Can you make an identity from this product?Lattice walks and a q-series curiosityCombinatorial Interpretation of an Extension of Gaussian Polynomialsverifying a combinatorial inequalityA Product Related to Unrestricted PartitionsElementary symmetric functions of reciprocals of monic polynomials in function fieldsAnother question related to the generating function for unrestricted partitionsA simplified version of an old problem about the generating function for unrestricted partitionsYet another question about unrestricted partitionsPartition the rationals with respect to a multivariate polynomial which sends classes to classesDirichlet series associated with polynomials
Can you make an identity from this product?
Lattice walks and a q-series curiosityCombinatorial Interpretation of an Extension of Gaussian Polynomialsverifying a combinatorial inequalityA Product Related to Unrestricted PartitionsElementary symmetric functions of reciprocals of monic polynomials in function fieldsAnother question related to the generating function for unrestricted partitionsA simplified version of an old problem about the generating function for unrestricted partitionsYet another question about unrestricted partitionsPartition the rationals with respect to a multivariate polynomial which sends classes to classesDirichlet series associated with polynomials
5
$begingroup$
Start with the product
$$(1+x+x^2) (1+x^2)(1+x^3)(1+x^4)cdots$$
(The first polynomial is a trinomial..The others are binomials..)
Is it possible by changing some of the signs to get a series all of whose coefficients are $ -1,0,$or $1$?
A simple computer search should suffice to answer the question if the answer is "no." I haven't yet done such a search myself.
This question is a takeoff on the well known partition identities like:
$$prod_n=1^infty (1-x^n)= 1-x-x^2+x^5+x^7-ldots$$
nt.number-theory co.combinatorics
edited 1 hour ago
kodlu
4,34022032
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
$endgroup$
|
show 2 more comments
5
$begingroup$
Start with the product
$$(1+x+x^2) (1+x^2)(1+x^3)(1+x^4)cdots$$
(The first polynomial is a trinomial..The others are binomials..)
Is it possible by changing some of the signs to get a series all of whose coefficients are $ -1,0,$or $1$?
A simple computer search should suffice to answer the question if the answer is "no." I haven't yet done such a search myself.
This question is a takeoff on the well known partition identities like:
$$prod_n=1^infty (1-x^n)= 1-x-x^2+x^5+x^7-ldots$$
nt.number-theory co.combinatorics
edited 1 hour ago
kodlu
4,34022032
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
$endgroup$
$begingroup$
The dots at the end lead me to think that the product goes on with an arbitrary number of terms, say $n$. But then you mention a "series", which is infinite; I interpret "coefficient" as the coefficient $a_i$ of a power series, $sum a_i x^i$. However, adding a term to the product, i.e. taking $n+1$, changes all the coefficients $a_i$. Can you please clarify? Also, the last sentence is unclear. Have you already performed the "computer search"? If so, what was the result? Or you are suggesting a method to the other members to answer your question?
$endgroup$
– Doriano Brogioli
8 hours ago
$begingroup$
@DorianoBrogioli The given product $(1+x+x^2)prod_n=2^infty(1+x^n)$ converges in the power series ring $mathbb Z[[x]]$, as does the product if one replaces any subset of the plus signs with minus signs. And since the $(1+x^n)$ terms with $nge N$ only affect the series coefficients starting with $x^N$, one could check all signs for, say, $N=4$, and if they fail to have $-1,0,1$ coefficients in one of their $x^i$ coefficients with $ile3$, then the answer is "No". I think that's what the OP means by "a simple computer search."
$endgroup$
– Joe Silverman
7 hours ago
$begingroup$
With $1-x-x^2$, up to N=7 the signs are -1, 0 and 1...
$endgroup$
– Doriano Brogioli
7 hours ago
1
$begingroup$
@DorianoBrogioli Nice, but $(1-x-x^2)prod_n=2^8 (1+x^n)=1 - x - x^4 - x^6 - x^7 - 2*x^8+O(x^9)$, so the $x^8$ term fails if you just change $1+x+x^2$ to $1-x-x^2$. In any case, you've shown that one needs to at least check up to $N=8$, which is quite a few possible sign choices, although certainly feasible.
$endgroup$
– Joe Silverman
6 hours ago
$begingroup$
Maybe: with $1+x+x^2$, changing the signs inside the binomianal: $1-x^2$, $1+x^3$, $1+x^4$, $1-x^5$ and so on. Just to check that I understood correctly.
$endgroup$
– Doriano Brogioli
5 hours ago
|
show 2 more comments
5
5
5
$begingroup$
Start with the product
$$(1+x+x^2) (1+x^2)(1+x^3)(1+x^4)cdots$$
(The first polynomial is a trinomial..The others are binomials..)
Is it possible by changing some of the signs to get a series all of whose coefficients are $ -1,0,$or $1$?
A simple computer search should suffice to answer the question if the answer is "no." I haven't yet done such a search myself.
This question is a takeoff on the well known partition identities like:
$$prod_n=1^infty (1-x^n)= 1-x-x^2+x^5+x^7-ldots$$
nt.number-theory co.combinatorics
edited 1 hour ago
kodlu
4,34022032
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
$endgroup$
Start with the product
$$(1+x+x^2) (1+x^2)(1+x^3)(1+x^4)cdots$$
(The first polynomial is a trinomial..The others are binomials..)
Is it possible by changing some of the signs to get a series all of whose coefficients are $ -1,0,$or $1$?
A simple computer search should suffice to answer the question if the answer is "no." I haven't yet done such a search myself.
This question is a takeoff on the well known partition identities like:
$$prod_n=1^infty (1-x^n)= 1-x-x^2+x^5+x^7-ldots$$
nt.number-theory co.combinatorics
nt.number-theory co.combinatorics
edited 1 hour ago
kodlu
4,34022032
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
edited 1 hour ago
kodlu
4,34022032
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
edited 1 hour ago
kodlu
4,34022032
edited 1 hour ago
kodlu
4,34022032
edited 1 hour ago
kodlu
4,34022032
4,34022032
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
asked 9 hours ago
David S. NewmanDavid S. Newman
970520
970520
$begingroup$
The dots at the end lead me to think that the product goes on with an arbitrary number of terms, say $n$. But then you mention a "series", which is infinite; I interpret "coefficient" as the coefficient $a_i$ of a power series, $sum a_i x^i$. However, adding a term to the product, i.e. taking $n+1$, changes all the coefficients $a_i$. Can you please clarify? Also, the last sentence is unclear. Have you already performed the "computer search"? If so, what was the result? Or you are suggesting a method to the other members to answer your question?
$endgroup$
– Doriano Brogioli
8 hours ago
$begingroup$
@DorianoBrogioli The given product $(1+x+x^2)prod_n=2^infty(1+x^n)$ converges in the power series ring $mathbb Z[[x]]$, as does the product if one replaces any subset of the plus signs with minus signs. And since the $(1+x^n)$ terms with $nge N$ only affect the series coefficients starting with $x^N$, one could check all signs for, say, $N=4$, and if they fail to have $-1,0,1$ coefficients in one of their $x^i$ coefficients with $ile3$, then the answer is "No". I think that's what the OP means by "a simple computer search."
$endgroup$
– Joe Silverman
7 hours ago
$begingroup$
With $1-x-x^2$, up to N=7 the signs are -1, 0 and 1...
$endgroup$
– Doriano Brogioli
7 hours ago
1
$begingroup$
@DorianoBrogioli Nice, but $(1-x-x^2)prod_n=2^8 (1+x^n)=1 - x - x^4 - x^6 - x^7 - 2*x^8+O(x^9)$, so the $x^8$ term fails if you just change $1+x+x^2$ to $1-x-x^2$. In any case, you've shown that one needs to at least check up to $N=8$, which is quite a few possible sign choices, although certainly feasible.
$endgroup$
– Joe Silverman
6 hours ago
$begingroup$
Maybe: with $1+x+x^2$, changing the signs inside the binomianal: $1-x^2$, $1+x^3$, $1+x^4$, $1-x^5$ and so on. Just to check that I understood correctly.
$endgroup$
– Doriano Brogioli
5 hours ago
|
show 2 more comments
$begingroup$
The dots at the end lead me to think that the product goes on with an arbitrary number of terms, say $n$. But then you mention a "series", which is infinite; I interpret "coefficient" as the coefficient $a_i$ of a power series, $sum a_i x^i$. However, adding a term to the product, i.e. taking $n+1$, changes all the coefficients $a_i$. Can you please clarify? Also, the last sentence is unclear. Have you already performed the "computer search"? If so, what was the result? Or you are suggesting a method to the other members to answer your question?
$endgroup$
– Doriano Brogioli
8 hours ago
$begingroup$
@DorianoBrogioli The given product $(1+x+x^2)prod_n=2^infty(1+x^n)$ converges in the power series ring $mathbb Z[[x]]$, as does the product if one replaces any subset of the plus signs with minus signs. And since the $(1+x^n)$ terms with $nge N$ only affect the series coefficients starting with $x^N$, one could check all signs for, say, $N=4$, and if they fail to have $-1,0,1$ coefficients in one of their $x^i$ coefficients with $ile3$, then the answer is "No". I think that's what the OP means by "a simple computer search."
$endgroup$
– Joe Silverman
7 hours ago
$begingroup$
With $1-x-x^2$, up to N=7 the signs are -1, 0 and 1...
$endgroup$
– Doriano Brogioli
7 hours ago
1
$begingroup$
@DorianoBrogioli Nice, but $(1-x-x^2)prod_n=2^8 (1+x^n)=1 - x - x^4 - x^6 - x^7 - 2*x^8+O(x^9)$, so the $x^8$ term fails if you just change $1+x+x^2$ to $1-x-x^2$. In any case, you've shown that one needs to at least check up to $N=8$, which is quite a few possible sign choices, although certainly feasible.
$endgroup$
– Joe Silverman
6 hours ago
$begingroup$
Maybe: with $1+x+x^2$, changing the signs inside the binomianal: $1-x^2$, $1+x^3$, $1+x^4$, $1-x^5$ and so on. Just to check that I understood correctly.
$endgroup$
– Doriano Brogioli
5 hours ago
$begingroup$
The dots at the end lead me to think that the product goes on with an arbitrary number of terms, say $n$. But then you mention a "series", which is infinite; I interpret "coefficient" as the coefficient $a_i$ of a power series, $sum a_i x^i$. However, adding a term to the product, i.e. taking $n+1$, changes all the coefficients $a_i$. Can you please clarify? Also, the last sentence is unclear. Have you already performed the "computer search"? If so, what was the result? Or you are suggesting a method to the other members to answer your question?
$endgroup$
– Doriano Brogioli
8 hours ago
$begingroup$
The dots at the end lead me to think that the product goes on with an arbitrary number of terms, say $n$. But then you mention a "series", which is infinite; I interpret "coefficient" as the coefficient $a_i$ of a power series, $sum a_i x^i$. However, adding a term to the product, i.e. taking $n+1$, changes all the coefficients $a_i$. Can you please clarify? Also, the last sentence is unclear. Have you already performed the "computer search"? If so, what was the result? Or you are suggesting a method to the other members to answer your question?
$endgroup$
– Doriano Brogioli
8 hours ago
$begingroup$
@DorianoBrogioli The given product $(1+x+x^2)prod_n=2^infty(1+x^n)$ converges in the power series ring $mathbb Z[[x]]$, as does the product if one replaces any subset of the plus signs with minus signs. And since the $(1+x^n)$ terms with $nge N$ only affect the series coefficients starting with $x^N$, one could check all signs for, say, $N=4$, and if they fail to have $-1,0,1$ coefficients in one of their $x^i$ coefficients with $ile3$, then the answer is "No". I think that's what the OP means by "a simple computer search."
$endgroup$
– Joe Silverman
7 hours ago
$begingroup$
@DorianoBrogioli The given product $(1+x+x^2)prod_n=2^infty(1+x^n)$ converges in the power series ring $mathbb Z[[x]]$, as does the product if one replaces any subset of the plus signs with minus signs. And since the $(1+x^n)$ terms with $nge N$ only affect the series coefficients starting with $x^N$, one could check all signs for, say, $N=4$, and if they fail to have $-1,0,1$ coefficients in one of their $x^i$ coefficients with $ile3$, then the answer is "No". I think that's what the OP means by "a simple computer search."
$endgroup$
– Joe Silverman
7 hours ago
$begingroup$
With $1-x-x^2$, up to N=7 the signs are -1, 0 and 1...
$endgroup$
– Doriano Brogioli
7 hours ago
$begingroup$
With $1-x-x^2$, up to N=7 the signs are -1, 0 and 1...
$endgroup$
– Doriano Brogioli
7 hours ago
1
1
$begingroup$
@DorianoBrogioli Nice, but $(1-x-x^2)prod_n=2^8 (1+x^n)=1 - x - x^4 - x^6 - x^7 - 2*x^8+O(x^9)$, so the $x^8$ term fails if you just change $1+x+x^2$ to $1-x-x^2$. In any case, you've shown that one needs to at least check up to $N=8$, which is quite a few possible sign choices, although certainly feasible.
$endgroup$
– Joe Silverman
6 hours ago
$begingroup$
@DorianoBrogioli Nice, but $(1-x-x^2)prod_n=2^8 (1+x^n)=1 - x - x^4 - x^6 - x^7 - 2*x^8+O(x^9)$, so the $x^8$ term fails if you just change $1+x+x^2$ to $1-x-x^2$. In any case, you've shown that one needs to at least check up to $N=8$, which is quite a few possible sign choices, although certainly feasible.
$endgroup$
– Joe Silverman
6 hours ago
$begingroup$
Maybe: with $1+x+x^2$, changing the signs inside the binomianal: $1-x^2$, $1+x^3$, $1+x^4$, $1-x^5$ and so on. Just to check that I understood correctly.
$endgroup$
– Doriano Brogioli
5 hours ago
$begingroup$
Maybe: with $1+x+x^2$, changing the signs inside the binomianal: $1-x^2$, $1+x^3$, $1+x^4$, $1-x^5$ and so on. Just to check that I understood correctly.
$endgroup$
– Doriano Brogioli
5 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
5
$begingroup$
Possible answers:
$$ left(1+x+x^2right) prod_i=2^+infty left(1 - (-1)^i x^i right) $$
and
$$ left(1-x+x^2right) prod_i=2^+infty left(1 - x^i right) $$
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
$endgroup$
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
4
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
add a comment |
4
$begingroup$
I made a mistake in my initial answer, apologies, I tested for all coefficients instead of stopping at the first $n_rm max$ and thought I found a violation at $n_rm max=9$. With the corrected code I find a solution for each $n_rm max$ I could check.
This Mathematica codes tests if the coefficients of $x^p$ with $pleq n_rm max$ of the polynomial
$$(1+sigma_1 x+sigma_2 x^2)prod_n=2^n_rm max(1+sigma_n+1x^n)$$
are in $-1,0,1$. The test is performed for each of the $2^n_rm max+1$ choices of signs $sigma_i=pm 1$.
nmax = 9;
s = Tuples[-1, 1, nmax + 1];
list = Table[CoefficientList[(1 + s[[i, 1]]*x + s[[i, 2]]*x^2)*
Product[(1 + s[[i, n + 1]]*x^n), n, 2, nmax], x],
i, 1, 2^(nmax + 1)];
Table[AllTrue[list[[i, 1;;nmax]], Abs[#] < 2 &], i, 1, 2^(nmax + 1)] // Sort
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
$endgroup$
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
1
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
Possible answers:
$$ left(1+x+x^2right) prod_i=2^+infty left(1 - (-1)^i x^i right) $$
and
$$ left(1-x+x^2right) prod_i=2^+infty left(1 - x^i right) $$
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
$endgroup$
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
4
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
add a comment |
5
$begingroup$
Possible answers:
$$ left(1+x+x^2right) prod_i=2^+infty left(1 - (-1)^i x^i right) $$
and
$$ left(1-x+x^2right) prod_i=2^+infty left(1 - x^i right) $$
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
$endgroup$
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
4
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
add a comment |
5
5
5
$begingroup$
Possible answers:
$$ left(1+x+x^2right) prod_i=2^+infty left(1 - (-1)^i x^i right) $$
and
$$ left(1-x+x^2right) prod_i=2^+infty left(1 - x^i right) $$
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
$endgroup$
Possible answers:
$$ left(1+x+x^2right) prod_i=2^+infty left(1 - (-1)^i x^i right) $$
and
$$ left(1-x+x^2right) prod_i=2^+infty left(1 - x^i right) $$
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
answered 4 hours ago
Doriano BrogioliDoriano Brogioli
1507
1507
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
4
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
add a comment |
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
4
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
This is correct up to 1000; I've checked it, Now it only remains to prove it.
$endgroup$
– David S. Newman
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
$begingroup$
indeed, apologies for my mistaken answer and for the confusion it created.
$endgroup$
– Carlo Beenakker
3 hours ago
4
4
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
Nice and easily proven from Euler's pentagonal number theorem: call $c(N)$ the $N$th coefficient, Then if $(m-1)(3m-2)/2le N-2<m(3m-1)/2$ we have $c(N)=(-1)^m-1$ except if $N-2=m(3m-1)/2-2$ in which case it is $0$, and if $m(3m-1)/2le N-2<m(3m+1)/2$ it is $0$ except if $N-2=m(3m+1)/2-2$ in which case it is $(-1)^m$.
$endgroup$
– Henri Cohen
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
@HenriCohen Could you post that as an answer, then it can be accepted?
$endgroup$
– Joe Silverman
3 hours ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
$begingroup$
Henri Cohen gives us two solutions . I believe it's still undecided if there are other solutions.
$endgroup$
– David S. Newman
1 hour ago
add a comment |
4
$begingroup$
I made a mistake in my initial answer, apologies, I tested for all coefficients instead of stopping at the first $n_rm max$ and thought I found a violation at $n_rm max=9$. With the corrected code I find a solution for each $n_rm max$ I could check.
This Mathematica codes tests if the coefficients of $x^p$ with $pleq n_rm max$ of the polynomial
$$(1+sigma_1 x+sigma_2 x^2)prod_n=2^n_rm max(1+sigma_n+1x^n)$$
are in $-1,0,1$. The test is performed for each of the $2^n_rm max+1$ choices of signs $sigma_i=pm 1$.
nmax = 9;
s = Tuples[-1, 1, nmax + 1];
list = Table[CoefficientList[(1 + s[[i, 1]]*x + s[[i, 2]]*x^2)*
Product[(1 + s[[i, n + 1]]*x^n), n, 2, nmax], x],
i, 1, 2^(nmax + 1)];
Table[AllTrue[list[[i, 1;;nmax]], Abs[#] < 2 &], i, 1, 2^(nmax + 1)] // Sort
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
$endgroup$
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
1
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
add a comment |
4
$begingroup$
I made a mistake in my initial answer, apologies, I tested for all coefficients instead of stopping at the first $n_rm max$ and thought I found a violation at $n_rm max=9$. With the corrected code I find a solution for each $n_rm max$ I could check.
This Mathematica codes tests if the coefficients of $x^p$ with $pleq n_rm max$ of the polynomial
$$(1+sigma_1 x+sigma_2 x^2)prod_n=2^n_rm max(1+sigma_n+1x^n)$$
are in $-1,0,1$. The test is performed for each of the $2^n_rm max+1$ choices of signs $sigma_i=pm 1$.
nmax = 9;
s = Tuples[-1, 1, nmax + 1];
list = Table[CoefficientList[(1 + s[[i, 1]]*x + s[[i, 2]]*x^2)*
Product[(1 + s[[i, n + 1]]*x^n), n, 2, nmax], x],
i, 1, 2^(nmax + 1)];
Table[AllTrue[list[[i, 1;;nmax]], Abs[#] < 2 &], i, 1, 2^(nmax + 1)] // Sort
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
$endgroup$
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
1
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
add a comment |
4
4
4
$begingroup$
I made a mistake in my initial answer, apologies, I tested for all coefficients instead of stopping at the first $n_rm max$ and thought I found a violation at $n_rm max=9$. With the corrected code I find a solution for each $n_rm max$ I could check.
This Mathematica codes tests if the coefficients of $x^p$ with $pleq n_rm max$ of the polynomial
$$(1+sigma_1 x+sigma_2 x^2)prod_n=2^n_rm max(1+sigma_n+1x^n)$$
are in $-1,0,1$. The test is performed for each of the $2^n_rm max+1$ choices of signs $sigma_i=pm 1$.
nmax = 9;
s = Tuples[-1, 1, nmax + 1];
list = Table[CoefficientList[(1 + s[[i, 1]]*x + s[[i, 2]]*x^2)*
Product[(1 + s[[i, n + 1]]*x^n), n, 2, nmax], x],
i, 1, 2^(nmax + 1)];
Table[AllTrue[list[[i, 1;;nmax]], Abs[#] < 2 &], i, 1, 2^(nmax + 1)] // Sort
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
$endgroup$
I made a mistake in my initial answer, apologies, I tested for all coefficients instead of stopping at the first $n_rm max$ and thought I found a violation at $n_rm max=9$. With the corrected code I find a solution for each $n_rm max$ I could check.
This Mathematica codes tests if the coefficients of $x^p$ with $pleq n_rm max$ of the polynomial
$$(1+sigma_1 x+sigma_2 x^2)prod_n=2^n_rm max(1+sigma_n+1x^n)$$
are in $-1,0,1$. The test is performed for each of the $2^n_rm max+1$ choices of signs $sigma_i=pm 1$.
nmax = 9;
s = Tuples[-1, 1, nmax + 1];
list = Table[CoefficientList[(1 + s[[i, 1]]*x + s[[i, 2]]*x^2)*
Product[(1 + s[[i, n + 1]]*x^n), n, 2, nmax], x],
i, 1, 2^(nmax + 1)];
Table[AllTrue[list[[i, 1;;nmax]], Abs[#] < 2 &], i, 1, 2^(nmax + 1)] // Sort
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
edited 3 hours ago
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
answered 5 hours ago
Carlo BeenakkerCarlo Beenakker
83.6k9196301
83.6k9196301
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
1
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
add a comment |
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
1
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
Thanks. I was going to write a PARI program when I got a chance, but you beat me to it. Seems that in general one can get the coefficients quite small, so the function giving the minimum over signs of the maximum coefficient, as a function of $n_max$, seems as if it might be an interesting sequence. Any chance you want to modify your code to list that sequence up to whatever value of $n_max$ can be computed in a reasonable amount of time. If it's long enough, would be interesting to see if it matches anything on the OEIS.
$endgroup$
– Joe Silverman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
$begingroup$
@Carlo Beenakker I'm going to have to check my computations. I found 40 coefficients which fit the conditions and no hint that this is all the end. I'll try reading the Mathematica code.
$endgroup$
– David S. Newman
4 hours ago
1
1
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
$begingroup$
If you have time, please check the answer I gave below.
$endgroup$
– Doriano Brogioli
4 hours ago
add a comment |
draft saved
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$begingroup$
The dots at the end lead me to think that the product goes on with an arbitrary number of terms, say $n$. But then you mention a "series", which is infinite; I interpret "coefficient" as the coefficient $a_i$ of a power series, $sum a_i x^i$. However, adding a term to the product, i.e. taking $n+1$, changes all the coefficients $a_i$. Can you please clarify? Also, the last sentence is unclear. Have you already performed the "computer search"? If so, what was the result? Or you are suggesting a method to the other members to answer your question?
$endgroup$
– Doriano Brogioli
8 hours ago
$begingroup$
@DorianoBrogioli The given product $(1+x+x^2)prod_n=2^infty(1+x^n)$ converges in the power series ring $mathbb Z[[x]]$, as does the product if one replaces any subset of the plus signs with minus signs. And since the $(1+x^n)$ terms with $nge N$ only affect the series coefficients starting with $x^N$, one could check all signs for, say, $N=4$, and if they fail to have $-1,0,1$ coefficients in one of their $x^i$ coefficients with $ile3$, then the answer is "No". I think that's what the OP means by "a simple computer search."
$endgroup$
– Joe Silverman
7 hours ago
$begingroup$
With $1-x-x^2$, up to N=7 the signs are -1, 0 and 1...
$endgroup$
– Doriano Brogioli
7 hours ago
1
$begingroup$
@DorianoBrogioli Nice, but $(1-x-x^2)prod_n=2^8 (1+x^n)=1 - x - x^4 - x^6 - x^7 - 2*x^8+O(x^9)$, so the $x^8$ term fails if you just change $1+x+x^2$ to $1-x-x^2$. In any case, you've shown that one needs to at least check up to $N=8$, which is quite a few possible sign choices, although certainly feasible.
$endgroup$
– Joe Silverman
6 hours ago
$begingroup$
Maybe: with $1+x+x^2$, changing the signs inside the binomianal: $1-x^2$, $1+x^3$, $1+x^4$, $1-x^5$ and so on. Just to check that I understood correctly.
$endgroup$
– Doriano Brogioli
5 hours ago