Take two positive integers $m$ and $n$ and consider the rational function $$G_{m,n}(x,t)=\frac{d}{dx}\left(\frac1{(1-x^m)(1-tx^n)}\right)$$ and the corresponding Taylor expansion as $$G_{m,n}(x,t)=u_0(t)+u_1(t)x+u_2(t)x^2+\cdots$$ where $u_j(t)$ itself is expanded as a polynomial in $t$ (note: coefficients in each $u_j(t)$ are all equal). Now, read-off the coefficients of $G_{m,n}(x,t)$ in the exact order they appear and be listed (including multiplicities) as $\beta_{\ell}(m,n)$.

QUESTION. Is this limit true? $$\lim_{\ell\rightarrow\infty}\frac{\beta_{\ell}^2(m,n)}{\ell}=2mn.$$

EXAMPLE 1. If $m=n=1$ then $$G_{1,1}(x,t)=1+t+(2+2t+2t^2)x+(3+3t+3t^2+3t^3)x^2+\cdots$$ and hence $\beta_{\ell}(1,1)$ starts with (keep in mind: $\beta_1=1, \beta_2=1, \beta_3=2$, etc) $$1,1,2,2,2,3,3,3,3,\dots.$$ Hence $$\beta_{\ell}(1,1)=\left\lfloor\frac{\sqrt{8\ell+1}-1}2\right\rfloor \qquad \Longrightarrow \qquad \lim_{\ell\rightarrow\infty}\frac{\beta^2_{\ell}(1,1)}{\ell}=2.$$

EXAMPLE 2. If $m=1$ and $n=2$ then $$G_{1,2}(x,t)=1+(2+2t)x+(3+3t)x^2+(4+4t+4t^2)x^3+\cdots$$ and hence $\beta_{\ell}(1,2)$ starts with (keep in mind: $\beta_1=1, \beta_2=2, \beta_3=2$, etc), $$1,2,2,3,3,4,4,4,\dots.$$ $$\beta_{\ell}(1,2)=\left\lfloor\sqrt{4\ell+1}-1\right\rfloor \qquad \Longrightarrow \qquad \lim_{\ell\rightarrow\infty}\frac{\beta^2_{\ell}(1,2)}{\ell}=4.$$

  • $\begingroup$ I take it $\beta_\ell^2(m,n)$ means $\beta_\ell(m,n)^2$? $\endgroup$ – Wojowu Jan 7 '20 at 22:46
  • $\begingroup$ Don't you need $n,m$ relatively prime? Because if $(n,m)=d>1$ then the power series expansion of $\frac{1}{1-x^m)(1-tx^n)}$ is a series in $x^d$, so that, deriving in $x$, $G_{m,n}(t,x)$ has $u_j\neq0$ only for $j=-1\mod d$, which implies the $\beta_\ell(m,n)$ are frequently zero... $\endgroup$ – Pietro Majer Jan 7 '20 at 22:55
  • $\begingroup$ Maybe in the definition of the $\beta_\ell(m,n)$ only the non-zero coefficients are listed? $\endgroup$ – Pietro Majer Jan 7 '20 at 22:59
  • $\begingroup$ @Wojowu: yes that is right. $\endgroup$ – T. Amdeberhan Jan 7 '20 at 23:04
  • $\begingroup$ @PietroMajer: yes, you are right, the non-zero terms are listed. Thanks. $\endgroup$ – T. Amdeberhan Jan 7 '20 at 23:05

Yes, the conjectured limit is true. Let $d=\gcd (m,n)$ and $m=m_1d, n=n_1d$. Suppose $a_k$ denotes the number of solutions to $k=m_1r+n_1s$ with $r,s\geq 0$, so that $$a_0+a_1x+a_2x^2+\cdots =\frac{1}{(1-x^{m_1})(1-x^{n_1})}.$$ We have $\beta_{\ell}(m,n)=Nd$ if and only if $\sum_{i=0}^{N-1}a_i< \ell\le \sum_{i=0}^{N}a_i$.

Notice that $a_i\in \{0,1\}$ for $0\le i\le m_1n_1-1$, and the rest satisfy $a_{mn+i}=1+a_i$.So we have $$\sum_{i=0}^{N-1}a_i\geq \sum_{i=0}^{m_1n_1\lfloor \frac{N-1}{m_1n_1}\rfloor}a_i\geq m_1n_1\binom{\lfloor \frac{N-1}{m_1n_1}\rfloor}{2}=\frac{N^2}{2m_1n_1}+O(N)$$ and similarly $$\sum_{i=0}^{N}a_i\le \sum_{i=0}^{m_1n_1\lceil \frac{N}{m_1n_1}\rceil}a_i\le m_1n_1\binom{1+\lceil \frac{N}{m_1n_1}\rceil}{2}=\frac{N^2}{2m_1n_1}+O(N).$$ So we have $$\lim_{\ell\to \infty} \frac{\ell}{\beta_{\ell}^2(m,n)}=\lim_{N\to \infty}\frac{\frac{N^2}{2m_1n_1}}{N^2d^2}=\frac{1}{2mn}$$ as desired.

  • $\begingroup$ Thank you for an elementary argument. $\endgroup$ – T. Amdeberhan Jan 8 '20 at 14:24

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