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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Discografie di artisti tedeschi Voce principale: Klaus Schulze . Discografia di Klaus Schulze Klaus Schulze e Lisa Gerrard durante un concerto Album in studio 56 ↙ Album dal vivo 15 ↙ Raccolte 37 ↙ Singoli 4 ↙ Album video 1 ↙ Colonne sonore 4 ↙ Indice 1 Album in studio 1.1 Solisti 1.1.1 Come Richard Wahnfried 1.2 Collaborazioni 1.2.1 Con Rainer Bloss 1.2.2 Con Rainer Bloss ed Ernst Fuchs 1.2.3 Con Lisa Gerrard 1.2.4 Con Andreas Grosser 1.2.5 Con Pete Namlook 1.2.6 Con Pete Namlook e Bill Laswell 1.2.7 Con Michael Shrieve 1.2.8 Con Solar Moon 1.2.9 Con artisti vari 2 Album dal vivo 3 Singoli 4 Antologie 5 Colonne sonore 6 Videografia 7 Collegamenti musica Album in studio | Solisti | 1972 – Irrlicht 1973 – Cyborg 1974 – Blackdance 1975 – Timewind 1975 – Picture Music 1976 – Moondawn 1977 – Mirage 1978 – "X" 1979 – Dune
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How to make inet_server_addr() return localhost in spite of ::1/128RETURN NEXT in Postgres FunctionConnect to...

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Potential client has a problematic employee I can't work with Can the "Friends" spell be used without making the target hostile? Does it take energy to move something in a circle? Why are carbons of Inositol chiral centers? Count repetitions of an array Which RAF squadrons and aircraft types took part in the bombing of Berlin on the 25th of August 1940? Memory usage: #define vs. static const for uint8_t Broad Strokes - missing letter riddle Why does 0.-5 evaluate to -5? Why is 'diphthong' pronounced the way it is? Why did Luke use his left hand to shoot?
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