General reverse-unimodality conjecture for beta-values of lists

About 24 years old · traced to 2002

Let $l$ be a positive integer, and let $i$ and $j$ be non-negative integers with $j>i$ . For $1\leq k\leq l$ , let $L_{i,j}^k$ be the list of length $l$ whose entries are $i$ except for $j$ in position $k$ , and define $a_{i,j}^k=\beta(L_{i,j}^k)$ . General reverse-unimodality conjecture. For fixed $i$ , $j$ , and $l$ , the sequence $\{a_{i,j}^k\}_{k=1}^l$ is reverse unimodal in $k$ . This generalizes the preceding reverse-unimodality result for the sequence $\beta(0^i,2,0^{n-i})$ ; the source gives no resolution of the conjecture.

References

Primary source

Swapneel Mahajan, “The cd-index of the Boolean lattice”, arXiv:math/0211390 (2002).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Open

A reader-provided calculation claims the conjecture is false in a five-entry example, but no independent verification was found.

The conjecture predicts a symmetry-shaped decrease toward the middle when one exceptional list entry is moved through all positions. No proposer or date is identified in the retrieved material.

Posted attempt

An unverified complete counterexample takes $i=2$ , $j=3$ , and $l=5$ , yielding the values $20{,}985{,}240{,}240$ , $18{,}526{,}714{,}104$ , $18{,}599{,}890{,}400$ , $18{,}526{,}714{,}104$ , and $20{,}985{,}240{,}240$ . Since $a_{2,3}^3-a_{2,3}^2=73{,}176{,}296>0$ , it claims the required middle decrease fails. The recurrence and arithmetic have not been independently verified.

Current status (as of August 2026): the conjecture has no verified proof, but a posted calculation claims a counterexample for $i=2$ , $j=3$ , and $l=5$ ; confirmation or refutation remains open.

CounterexampleThis solution needs a summarySee full solutionHide full solution

The proposed reverse-unimodality statement is false already for $cd$ -monomials of degree $19$ .

Take

i=2,\qquad j=3,\qquad \ell=5,

so $i<j$ . Form the five lists containing one entry $j=3$ and four entries $i=2$ , with the position of $3$ varying from first to fifth:

\begin{aligned} L_1&=(3,2,2,2,2),\\ L_2&=(2,3,2,2,2),\\ L_3&=(2,2,3,2,2),\\ L_4&=(2,2,2,3,2),\\ L_5&=(2,2,2,2,3). \end{aligned}

All five lists have degree

3+4\cdot2+2(5-1)=19.

The exact coefficients follow from Lemmas 3.2 and 4.4 of the source. For every nonempty list $a=(a_1,\ldots,a_r)$ ,

\begin{aligned} \beta(a) ={}&\sum_{\substack{1\leq t\leq r\\a_t>0}} \beta(a_1,\ldots,a_t-1,\ldots,a_r)\\ &+\sum_{t=1}^{r-1} \beta(a_1,\ldots,a_{t-1},a_t+a_{t+1}+1, a_{t+2},\ldots,a_r), \end{aligned}

with $\beta((0))=1$ . Every term on the right has degree one less, so this is a finite exact integer recurrence.

Its values are

\begin{aligned} \beta(L_1)&=20{,}985{,}240{,}240,\\ \beta(L_2)&=18{,}526{,}714{,}104,\\ \beta(L_3)&=18{,}599{,}890{,}400,\\ \beta(L_4)&=18{,}526{,}714{,}104,\\ \beta(L_5)&=20{,}985{,}240{,}240. \end{aligned}

In particular,

\beta(L_3)-\beta(L_2)=73{,}176{,}296>0.

A reverse-unimodal symmetric sequence of length five must decrease weakly from its first entry to its middle entry, requiring $\beta(L_2)\geq\beta(L_3)$ . Here the opposite strict inequality holds. Therefore the conjectured reverse unimodality fails.