Are the coefficients of certain product of Rogers-Ramanujan Continued Fraction non-negative?Estimating the...



Are the coefficients of certain product of Rogers-Ramanujan Continued Fraction non-negative?


Estimating the number of non-negative integer $n$-tuples satisfying two conditionsAbout two 'negative' continued fractions whose sum equals $1$Is any particular algebraic number known to have unbounded continued fraction coefficients?Linear functionsRogers-Ramanujan continued fraction $R(e^{-2 pi sqrt 5})$Can we use the Rogers-Ramanujan cfrac to parameterize the Fermat quintic $x^5+y^5=1$?Are the Fourier coefficients of $eta(q^m)^m / eta(q)$ non-negative?Reference request for function by which to compute coefficients of continued fraction of algebaic number













3












$begingroup$


Let $$R(q) = cfrac{q^{1/5}}{1 + cfrac{q}{1 + cfrac{q^{2}}{1 + cfrac{q^{3}}{1 + cdots}}}}$$



The following equality is famous:



$$cfrac{q^{1/5}}{R(q)} = prod_{k>0} cfrac{(1-q^{5k-2})(1-q^{5k-3})}{(1-q^{5k-1})(1-q^{5k-4})} ( = f(q))$$



The coefficients of $f(q)$ can be positive or negative. In fact,



$$f(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2 q^8 + cdots$$



Let



$$g(q) = prod_{k>0} f(q^k) = f(q)f(q^2)f(q^3) cdots$$



$g(q)$



$= (1 + q - q^3 + q^5 + cdots)(1 + q^2 - q^6 + q^{10} + cdots)cdots$



$= 1 + q + q^2 + q^3 + 2q^4 + 3q^5 + 3q^6 + 3q^7 + 4q^8 + 6q^9 + cdots$



The coefficients of $g(q)$ seem non-negative.
Are the coefficients of $g(q)$ non-negative?






nt.number-theory







share|cite|improve this question









$endgroup$










  • 1




    $begingroup$
    the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check
    $endgroup$
    – Carlo Beenakker
    6 hours ago
















3












$begingroup$


Let $$R(q) = cfrac{q^{1/5}}{1 + cfrac{q}{1 + cfrac{q^{2}}{1 + cfrac{q^{3}}{1 + cdots}}}}$$



The following equality is famous:



$$cfrac{q^{1/5}}{R(q)} = prod_{k>0} cfrac{(1-q^{5k-2})(1-q^{5k-3})}{(1-q^{5k-1})(1-q^{5k-4})} ( = f(q))$$



The coefficients of $f(q)$ can be positive or negative. In fact,



$$f(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2 q^8 + cdots$$



Let



$$g(q) = prod_{k>0} f(q^k) = f(q)f(q^2)f(q^3) cdots$$



$g(q)$



$= (1 + q - q^3 + q^5 + cdots)(1 + q^2 - q^6 + q^{10} + cdots)cdots$



$= 1 + q + q^2 + q^3 + 2q^4 + 3q^5 + 3q^6 + 3q^7 + 4q^8 + 6q^9 + cdots$



The coefficients of $g(q)$ seem non-negative.
Are the coefficients of $g(q)$ non-negative?






nt.number-theory







share|cite|improve this question









$endgroup$










  • 1




    $begingroup$
    the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check
    $endgroup$
    – Carlo Beenakker
    6 hours ago














3












3








3


1



$begingroup$


Let $$R(q) = cfrac{q^{1/5}}{1 + cfrac{q}{1 + cfrac{q^{2}}{1 + cfrac{q^{3}}{1 + cdots}}}}$$



The following equality is famous:



$$cfrac{q^{1/5}}{R(q)} = prod_{k>0} cfrac{(1-q^{5k-2})(1-q^{5k-3})}{(1-q^{5k-1})(1-q^{5k-4})} ( = f(q))$$



The coefficients of $f(q)$ can be positive or negative. In fact,



$$f(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2 q^8 + cdots$$



Let



$$g(q) = prod_{k>0} f(q^k) = f(q)f(q^2)f(q^3) cdots$$



$g(q)$



$= (1 + q - q^3 + q^5 + cdots)(1 + q^2 - q^6 + q^{10} + cdots)cdots$



$= 1 + q + q^2 + q^3 + 2q^4 + 3q^5 + 3q^6 + 3q^7 + 4q^8 + 6q^9 + cdots$



The coefficients of $g(q)$ seem non-negative.
Are the coefficients of $g(q)$ non-negative?






nt.number-theory







share|cite|improve this question









$endgroup$




Let $$R(q) = cfrac{q^{1/5}}{1 + cfrac{q}{1 + cfrac{q^{2}}{1 + cfrac{q^{3}}{1 + cdots}}}}$$



The following equality is famous:



$$cfrac{q^{1/5}}{R(q)} = prod_{k>0} cfrac{(1-q^{5k-2})(1-q^{5k-3})}{(1-q^{5k-1})(1-q^{5k-4})} ( = f(q))$$



The coefficients of $f(q)$ can be positive or negative. In fact,



$$f(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2 q^8 + cdots$$



Let



$$g(q) = prod_{k>0} f(q^k) = f(q)f(q^2)f(q^3) cdots$$



$g(q)$



$= (1 + q - q^3 + q^5 + cdots)(1 + q^2 - q^6 + q^{10} + cdots)cdots$



$= 1 + q + q^2 + q^3 + 2q^4 + 3q^5 + 3q^6 + 3q^7 + 4q^8 + 6q^9 + cdots$



The coefficients of $g(q)$ seem non-negative.
Are the coefficients of $g(q)$ non-negative?






nt.number-theory




nt.number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 9 hours ago









ManyamaManyama

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  • 1




    $begingroup$
    the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check
    $endgroup$
    – Carlo Beenakker
    6 hours ago














  • 1




    $begingroup$
    the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check
    $endgroup$
    – Carlo Beenakker
    6 hours ago








1




1




$begingroup$
the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check
$endgroup$
– Carlo Beenakker
6 hours ago




$begingroup$
the stronger conjecture is that the coefficients form a nondecreasing series, which seems to be the case as far as I could check
$endgroup$
– Carlo Beenakker
6 hours ago










1 Answer
1






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oldest

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6














$begingroup$

Notice that we can write
$$f(q)=prod_{ngeq 1} (1-q^n)^{-left(frac{n}{5}right)}$$
therefore
$$g(q)=prod_{kgeq 1} f(q^k)=prod_{ngeq 1} (1-q^n)^{-a(n)}$$
where $a(n)=sum_{d|n}left(frac{d}{5}right)$, where $left(frac{d}{5}right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $pequiv pm 1pmod{5}$, and $a(p^k)=frac{1+(-1)^k}{2}$ when $pequiv pm 2pmod{5}$. This means that $a(n)geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

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    6














    $begingroup$

    Notice that we can write
    $$f(q)=prod_{ngeq 1} (1-q^n)^{-left(frac{n}{5}right)}$$
    therefore
    $$g(q)=prod_{kgeq 1} f(q^k)=prod_{ngeq 1} (1-q^n)^{-a(n)}$$
    where $a(n)=sum_{d|n}left(frac{d}{5}right)$, where $left(frac{d}{5}right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $pequiv pm 1pmod{5}$, and $a(p^k)=frac{1+(-1)^k}{2}$ when $pequiv pm 2pmod{5}$. This means that $a(n)geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).






    share|cite|improve this answer











    $endgroup$




















      6














      $begingroup$

      Notice that we can write
      $$f(q)=prod_{ngeq 1} (1-q^n)^{-left(frac{n}{5}right)}$$
      therefore
      $$g(q)=prod_{kgeq 1} f(q^k)=prod_{ngeq 1} (1-q^n)^{-a(n)}$$
      where $a(n)=sum_{d|n}left(frac{d}{5}right)$, where $left(frac{d}{5}right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $pequiv pm 1pmod{5}$, and $a(p^k)=frac{1+(-1)^k}{2}$ when $pequiv pm 2pmod{5}$. This means that $a(n)geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).






      share|cite|improve this answer











      $endgroup$


















        6














        6










        6







        $begingroup$

        Notice that we can write
        $$f(q)=prod_{ngeq 1} (1-q^n)^{-left(frac{n}{5}right)}$$
        therefore
        $$g(q)=prod_{kgeq 1} f(q^k)=prod_{ngeq 1} (1-q^n)^{-a(n)}$$
        where $a(n)=sum_{d|n}left(frac{d}{5}right)$, where $left(frac{d}{5}right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $pequiv pm 1pmod{5}$, and $a(p^k)=frac{1+(-1)^k}{2}$ when $pequiv pm 2pmod{5}$. This means that $a(n)geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).






        share|cite|improve this answer











        $endgroup$



        Notice that we can write
        $$f(q)=prod_{ngeq 1} (1-q^n)^{-left(frac{n}{5}right)}$$
        therefore
        $$g(q)=prod_{kgeq 1} f(q^k)=prod_{ngeq 1} (1-q^n)^{-a(n)}$$
        where $a(n)=sum_{d|n}left(frac{d}{5}right)$, where $left(frac{d}{5}right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $pequiv pm 1pmod{5}$, and $a(p^k)=frac{1+(-1)^k}{2}$ when $pequiv pm 2pmod{5}$. This means that $a(n)geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 5 hours ago

























        answered 5 hours ago









        Gjergji ZaimiGjergji Zaimi

        64.8k4 gold badges173 silver badges323 bronze badges




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