Unimodality of the Boţ–Nguyen coefficient rows

Bauschke and Gao's Conjecture 6.5 in arXiv:2604.17084v2 concerns the coefficients $c_{n,k}$ defined by their Lemma 4.1, equations (44) and (45b). The conjecture excludes $c_{n,0}$. That exclusion matters: the source explains that including the constant coefficient can destroy unimodality. The case $\alpha=4$ is already proved in the source.

Theorem. For every real $\alpha>2$ and every integer $n\geq1$, the sequence

$$c_{n,1},c_{n,2},\ldots,c_{n,n}$$

is unimodal. More precisely, it is nondecreasing up to

$$\mu_n=\max\{1,\lfloor n/2\rfloor\}$$

and nonincreasing thereafter. The inequalities asserted here are weak; uniqueness of the mode is not needed.

Main idea. Normalize each row by its positive constant coefficient. The resulting recurrence has a nonnegative expansion indexed by matchings of a path. Grouping terms with all but the first marked edge fixed lets the decreasing edge weights produce sums of symmetric unimodal polynomials whose middle modes agree. Only the constant coefficient needs to be adjusted, so every coefficient in the conjectured range is unchanged.

1. A normalized recurrence

Put $\beta=\alpha/2>1$, and let

$$p_n(t)=\sum_{k=0}^n c_{n,k}t^k.$$

The source recurrence is

$$\begin{aligned} p_0(t)&=1,\\ p_1(t)&=\frac{\beta+1+\beta t}{2\beta+1}, \end{aligned}$$

and, for $n\geq2$,

$$\begin{aligned} p_n(t)&=\frac{n+\beta}{n+2\beta}(1+t)p_{n-1}(t)\\ &\quad-\frac{n}{n+2\beta}t p_{n-2}(t). \end{aligned} \tag{1}$$

For the rising factorial $(x)_n=x(x+1)\cdots(x+n-1)$, with $(x)_0=1$, the constant term of (1) gives

$$c_{n,0}=\frac{(\beta+1)_n}{(2\beta+1)_n}>0.$$

Consequently, the normalized polynomials $H_n=p_n/c_{n,0}$ satisfy

$$\begin{gathered} H_0=1,\qquad H_1=1+a t,\\ a=\frac{\beta}{\beta+1}\in(1/2,1), \end{gathered}$$ $$\begin{aligned} H_n&=(1+t)H_{n-1}\\ &\quad-(1-r_n)tH_{n-2},\\[3pt] r_n&=\frac{\beta(\beta-1)}{(n+\beta)(n+\beta-1)}. \end{aligned} \tag{2}$$

Indeed,

$$1-r_n=\frac{n(n+2\beta-1)}{(n+\beta)(n+\beta-1)}.$$

The quantities $r_n$ are positive and strictly decreasing in $n$. We will prove a recurrence lemma that only requires

$$\begin{gathered} \frac12\leq a\leq1,\\ r_2\geq r_3\geq\cdots\geq r_n\geq0. \end{gathered} \tag{3}$$

No upper bound on the $r_i$ is required for that lemma.

2. Symmetric unimodal polynomials

Write

$$S_L(t)=1+t+\cdots+t^L\qquad(L\geq0).$$

A polynomial is called symmetric of order $N$ if its coefficient array, padded by zeros to indices $0,\ldots,N$, is symmetric under $k\mapsto N-k$. Its actual degree may be less than $N$.

Nonnegative symmetric unimodal polynomials of order $N$ are precisely the nonnegative linear combinations of

$$\begin{gathered} t^jS_{N-2j}(t),\\ 0\leq j\leq\lfloor N/2\rfloor. \end{gathered} \tag{4}$$

To see this, the coefficient multiplying the term with index $j$ is the difference between the coefficients at $j$ and $j-1$, with the coefficient at $-1$ set to zero. The converse follows directly from the interval of ones in each summand.

Products of such polynomials are again nonnegative, symmetric and unimodal, with their orders added. One elementary proof uses

$$\begin{aligned} &S_b(t)S_c(t)\\ &\quad=\sum_{j=0}^{\min(b,c)}t^jS_{b+c-2j}(t), \end{aligned} \tag{5}$$

whose coefficients on both sides count the pairs of integers in $[0,b]\times[0,c]$ with prescribed sum, and then applies (4).

In particular, a nonnegative symmetric unimodal polynomial of order $N$, or of order $N-1$, has a mode at $\lfloor N/2\rfloor$ after padding to $0,\ldots,N$. For even $N$, an order-$N-1$ polynomial has equal middle coefficients at $N/2-1,N/2$. For odd $N$, an order-$N$ polynomial has equal middle coefficients at $(N-1)/2,(N+1)/2$. Thus nonnegative sums of these two types are unimodal with the same specified mode.

3. A prefix-sum identity

For integers $L\geq0$ and $0\leq h\leq L$, set

$$\begin{aligned} &U_{L,h}(t)\\ &\quad=\sum_{\ell=0}^h\bigl(S_\ell(t)+1\bigr)S_{L-\ell}(t). \end{aligned}$$

The following identity will supply the needed positivity:

$$\begin{aligned} U_{L,h}(t)&=(h+2)S_L(t)\\ &\quad+\sum_{j=1}^{\min(h,L-h)}t^jS_{L-2j}(t)\\ &\quad+\sum_{\ell=1}^h S_{\ell-1}(t)S_{L-\ell}(t). \end{aligned} \tag{6}$$

Empty sums are zero. The first two terms on the right together are symmetric unimodal of order $L$; the last sum is symmetric unimodal of order $L-1$. When $L=0$, the latter is zero.

Here is an algebraic verification valid for every $L,h$. Subtract the last sum in (6) from the definition of $U_{L,h}$, and use $S_\ell-S_{\ell-1}=t^\ell$. The result is

$$2S_L+\sum_{\ell=1}^h(1+t^\ell)S_{L-\ell}. \tag{7}$$

For $1\leq\ell\leq L$, define

$$E_\ell=(1+t^\ell)S_{L-\ell}-S_L.$$

If $2\ell\leq L$, then $E_\ell=t^\ell S_{L-2\ell}$. If $2\ell=L+1$, then $E_\ell=0$. The remaining terms obey

$$E_\ell=-E_{L+1-\ell}.$$

Pairing these terms in the prefix $\ell=1,\ldots,h$ leaves exactly

$$\begin{aligned} &\sum_{\ell=1}^h E_\ell\\ &\quad=\sum_{j=1}^{\min(h,L-h)}t^jS_{L-2j}. \end{aligned}$$

Substitution into (7) proves (6), including $h=0$, $h=L$, and both parities of $L$.

If $w_0\geq\cdots\geq w_L\geq0$, ordinary finite summation by parts gives

$$\begin{aligned} &\sum_{\ell=0}^L w_\ell(S_\ell+1)S_{L-\ell}\\ &\quad=w_L U_{L,L}\\ &\qquad+\sum_{h=0}^{L-1}(w_h-w_{h+1})U_{L,h}. \end{aligned} \tag{8}$$

All weights on the right are nonnegative, including the final boundary weight $w_L$. Equations (6) and (8) therefore decompose the left side into nonnegative symmetric unimodal parts of orders $L$ and $L-1$.

4. The path-matching expansion

Fix $n\geq1$, and consider a path with vertices $1,\ldots,n$. Label its edge $\{i-1,i\}$ by $i$, for $2\leq i\leq n$. A matching is a set $M\subseteq\{2,\ldots,n\}$ with no consecutive indices. Write

$$s=|M|,\qquad r_M=\prod_{i\in M}r_i,$$

where an empty product is one. After deleting the endpoints of its marked edges, let $L_0,L_1,\ldots,L_s$ be the lengths of the remaining consecutive vertex segments, in order; zero lengths are allowed. Thus

$$L_0+\cdots+L_s=n-2s.$$

The recurrence (2), for arbitrary $a$, has the expansion

$$\begin{aligned} H_n(t)&=\sum_M r_M t^s\\ &\quad\cdot\bigl(aS_{L_0}(t)+1-a\bigr)\\ &\quad\cdot\prod_{j=1}^s S_{L_j}(t). \end{aligned} \tag{9}$$

For completeness, (9) can be derived from a tridiagonal determinant. Take diagonal entries $1+at,1+t,\ldots,1+t$, superdiagonal entries one, and subdiagonal entry $(1-r_i)t$ at edge $i$. Expansion along the last row gives exactly (2), with the stated initial conditions. In the determinant's permutation expansion, every nontrivial cycle is an adjacent transposition, so the edge transpositions form a matching. Expand each factor $-(1-r_i)t$ into $-t+r_i t$, and fix the edges where the second term was chosen. These edges form the marked matching $M$. The remaining determinant factors over the undeleted path segments, with all remaining edge parameters equal to zero. A standard segment of length $L$ has determinant $S_L$, since it satisfies $S_L=(1+t)S_{L-1}-tS_{L-2}$; the initial segment has determinant $aS_L+1-a$. This proves (9).

Let $Q_n$ denote the same recurrence with $a=1$. Formula (9) shows that $Q_n$ is a nonnegative sum of products

$$t^s\prod_{j=0}^sS_{L_j},$$

each symmetric unimodal of order $n$. Hence $Q_n$ has that property.

It remains to treat $a=1/2$; denote the corresponding polynomial by $K_n$. In (9), the empty matching contributes $(S_n+1)/2$. For a nonempty matching, multiplication by two changes the initial factor into $S_{L_0}+1$.

Group the nonempty matchings by keeping all marked edges except the first fixed. There are two cases.

• If there are no other marked edges, the first edge ranges over $i=2,\ldots,n$. Put $L=n-2$, $\ell=i-2$, and $C(t)=1$.
• Otherwise, let $j$ be the first of the fixed later marked edges. The first edge ranges over $i=2,\ldots,j-2$. Put $L=j-4$, $\ell=i-2$, and let $C(t)$ be the product of the $S$-polynomials for the fixed segments after edge $j$.

Groups with no admissible first edge are omitted. In either case, $0\leq\ell\leq L$, the two initial segment lengths are $\ell,L-\ell$, and the group's contribution to $2K_n$ is

$$\begin{aligned} &r_{M'}t^s C(t)\\ &\quad\cdot\sum_{\ell=0}^L r_{\ell+2}(S_\ell+1)S_{L-\ell}, \end{aligned} \tag{10}$$

where $M'$ is the fixed set of later marked edges and $s=|M'|+1$. If $D$ is the degree of $C$, then

$$L+D=n-2s. \tag{11}$$

The weights $r_{\ell+2}$ in (10) are nonnegative and nonincreasing by (3). Apply (8) and then (6). The factor $C$ is a product of $S$-polynomials, and multiplying also by $t^s$ adds $2s$ to the symmetry order. By (11), every term obtained is symmetric unimodal of order $n$ or $n-1$, with a nonnegative scalar coefficient.

Adding the empty matching therefore proves that

$$\begin{aligned} 2K_n(t)-1&=S_n(t)\\ &\quad+A_n(t)+B_n(t), \end{aligned} \tag{12}$$

where $A_n$ and $B_n$ are nonnegative symmetric unimodal polynomials of orders $n$ and $n-1$, respectively. Either may be zero. Thus $2K_n-1$ is unimodal with a mode at $\lfloor n/2\rfloor$.

For $n=1$, there are no nonempty matchings and (12) simply reads $2K_1-1=S_1$. For $n=2$, the sole nonempty matching has $L=0$, and (10) is $2r_2t$, a nonnegative order-two term. These cases require no negative-index $S$-polynomial.

Finally, the recurrence is linear in its initial conditions. For every $a\in[1/2,1]$,

$$H_n=(2a-1)Q_n+2(1-a)K_n.$$

Consequently,

$$\begin{aligned} H_n-(1-a)&=(2a-1)Q_n\\ &\quad+(1-a)(2K_n-1). \end{aligned} \tag{13}$$

Both scalar weights are nonnegative, and both polynomials on the right are unimodal with a mode at $\lfloor n/2\rfloor$. This proves that the full coefficient sequence of $H_n-(1-a)$ has that mode.

5. Conclusion for the original coefficients

Equation (13) changes only the coefficient at index zero. Hence, for $n\geq2$, the coefficients of $H_n$ at indices $1,\ldots,n$ are nondecreasing up to $\lfloor n/2\rfloor$ and nonincreasing afterward. For $n=1$, the sequence has one term and is unimodal. Multiplication by the positive constant $c_{n,0}$ preserves every inequality. Applying (3) to the parameters in (2) proves the theorem for all real $\alpha>2$ and all $n\geq1$.