Integer but not Laurent sequencesIs A276175 integer-only?Rational powers of ideals in Noetherian ringsCan an...



Integer but not Laurent sequences


Is A276175 integer-only?Rational powers of ideals in Noetherian ringsCan an integer or rational sequence satisfy some bounded order recurrence $mod $ almost all primes but doesn't satisfy such in $mathbb{Q}$?Maximal Ideals in Formal Laurent Series Rings?Closed-form for modified formal power seriesDo Denominator Vectors Determine the Cluster VariableRegularizing the divergent sum $1^k + 2^k + cdots$``Occasional'' Laurent phenomenonOn one class of Somos-like sequencesInteger Gelfand-Kirillov dimensionA dense orbit of the logistic map













4












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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
    2 hours ago












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


















4












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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
    2 hours ago












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
















4












4








4


2



$begingroup$


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 1 hour ago









YCor

27.5k481134




27.5k481134










asked 2 hours ago









Wenze 'Sylvester' ZhangWenze 'Sylvester' Zhang

856




856








  • 1




    $begingroup$
    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
    2 hours ago












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
















  • 1




    $begingroup$
    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
    2 hours ago












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










1




1




$begingroup$
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
2 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
2 hours ago














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






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












2 Answers
2






active

oldest

votes


















3












$begingroup$

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
    1 hour ago










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










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    1 hour 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
    1 hour ago










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



















2












$begingroup$

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
    22 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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
    1 hour ago










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










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    1 hour 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
    1 hour ago










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
















3












$begingroup$

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
    1 hour ago










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










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    1 hour 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
    1 hour ago










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














3












3








3





$begingroup$

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 1 hour ago

























answered 1 hour ago









Sam HopkinsSam Hopkins

4,51012455




4,51012455












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










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










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    1 hour 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
    1 hour ago










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


















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










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










  • $begingroup$
    Nice counterexample!!
    $endgroup$
    – darij grinberg
    1 hour 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
    1 hour ago










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
















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




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












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




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












$begingroup$
Nice counterexample!!
$endgroup$
– darij grinberg
1 hour ago




$begingroup$
Nice counterexample!!
$endgroup$
– darij grinberg
1 hour 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
1 hour 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
1 hour ago












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




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











2












$begingroup$

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
    22 mins ago
















2












$begingroup$

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
    22 mins ago














2












2








2





$begingroup$

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 29 mins ago









John MachacekJohn Machacek

4,43211028




4,43211028












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


















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
















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




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


















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Armoriali delle famiglie italiane Blasone Cesenate Famiglie nobili di SiciliaFamiglie nobili di SiciliaTropea MagazineTropea MagazineNobili napoletani Nobili napoletani[1]Nobili napoletaniNobili napoletaniTropea Magazine Nobili napoletaniNobili napoletani Famiglie nobili di SiciliaBlasonario CremoneseFamiglie nobili di Sicilia Archivio Storico CapitolinoRaccolta stemmi TrippiniFamiglie nobili di SiciliaNobili napoletaniArmas de los cavalleros de Veneçia Tropea Magazine Famiglie nobili di Sicilia Nobili napoletaniFamiglie nobili di SiciliaRaccolta stemmi Topoguz Famiglie nobili di SiciliaBlasone CesenateRaccolta stemmi TrippiniArchivio Storico Capitolino Famiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Famiglie nobili di SiciliaFamiglie nobili di SiciliaBlasone Cesenate Raccolta stemmi TrippiniFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Blasonario CremoneseBlasonario Cr...
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Lupi Siderali Indice Storia | Organizzazione | La Tredicesima Compagnia | Aspetto | Membri Importanti...

Capitoli degli Space Marine Space MarineImperium della Terrawargame tridimensionaleWarhammer 40.000PrimarcalupiuroImperatoreRune di DissimulazioneAngeli OscuriEresia di HorusStirpe dei MilleOcchio del TerroreCaosTzeentchHorusCodex AstartesSeconda FondazioneFratelli del LupoOcchio del TerroreAlbero della VitareliquieGuardia ImperialeLogan Grimnartruppe d'assaltoImperium dell'UomoveteraniStirpe dei MilleCaosOcchio del TerroreAngeli SanguinariEresia di Horus Lupi Siderali Nome originale Space Wolves Primarca Leman Russ Grido di battaglia Per Russ e per il Padre di Tutti (un feroce ululato) Colori Grigio Altri capitoli I Lupi Siderali sono una delle venti Legioni della Prima Fondazione degli Space Marine che servono l'Imperium della Terra nel wargame tridimensionale Warhammer 40.000. Sono formidabili nel corpo a corpo, i Confratelli desiderano solo la gloria sul campo di battaglia. Indice 1 Storia 1.1 Leman Russ 1.2 La ...
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