Exceptional negative coefficients conjecture for $D'(2,3,n)$

Exact classification of the negative coefficients of $D'(2,3,n)$

George E. Andrews and Mohamed El Bachraoui formulate this assertion as Conjecture 3.6 of Certain positive $q$-series and inequalities for two-color partitions, Arabian Journal of Mathematics 15 (2026), 551–560, and as Conjecture 4 in the original preprint.

We prove

$$D'(2,3,n)<0 \quad\Longleftrightarrow\quad n\in\{10,22\}, \qquad D'(2,3,10)=D'(2,3,22)=-1. \tag{1}$$

1. Exact source identities

Put

$$F(q)=q^{-3}\sum_{n\geq0}C'(2,3,n)q^n, \qquad H(q)=\frac{(q^2,q^4;q^2)_\infty}{(q;q^2)_\infty^2}. \tag{2}$$

The published Theorem 3.2 gives

$$\sum_{n\geq0}D'(2,3,n)q^n=F(q)-H(q). \tag{3}$$

Specializing the published Theorem 2.2 to $(k,m)=(2,3)$ gives

$$F(q)=\sum_{j\geq0}q^jR_j(q), \qquad R_j(q)= \frac{(1-q^{2j+2})(1-q^{2j+4})} {(1-q)(1-q^{2j+3})(1-q^{2j+5})}. \tag{4}$$

The factor $q^j$ is essential: without it, the infinite sum would not define a formal power series. For $a=2j+2$, source Lemma 5.3 gives the partial-fraction identity

$$q^3R_j(q) =\frac{q}{1-q} -\frac{q^2}{(1+q)(1-q^{a+1})} -\frac{q}{1-q^{a+3}} -\frac{q^3}{(1+q)(1-q^{a+3})}. \tag{5}$$

Multiplication by $q^j$ and summation yield

$$q^3F(q) =\frac{q}{(1-q)^2} -\frac{L_1(q)+L_3(q)}{1+q} -L_2(q), \tag{6}$$

where

$$\begin{aligned} L_1(q)&=\sum_{j\geq0}\frac{q^{j+2}}{1-q^{2j+3}},\\ L_2(q)&=\sum_{j\geq0}\frac{q^{j+1}}{1-q^{2j+5}},\\ L_3(q)&=\sum_{j\geq0}\frac{q^{j+3}}{1-q^{2j+5}}. \end{aligned} \tag{7}$$

2. Divisor formulas and a square-root bound

Let $\tau(N)$ denote the number of positive divisors of $N$, and put $\ell_i(M)=[q^M]L_i(q)$. For every $M\geq1$, factoring the corresponding odd integers gives

$$\begin{aligned} \ell_1(M) &=\tau(2M-1)-1,\\ \ell_3(M) &=\tau(2M-1)-1-\mathbf 1_{3\mid(2M-1)},\\ \ell_2(M) &=\tau(2M+3)-1-\mathbf 1_{3\mid M}. \end{aligned} \tag{8}$$

Indeed, the first two progressions in question correspond to factorizations

$$2M-1=(2j+3)(2t+1), \qquad 2M-1=(2j+5)(2t+1), \tag{9}$$

respectively, while the third corresponds to

$$2M+3=(2j+5)(2t+1), \qquad j,t\geq0. \tag{10}$$

The excluded first factors are $1$, and additionally $3$ whenever it divides the relevant integer.

Set $u_0=0$. Equations (6) and (8) give the exact integer recurrences

$$u_M =2\tau(2M-1)-2-\mathbf 1_{3\mid(2M-1)}-u_{M-1} \qquad(M\geq1), \tag{11}$$

and, writing $F_n=[q^n]F(q)$,

$$F_n =n+4-\tau(2n+9)+\mathbf 1_{3\mid n}-u_{n+3} \qquad(n\geq0). \tag{12}$$

To bound $u_M$, let $\chi_4$ be the nonprincipal character modulo $4$, and define

$$S(x)=\sum_{d\leq x}\chi_4(d), \qquad A(X)=\sum_{d\leq X}\chi_4(d)\tau(d). \tag{13}$$

Because $S(x)\in\{0,1\}$ and $\chi_4$ is completely multiplicative, the Dirichlet hyperbola identity gives

$$A(X) =2\sum_{a\leq\sqrt X}\chi_4(a) S\left(\frac Xa\right) -S(\sqrt X)^2. \tag{14}$$

Only odd $a$ contribute. Consequently,

$$|A(X)| \leq 2\left\lceil\frac{\lfloor\sqrt X\rfloor}{2}\right\rceil+1 \leq\sqrt X+2. \tag{15}$$

Since

$$(-1)^t=-\chi_4(2t-1), \tag{16}$$

the alternating sum of the divisor terms in (11) is, up to sign, $2A(2M-1)$. The constant terms contribute at most $2$ in absolute value, and the terms with $3\mid(2t-1)$ alternate because their indices differ by $3$, so they contribute at most $1$. Therefore

$$|u_M|\leq2\sqrt{2M-1}+7 \qquad(M\geq1). \tag{17}$$

Moreover, $\tau(N)\leq2\sqrt N$. Combining (8), (12), and (17) yields

$$F_n \geq n-3-2\sqrt{2n+5}-2\sqrt{2n+9} \geq n-3-4\sqrt{2n+9}. \tag{18}$$

3. The triangular-number contribution

Gauss's triangular-number identity gives

$$H(q) =\frac{1}{1-q^2} \left(\sum_{s\geq0}q^{s(s+1)/2}\right)^2. \tag{19}$$

Thus, with $T_s=s(s+1)/2$,

$$H_n=[q^n]H(q) =\#\left\{(s,t)\in\mathbb Z_{\geq0}^2: T_s+T_t\leq n,\quad T_s+T_t\equiv n\pmod2 \right\}. \tag{20}$$

The substitution $X=2s+1$, $Y=2t+1$ gives

$$X^2+Y^2\leq8n+2, \qquad X^2+Y^2\equiv8n+2\pmod{16}. \tag{21}$$

Exactly eight of the sixteen odd residue pairs modulo $8$ satisfy this congruence, for either parity of $n$. Partition the first quadrant into half-open $8\times8$ squares. Each contains exactly eight admissible points, and any square containing a point of norm at most $R=\sqrt{8n+2}$ lies inside the quarter disk of radius $R+8\sqrt2$. Comparing their areas and using $\pi<22/7$ gives

$$\begin{aligned} H_n &\leq \frac{\pi}{32}\left(\sqrt{8n+2}+8\sqrt2\right)^2\\ &< \frac{11n}{14} +\frac{44}{7}\sqrt{n+1} +\frac{715}{56}. \end{aligned} \tag{22}$$

Combining (18) and (22), and using $\sqrt2<10/7$, shows that $F_n>H_n$ whenever

$$12n-883>672\sqrt{n+5}. \tag{23}$$

Indeed,

$$4\sqrt{2n+9} +\frac{44}{7}\sqrt{n+1} < 12\sqrt{n+5}. \tag{24}$$

For

$$P(n)=(12n-883)^2-672^2(n+5), \tag{25}$$

direct evaluation gives

$$P(3286)=-137543, \qquad P(3287)=336193, \tag{26}$$

and

$$P(n+1)-P(n)=288n-472632>0 \qquad(n\geq3287). \tag{27}$$

Since $12n-883>0$ in the same range, (23) follows for every $n\geq3287$. Hence

$$D'(2,3,n)=F_n-H_n>0 \qquad(n\geq3287). \tag{28}$$

4. Complete finite arithmetic certificate

The classical sum-of-two-squares formula gives

$$b_n =\#\{(s,t)\in\mathbb Z_{\geq0}^2:T_s+T_t=n\} =\sum_{d\mid(4n+1)}\chi_4(d). \tag{29}$$

Indeed, $b_n$ is one quarter of the number of signed representations of $8n+2$ as a sum of two squares, and its odd divisors are precisely the divisors of $4n+1$. Equation (19) now gives

$$H_0=b_0, \qquad H_1=b_1, \qquad H_n=b_n+H_{n-2}\quad(n\geq2). \tag{30}$$

Thus (11), (12), (29), and (30) determine every coefficient using only ordinary divisor counts and the character $\chi_4$. Put

$$d_n=F_n-H_n. \tag{31}$$

Exact evaluation over the entire remaining interval gives the following finite certificate; the final column provides an independent block-total check.

$$\begin{array}{c|r|l|r} \text{interval }I & \min_{n\in I}d_n & \{n\in I:d_n=\min_{k\in I}d_k\} & \sum_{n\in I}d_n \\ \hline [0,9]&0&0,1,2,3,4,5,7,8&3\\ [10,10]&-1&10&-1\\ [11,21]&0&13,15,17,20&22\\ [22,22]&-1&22&-1\\ [23,99]&0&29&790\\ [100,199]&9&110,112&2829\\ [200,499]&29&220&21146\\ [500,999]&92&515&77765\\ [1000,1499]&196&1030&131141\\ [1500,1999]&295&1520&184631\\ [2000,2499]&407&2012&238170\\ [2500,2999]&520&2530&291697\\ [3000,3286]&617&3008,3010&191600 \end{array} \tag{32}$$

The intervals in (32) partition all integers from $0$ through $3286$. The only negative entries occur at $n=10$ and $n=22$, and in both cases the exact coefficient is $-1$. Combining this complete finite certificate with the unconditional infinite-range inequality (28) proves (1).