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$ \newcommand{mset}[2]{\left\{ #1 \;\middle|\; #2 \right\}} \newcommand{PPP}{\mathcal{P}} \newcommand{CCC}{\mathcal{C}} \newcommand{FFF}{\mathcal{F}} \newcommand{RR}{\mathbb{R}} \newcommand{ZZ}{\mathbb{Z}} \newcommand{NN}{\mathbb{N}} \newcommand{comp}{\operatorname{com}} \newcommand{mvec}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{mmat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{mmod}{\; \operatorname{mod} \;} \newcommand{lcm}{\operatorname{lcm}} \newcommand{ehr}{\operatorname{ehr}} $
Combinatorial Witnesses for Congruences of Restricted Partition Functions via Ehrhart Theory
Felix Breuer
RISC, JKU, Linz
FWF SFB Algorithmic and Enumerative Combinatorics
joint work with
Dennis Eichhorn (UC Irvine) and Brandt Kronholm (RISC)
Partition of 18 with 3 parts: $ 18 = 9 + 5 + 4 $
\begin{eqnarray*} p(n,d) &=\;\; \#&&\text{partitions of $n$} \\ &&&\text{into exactly $d$ parts} \end{eqnarray*}
\begin{eqnarray*} p(s\cdot k + r,d) &\equiv& 0 \mmod m \text{ for all $k\in\ZZ_{\geq 0}$} \end{eqnarray*}
\begin{eqnarray*} p(6k,3) &\equiv& 0 \mmod 3 \text{ for all $k\in\ZZ_{\geq 0}$} \end{eqnarray*}
Witnesses?
partitions $\leftrightarrow$ integer points
$$9 + 5 + 4 \leftrightarrow (9,5,4) $$
$p(18,3)$
$p(18,3)$
$p(24,3)$
$p(30,3)$
\begin{align*} p(n,d) = \#\ZZ^d \cap n\cdot\PPP_d =: \ehr_{\PPP_d}(n) \end{align*}
\begin{align*} \color{white}{n\cdot}\PPP_d = \left\{ x\in\RR^d \; \middle| \right. \; & \sum x_i = 1, \\ & \left. x_1\geq x_2 \geq \ldots \geq x_d > 0 \right\} \end{align*}
\begin{align*} n\cdot\PPP_d = \left\{ x\in\RR^d \; \middle| \right. \; & \sum x_i = n, \\ & \left. x_1\geq x_2 \geq \ldots \geq x_d > 0 \right\} \end{align*}
\begin{align*} \PPP_3 = \left\{ x\in\RR^3_{\geq 0} \middle| \right. \; & x_1+x_2+x_3 = 1, \end{align*}
\begin{align*} \PPP_3 = \left\{ x\in\RR^3_{\geq 0} \middle| \right. \; & x_1+x_2+x_3 = 1, \\ & \left. \color{red}{x_1\geq x_2}, \color{green}{x_2\geq x_3}, \color{blue}{x_3 > 0} \right\} \end{align*}
Example
\begin{align*} \CCC_d &= \mset{x\in\RR^d}{x_1 \geq \ldots \geq x_d > 0} \\ \end{align*}
\begin{align*} n\PPP_d &= \mset{x\in\RR^d}{\sum x_i = n} \cap \CCC_d \end{align*}
\begin{align*} \CCC_d &= V \left( \RR_{\geq 0}^{d-1} \times \RR_{> 0} \right) \end{align*}
$V = \mmat{ \color{red}{\lcm(d)} & \color{green}{\frac{\lcm(d)}{2}} & \ldots & \color{blue}{\frac{\lcm(d)}{d}} \\ \color{red}{0} & \color{green}{\frac{\lcm(d)}{2}} & \ldots & \color{blue}{\frac{\lcm(d)}{d}} \\ \color{red}{0} & \color{green}{0} & & \color{blue}{\frac{\lcm(d)}{d}} \\ \color{red}{\vdots} & \color{green}{\vdots} & & \color{blue}{\vdots} \\ \color{red}{0} & \color{green}{0} & & \color{blue}{\frac{\lcm(d)}{d}} }$
$\lcm(d) := \lcm(1,2,\ldots,d)$
For $d=3$: $\lcm(3)=6$ and $V = \mmat{ \color{red}{6} & \color{green}{3} & \color{blue}{2} \\ \color{red}{0} & \color{green}{3} & \color{blue}{2} \\ \color{red}{0} & \color{green}{0} & \color{blue}{2} }$
\begin{align*} \CCC_d &= V \left( \RR_{\geq 0}^{d-1} \times \RR_{> 0} \right) \end{align*}
\begin{align*} \FFF_d &= V\left( [0,1)^{d-1}\times (0,1]\right) \end{align*}
\begin{align*} \CCC_d &= V\left(\RR_{\geq 0}^{d-1} \times \RR_{> 0} \right) \\ \FFF_d &= V\left( [0,1)^{d-1}\times (0,1]\right) \end{align*}
\begin{align*} \CCC_d &= V\ZZ_{\geq 0}^3 + \FFF_d \end{align*}
\begin{align*} \CCC_d &= V\left(\RR_{\geq 0}^{d-1} \times \RR_{> 0} \right) \\ \FFF_d &= V\left( [0,1)^{d-1}\times (0,1]\right) \end{align*}
\begin{align*} \CCC_d &= V\ZZ_{\geq 0}^3 + \FFF_d \end{align*}
$$ \frac{ \sum_{v\in \FFF_3} x^v } { (1-x^{(6,0,0)})(1-x^{(3,3,0)})(1-x^{(2,2,2)}) } \;\; \longleftrightarrow \;\; \frac{\cdots}{(1-q^6)^3} $$
\begin{align*} &\sum_{n\geq 0} p(n,3) q^n = \frac{q^3}{(1-q)(1-q^2)(1-q^3)} = \frac{q^3(1+q+q^2+q^3+q^4+q^5)(1+q^2+q^4)(1+q^3)}{(1-q^6)^3} \\ = &\frac{0q^0 + 0q^1 + 0q^2 + 1 q^3 + 1q^4 + 2q^5 + 3q^6 + 4 q^7 + 5q^8 + 4 q^9 + 5q^{10} + 4q^{11} + 3q^{12} + 2q^{13} + 1q^{14} + 1q^{15} + 0q^{16} + 0q^{17} }{(1-q^6)^3} \end{align*}
\begin{align*} p(6k+0,3) =& 0\binom{k+2}{2} + 3\binom{k+1}{2} + 3\binom{k}{2} \\ p(6k+1,3) =& 0\binom{k+2}{2} + 4\binom{k+1}{2} + 2\binom{k}{2} \\ p(6k+2,3) =& 0\binom{k+2}{2} + 5\binom{k+1}{2} + 1\binom{k}{2} \\ p(6k+3,3) =& 1\binom{k+2}{2} + 4\binom{k+1}{2} + 1\binom{k}{2} \\ p(6k+4,3) =& 1\binom{k+2}{2} + 5\binom{k+1}{2} + 0\binom{k}{2} \\ p(6k+5,3) =& 2\binom{k+2}{2} + 4\binom{k+1}{2} + 0\binom{k}{2} \end{align*}
Theorem. [B-Eichhorn-Kronholm]
There are explicit cycles witnessing the congruences $$ p(\ell\lcm(\ell)k+r,\ell) \equiv 0 \mod \ell $$ for all $\ell\in\NN$ prime, $-\binom{\ell}{2} < r < \ell$ and $1\leq k \in \ZZ$.
Example: $r=1$, $\ell=3$. Here $3 \lcm(3) = 18$. $$ p(18\cdot k + 1,3) \equiv 0 \mmod 3 $$ In particular $ p(19,3) \equiv 0 \mmod 3 $.
Theorem. [B-Eichhorn-Kronholm]
There are explicit cycles witnessing the congruences $$ p(\pi_{d,\ell}k+r,d) \equiv 0 \mod \ell $$ for all $\ell\in\NN$ prime, $d\in\NN$, $-\binom{d}{2} < r < d$ and $1\leq k \in \ZZ$.
Example: $d=3$, $r=1$, $\ell=5$. Here $\pi_{d,\ell} = 5\lcm(3) = 30$. $$ p(30\cdot k + 1,3) \equiv 0 \mmod 5 $$ In particular $ p(31,3) \equiv 0 \mmod 5 $.
Theorem. [B-Eichhorn-Kronholm]
There are explicit cycles witnessing the congruences $$ p(\lcm(\ell)k - t\ell,\ell) \equiv 0 \mod \ell $$ for all $\ell\in\NN$ an odd prime, $0 \leq t \leq \frac{\ell-3}{2}$ and $1\leq k \in \ZZ$.
Example: $p(6k,3) \equiv 0 \mod 3$
Example: $p(115,5) \equiv 0 \mod 5$. $\lcm(4) = 12$, $\lcm(5)=60$.
A crank for a given artihmetic progression of divisibilities modulo $m$ is a function $c_n:P(n,d) \rightarrow \{ 0,\ldots,m-1 \}$ such that $$ |c^{-1}(0)| = \ldots = |c^{-1}(m-1)| $$ for all $n$ in the arithmetic progression.
A supercrank is a crank that witnesses divisibility for any $n$ such that $m|p(n,d)$.
Theorem. [B-Eichhorn-Kronholm]
If $m$ is a prime of the form $6j-1$, then \[ p(n,3) \equiv 0 \mod m \] if and only if $n = \pm 0,1,2,2m-2,2m+1 \mod 6m$. Moreover, $$ c(\lambda) = (\lambda_1 - \lambda_3) \mod m$$ is a supercrank witnessing all of these divisibilities.
How many constituents of $p(\operatorname{lcm}(d)k+r,d)$ have coefficients with non-trivial greatest common divisor?
#parts | #constituents | #non-trival gcds | % non-trivial gcds | examples (gcd, frequency) | ||
3 | 6 | 3 | 50% | [(3, 1), (2, 2)] | ||
4 | 12 | 6 | 50% | [(6, 2), (3, 4)] | ||
5 | 60 | 4 | 7% | [(5, 4)] | ||
6 | 60 | 22 | 37% | [(10, 4), (5, 6), (2, 12)] | ||
7 | 420 | 203 | 48% | [(196, 1), (14, 4), (7, 8), (4, 12), (2, 178)] | ||
8 | 840 | 118 | 14% | [(14, 3), (7, 3), (343, 4), (49, 6), (2, 102)] | ||
9 | 2520 | 460 | 18% | [(10, 20), (5, 100), (2, 340)] | ||
10 | 2520 | 118 | 5% | [(5, 118)] | ||
11 | 27720 | 7277 | 26% | [(1815, 3), (33, 6), (605, 6), (22, 28), (30, 128), (10, 156), (11, 188), (15, 228), (6, 360), (5, 826), (3, 2510), (2, 2838)] | ||
12 | 27720 | 6468 | 23% | [(66, 4), (165, 4), (110, 20), (30, 20), (121, 40), (33, 68), (22, 74), (10, 108), (6, 112), (15, 150), (55, 152), (11, 822), (2, 1432), (3, 1690), (5, 1772)] |
$$\pi_{d,\ell^a} = \ell^{r_\ell(d)+a-1}L_d$$ $$L = \text{the $\ell$-free part of $\lcm(d)$}$$ $$ r_\ell(d) = \text{the least integer such that $\ell^{r_\ell(d)} \geq \sum_{i=1}^d \ell^{\nu_\ell(i)}$}$$ $$ \nu_\ell(m) = \text{the largest integer such that $\ell^{\nu_\ell(m)} | m$} $$
Example: $\pi_{\ell,\ell}$ $$L = \lcm(\ell-1)$$ $$ \sum = \underbrace{1+1+\ldots+1}{\ell-1} + \ell = 2\ell -1 \leq \ell^2$$ $$\pi_{\ell,\ell} = \ell^2 \lcm(\ell-1) = \ell\lcm(\ell)$$
Example: $\pi_{14,3}$ $$\lcm(14) = 1 \cdot 2 \cdot 3 \cdot 2 \cdot 5 \cdot 1 \cdot 7 \cdot 2 \cdot 3 \cdot 1 \cdot 11 \cdot 1 \cdot 13 \cdot 1$$ $$L = 1 \cdot 2 \cdot 1 \cdot 2 \cdot 5 \cdot 1 \cdot 7 \cdot 2 \cdot 1 \cdot 1 \cdot 11 \cdot 1 \cdot 13 \cdot 1$$ $$ \sum = 1+1+3+1+1+3+1+1+9+1+1+3+1+1 = 28 \leq 3^4$$ $$\pi_{14,3} = 3^4 L = 3^2 \lcm(14) $$