Both proposed hook-length inequalities for 5-core partitions are false

Problem. MathDB #373536, the hook-length bias conjecture for 5-core partitions.

Primary source. Nayandeep Deka Baruah, Hirakjyoti Das, Pankaj Jyoti Mahanta, and Manjil P. Saikia, Hook Length Biases in $t$-Core Partitions, arXiv:2603.10140v2, Conjecture 1.5.

For a partition $\lambda$, let $H_h(\lambda)$ denote its number of cells with hook length $h$. The source defines

$$a_{5,h}(n) =\sum_{\substack{\lambda\vdash n\\ \lambda\text{ is a }5\text{-core}}} H_h(\lambda). \tag{1}$$

Here a partition is a $5$-core precisely when none of its hook lengths is divisible by $5$. Conjecture 1.5 asserts, for every $n\geq0$, that

$$a_{5,1}(n)\geq a_{5,3}(n)\geq a_{5,6}(n). \tag{2}$$

We show that both inequalities fail separately. Their sharp first counterexamples are

$$\begin{aligned} \bigl(a_{5,1}(93),a_{5,3}(93),a_{5,6}(93)\bigr) &=(382,384,284),\\ \bigl(a_{5,1}(793),a_{5,3}(793),a_{5,6}(793)\bigr) &=(10701,9732,9746). \end{aligned} \tag{3}$$

Thus

$$382<384, \qquad 9732<9746. \tag{4}$$

The first inequality holds at every smaller $n<93$, while the second holds at every smaller $n<793$.

1. A complete finite parametrization of 5-core partitions

Write the beta set of a partition as

$$\mathcal B(\lambda) =\{\lambda_i-i:i\geq1\}\subseteq\mathbb Z. \tag{5}$$

The empty partition has beta set consisting of all negative integers. The usual bead-removal criterion says that $\lambda$ is a $5$-core if and only if

$$b\in\mathcal B(\lambda) \quad\Longrightarrow\quad b-5\in\mathcal B(\lambda). \tag{6}$$

Consequently, on each residue class $r\in\{0,1,2,3,4\}$, its occupied positions form a lower ray. There are unique integers $c_0,\ldots,c_4$ such that

$$r+5j\in\mathcal B(\lambda) \quad\Longleftrightarrow\quad j<c_r. \tag{7}$$

Because the beta set comes from a partition, the number of beads moved above the empty-partition boundary equals the number of holes created below it. Equivalently,

$$c_0+c_1+c_2+c_3+c_4=0. \tag{8}$$

Conversely, every integer vector satisfying (8) defines a balanced beta set by (7), hence exactly one $5$-core partition.

Its size is

$$|\lambda| =\frac52\sum_{r=0}^4c_r^2 +\sum_{r=0}^4r c_r. \tag{9}$$

For completeness, moving the runner boundary from $0$ to $c_r$ changes the beta-set energy by

$$5\binom{c_r}{2}+r c_r,$$

where the same polynomial also applies when $c_r<0$. Summing over runners and using (8) gives (9).

Completing squares turns the size condition into the particularly useful finite enumeration criterion

$$\boxed{ \sum_{r=0}^4(5c_r+r)^2=10|\lambda|+30. } \tag{10}$$

Thus every $5$-core partition of $n$ is represented once and only once by a vector in the explicit finite set

$$\mathcal C_n =\left\{ (c_0,\ldots,c_4)\in\mathbb Z^5: \sum_{r=0}^4c_r=0, \quad \sum_{r=0}^4(5c_r+r)^2=10n+30 \right\}. \tag{11}$$

In particular, no unbounded search or presumed cutoff is involved:

$$|5c_r+r|\leq\left\lfloor\sqrt{10n+30}\right\rfloor \qquad(0\leq r\leq4). \tag{12}$$

2. Counting actual hooks from the balanced beta set

A cell of hook length $h$ corresponds to an occupied beta position $b$ for which $b-h$ is unoccupied. Fix its runner $r$, put

$$s_r(h)=(r-h)\bmod5\in\{0,1,2,3,4\}, \qquad q_r(h)=\frac{r-h-s_r(h)}5, \tag{13}$$

and write $b=r+5j$. By (7), the position $b$ is occupied exactly when $j<c_r$, whereas

$$b-h=s_r(h)+5\bigl(j+q_r(h)\bigr)$$

is unoccupied exactly when $j+q_r(h)\geq c_{s_r(h)}$. The number of such integers $j$ is therefore

$$\max\left\{0,c_r-c_{s_r(h)}+q_r(h)\right\}.$$

Summing independently over the five runners gives the exact hook-count formula

$$\boxed{ H_h(c) =\sum_{r=0}^4 \max\left\{0,c_r-c_{s_r(h)}+q_r(h)\right\}. } \tag{14}$$

Combining (11) and (14) now gives a finite, completely explicit expression for every quantity in the conjecture:

$$a_{5,h}(n)=\sum_{c\in\mathcal C_n}H_h(c). \tag{15}$$

This formula counts the individual Young-diagram cells from the source; it does not replace hooks by distinct part sizes, residue classes, or a different partition statistic.

3. First inequality: the smallest counterexample is 93

For $n=93$, equation (10) becomes

$$\sum_{r=0}^4(5c_r+r)^2=960. \tag{16}$$

There are exactly $46$ solutions satisfying (8). The following table groups them by the first runner coordinate $c_0$; the final three columns are the sums of their actual $1$-, $3$-, and $6$-hook counts from (14).

$c_0$	Number of $5$-cores	$1$-hooks	$3$-hooks	$6$-hooks
$-5$	$1$	$7$	$7$	$5$
$-3$	$9$	$70$	$80$	$51$
$-2$	$2$	$18$	$16$	$14$
$-1$	$10$	$83$	$82$	$62$
$1$	$7$	$61$	$55$	$48$
$3$	$12$	$101$	$101$	$74$
$5$	$5$	$42$	$43$	$30$
Total	$46$	$382$	$384$	$284$

Hence

$$a_{5,1}(93)-a_{5,3}(93)=382-384=-2<0. \tag{17}$$

The second inequality does hold at this size, since $384>284$.

4. Second inequality: the smallest counterexample is 793

For $n=793$, the corresponding complete enumeration condition is

$$\sum_{r=0}^4(5c_r+r)^2=7960. \tag{18}$$

This time there are exactly $396$ $5$-core partitions. Grouping them by the same runner coordinate gives the complete certificate below.

$c_0$	Number of $5$-cores	$1$-hooks	$3$-hooks	$6$-hooks
$-15$	$5$	$126$	$103$	$113$
$-14$	$10$	$256$	$244$	$234$
$-13$	$17$	$437$	$449$	$397$
$-11$	$16$	$472$	$375$	$432$
$-10$	$8$	$220$	$200$	$200$
$-9$	$16$	$458$	$383$	$417$
$-7$	$12$	$363$	$249$	$333$
$-6$	$24$	$632$	$597$	$571$
$-5$	$35$	$985$	$803$	$900$
$-3$	$28$	$744$	$700$	$677$
$-2$	$10$	$246$	$266$	$226$
$-1$	$16$	$443$	$385$	$404$
$1$	$27$	$701$	$687$	$639$
$2$	$20$	$542$	$492$	$491$
$3$	$32$	$864$	$777$	$790$
$5$	$15$	$395$	$385$	$358$
$6$	$4$	$108$	$98$	$99$
$7$	$20$	$572$	$461$	$522$
$9$	$16$	$423$	$409$	$382$
$10$	$20$	$536$	$506$	$489$
$11$	$18$	$467$	$482$	$426$
$13$	$19$	$517$	$480$	$473$
$15$	$8$	$194$	$201$	$173$
Total	$396$	$10701$	$9732$	$9746$

Therefore

$$a_{5,3}(793)-a_{5,6}(793) =9732-9746 =-14<0. \tag{19}$$

At this size, the other inequality holds, since $10701>9732$. The failures are therefore genuinely independent: neither adjacent comparison in the proposed chain is valid for all $n$.

5. Completeness, minimality, and independent cross-checks

To verify minimality, enumerate all balanced vectors satisfying

$$\sum_{r=0}^4(5c_r+r)^2\leq7960. \tag{20}$$

The finite bound (12) makes this exhaustive. Evaluate (14) for each vector and group by its exact size (9). The resulting complete coefficient lists satisfy

$$\begin{aligned} a_{5,1}(n)&\geq a_{5,3}(n) &&\text{for every }0\leq n<93,\\ a_{5,3}(n)&\geq a_{5,6}(n) &&\text{for every }0\leq n<793. \end{aligned} \tag{21}$$

As an independent completeness check, the number of reconstructed $5$-cores at every size agrees with the classical generating function already cited in the primary source:

$$\sum_{n\geq0}\#\mathcal C_n\,q^n =\prod_{m\geq1}\frac{(1-q^{5m})^5}{1-q^m}. \tag{22}$$

Moreover, at both counterexample sizes the beta sets were converted back into all $46$ and $396$ actual Young diagrams. Their individual cell hooks were counted directly, and every diagram was independently checked to have no hook divisible by $5$. Direct enumeration of every ordinary partition through size $25$ provides another independent verification of both the core criterion and formula (14).

Thus the first failure of the first inequality occurs at $n=93$, the first failure of the second occurs at $n=793$, and Conjecture 1.5 is false in both of its proposed comparisons.