Integer but not Laurent sequences












7












$begingroup$


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?






ra.rings-and-algebras sequences-and-series cluster-algebras laurent-polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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
    7 hours ago












  • $begingroup$
    @darij grinberg, thanks, I forgot the positivity
    $endgroup$
    – Wenze 'Sylvester' Zhang
    6 hours ago


















7












$begingroup$


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?






ra.rings-and-algebras sequences-and-series cluster-algebras laurent-polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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
    7 hours ago












  • $begingroup$
    @darij grinberg, thanks, I forgot the positivity
    $endgroup$
    – Wenze 'Sylvester' Zhang
    6 hours ago
















7












7








7


3



$begingroup$


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?






ra.rings-and-algebras sequences-and-series cluster-algebras laurent-polynomials







share|cite|improve this question











$endgroup$




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?






ra.rings-and-algebras sequences-and-series cluster-algebras laurent-polynomials




ra.rings-and-algebras sequences-and-series cluster-algebras laurent-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









YCor

27.5k481134




27.5k481134










asked 7 hours ago









Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

1006




1006








  • 2




    $begingroup$
    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
    7 hours ago












  • $begingroup$
    @darij grinberg, thanks, I forgot the positivity
    $endgroup$
    – Wenze 'Sylvester' Zhang
    6 hours ago
















  • 2




    $begingroup$
    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
    7 hours ago












  • $begingroup$
    @darij grinberg, thanks, I forgot the positivity
    $endgroup$
    – Wenze 'Sylvester' Zhang
    6 hours ago










2




2




$begingroup$
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
7 hours ago






$begingroup$
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
7 hours ago














$begingroup$
@darij grinberg, thanks, I forgot the positivity
$endgroup$
– Wenze 'Sylvester' Zhang
6 hours ago






$begingroup$
@darij grinberg, thanks, I forgot the positivity
$endgroup$
– Wenze 'Sylvester' Zhang
6 hours ago












2 Answers
2






active

oldest

votes


















5












$begingroup$

I think you're gonna want to add the assumption that $x_{n}$ grows unboundedly. Even with the positive coefficient condition that Darij suggested, you have examples like $x_4=frac{x_1+x_2}{2x_3}$ that give the constant sequence $1,1,1,ldots$.



(By the way, in class Pasha did suggest that in all "known" or "studied" examples of integrability/non-integrability it suffices to check what happens when you plug in all $1$'s for the initial variables.)



EDIT:



Actually, how about $x_4 = frac{x_1+x_2}{x_3}$?



To me 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 $x_5=frac{x_2x_3+x_3^2}{x_1+x_2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    @darijgrinberg: depends on your definition, I guess, but anyways see my edit.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
    $endgroup$
    – Sam Hopkins
    6 hours ago



















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great example! Suggests no general result here...
    $endgroup$
    – Sam Hopkins
    5 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

I think you're gonna want to add the assumption that $x_{n}$ grows unboundedly. Even with the positive coefficient condition that Darij suggested, you have examples like $x_4=frac{x_1+x_2}{2x_3}$ that give the constant sequence $1,1,1,ldots$.



(By the way, in class Pasha did suggest that in all "known" or "studied" examples of integrability/non-integrability it suffices to check what happens when you plug in all $1$'s for the initial variables.)



EDIT:



Actually, how about $x_4 = frac{x_1+x_2}{x_3}$?



To me 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 $x_5=frac{x_2x_3+x_3^2}{x_1+x_2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    @darijgrinberg: depends on your definition, I guess, but anyways see my edit.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
    $endgroup$
    – Sam Hopkins
    6 hours ago
















5












$begingroup$

I think you're gonna want to add the assumption that $x_{n}$ grows unboundedly. Even with the positive coefficient condition that Darij suggested, you have examples like $x_4=frac{x_1+x_2}{2x_3}$ that give the constant sequence $1,1,1,ldots$.



(By the way, in class Pasha did suggest that in all "known" or "studied" examples of integrability/non-integrability it suffices to check what happens when you plug in all $1$'s for the initial variables.)



EDIT:



Actually, how about $x_4 = frac{x_1+x_2}{x_3}$?



To me 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 $x_5=frac{x_2x_3+x_3^2}{x_1+x_2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    @darijgrinberg: depends on your definition, I guess, but anyways see my edit.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
    $endgroup$
    – Sam Hopkins
    6 hours ago














5












5








5





$begingroup$

I think you're gonna want to add the assumption that $x_{n}$ grows unboundedly. Even with the positive coefficient condition that Darij suggested, you have examples like $x_4=frac{x_1+x_2}{2x_3}$ that give the constant sequence $1,1,1,ldots$.



(By the way, in class Pasha did suggest that in all "known" or "studied" examples of integrability/non-integrability it suffices to check what happens when you plug in all $1$'s for the initial variables.)



EDIT:



Actually, how about $x_4 = frac{x_1+x_2}{x_3}$?



To me 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 $x_5=frac{x_2x_3+x_3^2}{x_1+x_2}$.






share|cite|improve this answer











$endgroup$



I think you're gonna want to add the assumption that $x_{n}$ grows unboundedly. Even with the positive coefficient condition that Darij suggested, you have examples like $x_4=frac{x_1+x_2}{2x_3}$ that give the constant sequence $1,1,1,ldots$.



(By the way, in class Pasha did suggest that in all "known" or "studied" examples of integrability/non-integrability it suffices to check what happens when you plug in all $1$'s for the initial variables.)



EDIT:



Actually, how about $x_4 = frac{x_1+x_2}{x_3}$?



To me 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 $x_5=frac{x_2x_3+x_3^2}{x_1+x_2}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 6 hours ago

























answered 6 hours ago









Sam HopkinsSam Hopkins

4,53012455




4,53012455












  • $begingroup$
    Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    @darijgrinberg: depends on your definition, I guess, but anyways see my edit.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
    $endgroup$
    – Sam Hopkins
    6 hours ago


















  • $begingroup$
    Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    @darijgrinberg: depends on your definition, I guess, but anyways see my edit.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    6 hours ago










  • $begingroup$
    You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
    $endgroup$
    – Sam Hopkins
    6 hours ago










  • $begingroup$
    For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
    $endgroup$
    – Sam Hopkins
    6 hours ago
















$begingroup$
Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
$endgroup$
– darij grinberg
6 hours ago




$begingroup$
Uhm, I wouldn't allow a $2$ in the denominator for a Laurent polynomial -- we're talking integers, right?
$endgroup$
– darij grinberg
6 hours ago












$begingroup$
@darijgrinberg: depends on your definition, I guess, but anyways see my edit.
$endgroup$
– Sam Hopkins
6 hours ago




$begingroup$
@darijgrinberg: depends on your definition, I guess, but anyways see my edit.
$endgroup$
– Sam Hopkins
6 hours ago












$begingroup$
Nice counterexample!!
$endgroup$
– darij grinberg
6 hours ago




$begingroup$
Nice counterexample!!
$endgroup$
– darij grinberg
6 hours ago












$begingroup$
You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
$endgroup$
– Sam Hopkins
6 hours ago




$begingroup$
You can still ask for "interesting" examples where starting with all $1$'s gives you integers forever but it doe not exhibit the Laurent phenomenon.
$endgroup$
– Sam Hopkins
6 hours ago












$begingroup$
For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
$endgroup$
– Sam Hopkins
6 hours ago




$begingroup$
For instance, you can ask that for every $M$, eventually we have all $x_n > M$.
$endgroup$
– Sam Hopkins
6 hours ago











3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great example! Suggests no general result here...
    $endgroup$
    – Sam Hopkins
    5 hours ago
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great example! Suggests no general result here...
    $endgroup$
    – Sam Hopkins
    5 hours ago














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









John MachacekJohn Machacek

4,44211028




4,44211028












  • $begingroup$
    Great example! Suggests no general result here...
    $endgroup$
    – Sam Hopkins
    5 hours ago


















  • $begingroup$
    Great example! Suggests no general result here...
    $endgroup$
    – Sam Hopkins
    5 hours ago
















$begingroup$
Great example! Suggests no general result here...
$endgroup$
– Sam Hopkins
5 hours ago




$begingroup$
Great example! Suggests no general result here...
$endgroup$
– Sam Hopkins
5 hours ago


















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