Eventual positivity conjecture for $D'(k,m,n)$

Complete asymptotic proof of eventual positivity for every $k>m$

George E. Andrews and Mohamed El Bachraoui conjecture that the coefficients $D'(k,m,n)$ are eventually nonnegative whenever $k>m\geq1$. This is Conjecture 3.7 of Certain positive $q$-series and inequalities for two-color partitions, Arabian Journal of Mathematics 15 (2026), 551–560, and Conjecture 5 of the original preprint.

We prove the stronger precise asymptotic: for every fixed $k>m\geq1$, put $r=k-1$. Then

$$D'(k,m,n) =\left( \frac{(2r-2)!!}{r!\big((2r-1)!!\big)^2} -\frac{\pi}{2^{r+1}(r!)^2} \right)n^r+o_{k,m}(n^r). \tag{1}$$

The parenthesized constant is strictly positive for every $r\geq1$. In particular,

$$D'(k,m,n)>0 \qquad\text{for every sufficiently large }n, \tag{2}$$

which proves the full conjecture with strict eventual positivity. The threshold may depend on the fixed pair $(k,m)$.

1. An exact rational function for the first source term

Write

$$(a;q)_j=\prod_{i=0}^{j-1}(1-aq^i), \qquad (a;q)_\infty=\prod_{i\geq0}(1-aq^i), \tag{3}$$

and define

$$F_{k,m}(q)=q^{-m}\sum_{n\geq0}C'(k,m,n)q^n, \qquad H_k(q)=\frac{(q^2,q^{2k};q^2)_\infty}{(q;q^2)_\infty^2}. \tag{4}$$

The published Theorem 3.2 gives the exact formal-series identity

$$\sum_{n\geq0}D'(k,m,n)q^n=F_{k,m}(q)-H_k(q). \tag{5}$$

Since $m<k$, the published Theorem 2.2 gives

$$F_{k,m}(q) =\frac{1}{(q;q^2)_r} \sum_{j\geq0}q^j (q^{2j+2};q^2)_{m-1} (q^{2j+2m+1};q^2)_{k-m-1}. \tag{6}$$

Let $\mathcal C_{k,m}$ consist of the following $r-1$ positive integers, with either progression allowed to be empty:

$$\mathcal C_{k,m} =\{2,4,\ldots,2m-2\} \cup \{2m+1,2m+3,\ldots,2k-3\}. \tag{7}$$

For $S\subseteq\mathcal C_{k,m}$, write $c(S)=\sum_{c\in S}c$. Expanding the finite products in (6) and summing the resulting geometric series yields the exact rational function

$$F_{k,m}(q) =\frac{1}{\prod_{i=0}^{r-1}(1-q^{2i+1})} \sum_{S\subseteq\mathcal C_{k,m}} \frac{(-1)^{|S|}q^{c(S)}}{1-q^{2|S|+1}}. \tag{8}$$

At $q=1$, the coefficient of $(1-q)^{-r-1}$ in (8) is

$$\begin{aligned} L_r &=\frac{1}{(2r-1)!!} \sum_{j=0}^{r-1} \frac{(-1)^j}{2j+1}\binom{r-1}{j}\\ &=\frac{1}{(2r-1)!!} \int_0^1(1-x^2)^{r-1}\,dx\\ &=\frac{(2r-2)!!}{\big((2r-1)!!\big)^2}>0. \end{aligned} \tag{9}$$

The last equality follows by the elementary beta integral, and includes $r=1$ with $0!!=1$.

Every pole of (8) is a root of unity. At any root of unity different from $1$, the factor $1-q$ in the first denominator does not vanish, so at most $r-1$ of its $r$ factors vanish. Each summand in the second factor contributes at most one additional simple pole. Hence every nonprincipal pole has order at most $r$, whereas (9) shows that $q=1$ has order exactly $r+1$. Partial fractions therefore give

$$[q^n]F_{k,m}(q) =\frac{L_r}{r!}n^r+O_{k,m}(n^{r-1}) =\frac{(2r-2)!!}{r!\big((2r-1)!!\big)^2}n^r +O_{k,m}(n^{r-1}). \tag{10}$$

In particular, the leading coefficient is independent of $m$.

2. A weighted triangular-number lattice asymptotic

Gauss's identity gives

$$\psi(q) =\sum_{s\geq0}q^{s(s+1)/2} =\frac{(q^2;q^2)_\infty}{(q;q^2)_\infty}. \tag{11}$$

Consequently,

$$H_k(q)=\frac{\psi(q)^2}{(q^2;q^2)_r}. \tag{12}$$

Put

$$T_s=\frac{s(s+1)}2, \qquad p_r(L)=[z^L]\prod_{i=1}^{r}\frac1{1-z^i}. \tag{13}$$

Extracting coefficients from (12) yields

$$[q^n]H_k(q) =\sum_{\substack{s,t\geq0\\ T_s+T_t\leq n\\ T_s+T_t\equiv n\pmod2}} p_r\left(\frac{n-T_s-T_t}{2}\right). \tag{14}$$

For $r\geq2$, the rational function defining $p_r$ has an order-$r$ pole at $z=1$, whose principal coefficient is $1/r!$. Every other root of unity has pole order at most $r-1$, because its denominator contains the nonvanishing factor $1-z$. Thus

$$p_r(L) =\frac{L^{r-1}}{r!(r-1)!} +O_r\big((L+1)^{r-2}\big). \tag{15}$$

For $r=1$, the exact corresponding identity is simply

$$p_1(L)=1. \tag{16}$$

There are $O(n)$ pairs $(s,t)$ with $T_s+T_t\leq n$. Therefore, for $r\geq2$, summing the error in (15) across (14) gives $O_r(n^{r-1})$, and hence

$$[q^n]H_k(q) =\frac{1}{2^{r-1}r!(r-1)!} \sum_{\substack{s,t\geq0\\ T_s+T_t\leq n\\ T_s+T_t\equiv n\pmod2}} (n-T_s-T_t)^{r-1} +O_r(n^{r-1}). \tag{17}$$

When $r=1$, the same main-term formula is exact if each admissible summand is interpreted as $1$.

The parity pattern of triangular numbers is

$$(T_0,T_1,T_2,T_3)\equiv(0,1,1,0)\pmod2, \tag{18}$$

and repeats modulo $4$ in the index. Consequently, for either parity of $n$, exactly eight of the sixteen classes

$$(s,t)\pmod4 \tag{19}$$

satisfy the congruence in (17).

Fix one of these residue classes. Under the rescaling

$$x=\frac{s}{\sqrt n}, \qquad y=\frac{t}{\sqrt n}, \tag{20}$$

its lattice has asymptotic density $1/16$. Moreover,

$$\frac{T_s+T_t}{n} =\frac{x^2+y^2}{2} +\frac{x+y}{2\sqrt n}. \tag{21}$$

For $r\geq2$, the resulting bounded, compactly supported weight converges uniformly to

$$\left(1-\frac{x^2+y^2}{2}\right)_+^{r-1}. \tag{22}$$

The ordinary two-dimensional Riemann-sum theorem therefore gives, for that residue class,

$$\begin{aligned} &\frac1{n^r} \sum_{\substack{s,t\geq0\\ T_s+T_t\leq n\\ (s,t)\text{ in the fixed class}}} (n-T_s-T_t)^{r-1}\\ &\hspace{15mm}\longrightarrow \frac1{16} \int_{\substack{x,y\geq0\\x^2+y^2\leq2}} \left(1-\frac{x^2+y^2}{2}\right)^{r-1}\,dx\,dy. \end{aligned} \tag{23}$$

For $r=1$, replace the weight in (22) by the indicator of the quarter disk. Its boundary has area zero, so the same Riemann-sum limit holds by Jordan measurability; the vanishing linear term in (21) changes neither domain nor limit.

In polar coordinates the integral is

$$\begin{aligned} \int_{\substack{x,y\geq0\\x^2+y^2\leq2}} \left(1-\frac{x^2+y^2}{2}\right)^{r-1}\,dx\,dy &=\frac\pi2\int_0^{\sqrt2} \rho\left(1-\frac{\rho^2}{2}\right)^{r-1}\,d\rho\\ &=\frac\pi{2r}. \end{aligned} \tag{24}$$

Summing (23) over the eight admissible residue classes and substituting (24) into (17) gives, for every $r\geq1$,

$$[q^n]H_k(q) \sim\frac{1}{2^{r-1}r!(r-1)!} \cdot\frac12\cdot\frac\pi{2r}\,n^r =\frac\pi{2^{r+1}(r!)^2}\,n^r. \tag{25}$$

Both parities of $n$ have the same eight-class density, so (25) holds along the full integer sequence, not merely along a parity subsequence.

3. The leading gap is strictly positive

Combining (5), (10), and (25) proves the asymptotic formula (1). To compare its two constants, their ratio simplifies to

$$\frac{\displaystyle \frac{(2r-2)!!}{r!\big((2r-1)!!\big)^2}} {\displaystyle \frac\pi{2^{r+1}(r!)^2}} =\frac{16^r}{\pi r\binom{2r}{r}^2}. \tag{26}$$

For $0<x<\pi/2$, one has

$$\sin^{2r}x<\sin^{2r-1}x. \tag{27}$$

Integrating and using the elementary Wallis integrals yields

$$\frac\pi2\frac{(2r-1)!!}{(2r)!!} <\frac{(2r-2)!!}{(2r-1)!!}. \tag{28}$$

Equivalently,

$$\pi r\left(\frac{(2r-1)!!}{(2r)!!}\right)^2<1. \tag{29}$$

Since

$$\frac{(2r-1)!!}{(2r)!!} =\frac1{4^r}\binom{2r}{r}, \tag{30}$$

inequality (29) is exactly

$$\pi r\binom{2r}{r}^2<16^r. \tag{31}$$

Thus the ratio (26) is strictly greater than $1$ for every positive integer $r$. The leading constant in (1) is strictly positive, proving (2) for every fixed pair $k>m\geq1$.

The result concerns precisely the two-color statistic $D'(k,m,n)$ of Andrews and El Bachraoui. A separately indexed 2017 conjecture involving the different bounded-partition rational function $H_{L,s,k}(q)$ is not the same statement and is not claimed here.