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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19世紀

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この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
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