Unimodality conjecture for products of q-analogs

Infinitely many counterexamples to the claimed necessity, beginning immediately beyond the tested range. The conjecture asserts that if $k\le3$ or $r\le3$, then unimodality of

$$F(q)=\prod_{i=1}^{k}[a_i]_q\,[b]_{q^r}$$

requires either $r\mid a_i$ for some $i$, or

$$b\le1+\sum_{i=1}^{k}\left\lfloor\frac{a_i}{r}\right\rfloor.$$

Set

$$r=3,\qquad b=2,\qquad k=6,\qquad a_1=a_2=a_3=a_4=a_5=a_6=2.$$

No $a_i$ is divisible by $3$, and

$$1+\sum_{i=1}^6\left\lfloor\frac23\right\rfloor=1<2=b.$$

Thus both supposedly necessary alternatives fail. Nevertheless,

$$\begin{aligned} F(q)&=(1+q)^6(1+q^3)\\ &=1+6q+15q^2+21q^3+21q^4+21q^5\\ &\hspace{2em}+21q^6+15q^7+6q^8+q^9, \end{aligned}$$

whose coefficient sequence

$$(1,6,15,21,21,21,21,15,6,1)$$

is symmetric and unimodal. Since $r=3$, this lies directly within the conjectured necessity range.

More generally, for every $k\ge6$, take $r=3$, $b=2$, and $a_1=\cdots=a_k=2$. The two numerical alternatives still both fail, while

$$F(q)=(1+q)^{k-6}\bigl((1+q)^6(1+q^3)\bigr)$$

remains symmetric and unimodal. Indeed, if $(u_0,\ldots,u_d)$ is symmetric, unimodal, and nonnegative, the coefficients of $(1+q)\sum_j u_jq^j$ are $v_j=u_j+u_{j-1}$; they are symmetric and increase up to the midpoint because

$$v_{j+1}-v_j=u_{j+1}-u_{j-1}\ge0.$$

Induction proves the claim for every $k\ge6$.

The authors explicitly tested only $k\le5$, explaining why this entire infinite family was missed. This disproves the asserted necessity when $r\le3$; the separate sufficiency assertion and the $k\le3$ necessity assertion are not decided here.

Source: Connelly, Ito, Martinez, Shevchenko, and Yang, Unimodality of $q$-Fibonomial coefficients for small cases, Conjecture 5.4.

Proof of the sufficient condition in Conjecture 5.4

The sufficient condition in Conjecture 5.4 of Connelly–Ito–Martinez–Shevchenko–Yang holds for all the stated parameters. The necessity assertion for r ≤ 3 has already been refuted in Shivam Patel's solution on this MathDB entry. The proof below addresses the sufficient direction separately.

Write Q(n; z) = 1 + z + … + z^(n−1). For integers r ≥ 2, k ≥ 1, and positive integers a₁, …, aₖ, b, set

P(q) = Q(a₁; q) · … · Q(aₖ; q) · Q(b; q^r),
T = floor(a₁/r) + … + floor(aₖ/r).

Claim. If r divides some aᵢ, or b ≤ T + 1, then P is symmetric and unimodal.

We use two elementary closure facts for polynomials with nonnegative coefficients: products of symmetric unimodal polynomials are symmetric unimodal; sums of such polynomials with the same center are symmetric unimodal. The product fact follows by decomposing each symmetric unimodal polynomial of degree d into nonnegative linear combinations of centered intervals q^j Q(d−2j+1; q). Products of two interval polynomials have triangular or trapezoidal coefficients. When comparing centers, extend coefficient sequences by zero outside their supports.

If aᵢ = rt, then

Q(aᵢ; q) Q(b; q^r) = Q(r; q) Q(t; q^r) Q(b; q^r).

This repeats each coefficient of the symmetric unimodal polynomial Q(t; x) Q(b; x) exactly r times. Multiplying by the remaining factors preserves the property. This divisibility case is already established in Corollary 4.3 of the original paper.

It remains to prove the numerical condition when no aᵢ is divisible by r. Induct on b. For b = 1, the product consists entirely of ordinary q-integers. For b ≥ 2, the inequality b ≤ T + 1 implies T ≥ 1. Hence some aⱼ > r, since equality aⱼ = r is excluded. Put a = aⱼ and let R(q) be the product of all ordinary factors except Q(aⱼ; q). Expanding geometric sums gives

Q(a; q) Q(b; q^r) = Q(a+r(b−1); q) + q^r Q(a−r; q) Q(b−1; q^r).

Replacing a by a − r > 0 preserves nondivisibility and reduces T by exactly one. Therefore b − 1 ≤ 1 + (T − 1), so the induction hypothesis applies to

B(q) = R(q) Q(a−r; q) Q(b−1; q^r).

Let D = deg P = (a₁−1) + … + (aₖ−1) + r(b−1). After multiplication by R(q), the identity expresses P as the sum of two polynomials. The first, R(q) Q(a+r(b−1); q), is symmetric unimodal of degree D. The second is q^r B(q). Since deg B = D−2r, this second summand has exponents r through D − r and center D/2. Both summands have the same center D/2, so their sum P is symmetric and unimodal. This completes the induction.

Existing counterexample. Patel's parameters r = 3, k = 6, all aᵢ = 2, b = 2 give (1+q)^6(1+q^3) with coefficients 1, 6, 15, 21, 21, 21, 21, 15, 6, 1, despite neither condition holding. Its unimodality also follows from Handelman's 2011 theorem. No priority is claimed for this counterexample or its unimodality.

Verification and scope. This note was developed with AI assistance. Local verification with Lean 4.33.1 checked the explicit counterexample and the displayed algebraic identity; the sources and verification record are attached. The complete sufficient-direction proof above has not been formalized in Lean. No historical priority is claimed for this proof. Necessity for k ≤ 3 and the general q-Fibonomial conjecture are not settled here.