{n-1} + 1)(a_{n-2}+1)(a_{n-3} + 1)}{a_{n-4}}$$ with $a_0 = a_1 = a_2 = a_3 = 1$. The OEIS page conjectures it to be an integer for all $n$. The MSE question contains a proof the all $a_n$ are integer (though I haven't read the proof). In the comments of the MO question it is observed $a_8$ is not Laurent.

Added in edit: I offer another example which does not exhibit the Laurent phenomenon, but conjecturally is an integer sequence. 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 $b_0 = b_1 = b_2 = b_3 = b_4 = 1$. I computed and after reducing I found the denominator of $b_{10}$ to be $b_0^14(b_1 + 1)b_1^8(b_2 + 1)b_2^4b_3^2b_4$ (not a monomial). I verified this sequence to be integer up to $n=36$, and Kevin O'Bryant later verified up to $n=41$. I asked if the sequence is integer for all $n \geq 0$ in a separate question.

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    $\begingroup$ Great example! Suggests no general result here... $\endgroup$
    – Sam Hopkins
    Commented Feb 22, 2019 at 21:50

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