# A lozenge triangulation of the plane with integers

Raghavendra N. Bhat  Department of Mathematics, University of Illinois at
Urbana-Champaign, 1409 West Green Street, Urbana, IL 61801, USA
<EMAIL_ADDRESS>,  Cristian Cobeli* “Simion Stoilow” Institute of
Mathematics of the Romanian Academy, 21 Calea Grivitei Street, P. O. Box
1-764, Bucharest 014700, Romania<EMAIL_ADDRESS>and  Alexandru
Zaharescu  Department of Mathematics, University of Illinois at Urbana-
Champaign, 1409 West Green Street, Urbana, IL 61801, USA, and “Simion Stoilow”
Institute of Mathematics of the Romanian Academy, 21 Calea Grivitei Street, P.
O. Box 1-764, Bucharest 014700, Romania<EMAIL_ADDRESS>
###### Abstract.

We introduce and study a three-folded linear operator depending on three
parameters that has associated a triangular number tilling of the plane. As a
result the set of all triples of integers is decomposed in classes of
equivalence organized in four towers of two-dimensional triangulations. We
provide the full characterization of the represented integers belonging to
each network as families of certain quadratic forms. We note that one of the
towers is generated by a germ that produces a covering of the plane with
Löschian numbers.

###### 2020 Mathematics Subject Classification:

Primary 11B37; Secondary 11B50, 52C20

*Corresponding author: Cristian Cobeli<EMAIL_ADDRESS>
Key words and phrases: dynamical lozenge tiling with integers, Löschian
numbers, rhombus number tiling, modular prime covering, lattice path

##  1\. Introduction

The study of discrete dynamical systems has gained attention because of their
capacity to model complex phenomena through the iterative application of
simple rules. Understanding the common features and applications of these
systems can provide valuable insights into the broader field and their
practical implications.

A few examples that have been discussed again in recent times are related to
various aspects of the Ducci game rule (see [14, 17, 15]), the study of
phenomena occurring in Pascal-like triangles (see [31, 16]) and to the
contrasting patterns produced by the iteration of the PG [4, 6] operator that
calculates the gaps between the neighbor elements of a sequence in relation to
the Proth-Gilbreath Conjecture [30, 24, 23, 20].

In this context, we introduce here the following three-folded linear operator
that has nonlinear characteristics. Let $H=H(x,y,x)$ be the set
$H=\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$ of
transformations that each leaves two components unchanged while to the
negative third adds $1$ and the other two, that is,

$\begin{split}H^{\prime}(x,y,z)&=(-x+1+y+z,y,z),\\\
H^{\prime\prime}(x,y,z)&=(x,-y+1+z+x,z),\\\
H^{\prime\prime\prime}(x,y,z)&=(x,y,-z+1+x+y).\end{split}$ (1)

As we will show, the repeated application of any combination of these
operators produce a set of interconnected integers with remarkable properties.
A special instance of these are the Löschian numbers. Introduced in an
economic model [26, Chapter 10], in which the producers and consumers are
organized in a convenient hexagonal network, Löschian numbers appear also in
the more general geographical Central Place theory [3, 27, 2]. Their special
properties make them useful in other theoretical [18, 21, 22, 19, 25] or
practical contexts (see [29, Chapter 3], [1, 29, 32] and the references
mentioned there).

We denote by $H^{[n]}$, $n\geq 0$, the iterations of $H$, which are the
successive compositions of $H$, where at any step any of
$H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}$ is applied. Thus, if
$(a,b,c)\in\mathbb{Z}^{3}$, then $H^{[n]}(a,b,c)$ contains $3^{n}$ triples of
integers, not necessarily distinct. For example,
$H^{[0]}(1,2,3)=\\{(1,2,3)\\}$ and $(5,9,5)$ is one of the $243$ elements of
$H^{[5]}(1,2,3)$, because

$\begin{split}(1,2,3)&\xrightarrow{\text{ $H^{\prime}$
}}(5,2,3)\xrightarrow{\text{ $H^{\prime\prime\prime}$
}}(5,2,5)\xrightarrow{\text{ $H^{\prime\prime}$ }}(5,9,5)\\\
&\xrightarrow{\text{ $H^{\prime}$ }}(10,9,5)\xrightarrow{\text{ $H^{\prime}$
}}(5,9,5).\end{split}$ (2)

Let $\mathcal{T}_{H}(a,b,c)$ denote the union of all these triples, that is,

$\mathcal{T}_{H}(a,b,c):=\bigcup_{n\geq 0}H^{[n]}(a,b,c)\,.$ (3)

We note that if the union were disjoint, then it would be possible for a
triple to repeat, because for example if the sum of two components, let’s say
$a$ and $b$, is odd and the other is $c=\frac{a+b+1}{2}$, then
$H^{\prime\prime\prime}(a,b,c)=(a,b,c)$. Other type of repetitions also occur,
all due to the symmetries created by all possible orders in which operators
(1) are applied.

We say that an integer is represented by $H^{[n]}(a,b,c)$ if it appears as a
component of a triple in $\mathcal{T}_{H}(a,b,c)$. Since a represented integer
can appear multiple times in the triples from $\mathcal{T}_{H}(a,b,c)$, we
call the length of $m$ as the smallest possible number of compositions
required to go from $(a,b,c)$ to a triple that has $m$ as a component. Thus,
following the sequence in (2), we see that $1,2$ and $3$ have length $0$, then
$5$ has length $1$, and it can be checked that $9$ has length $3$, as there is
no shorter path to obtain it. However, the length of $10$ is not $4$, but also
$3$, since $10$ is represented in this shorter branch:
$(5,2,3)\xrightarrow{\text{ $H^{\prime\prime}$ }}(5,7,3)\xrightarrow{\text{
$H^{\prime\prime\prime}$ }}(5,7,10)$.

Let $\mathcal{R}_{H}(a,b,c)$ denote the set of all integers represented by
$H(a,b,c)$, that is,

$\mathcal{R}_{H}(a,b,c):=\big{\\{}m\in\\{x,y,z\\}:(x,y,z)\in\mathcal{T}_{H}(a,b,c)\big{\\}}\,.$

Given a triple $(a,b,c)\in\mathbb{Z}^{3}$, from the way definitions (1) are
introduced, one should not expect that all or at least most integers are
represented in $\mathcal{R}_{H}(a,b,c)$. And the reason is not the fact that
two variables are added and only one subtracted in formulas (1), and therefore
depending on the signs of $a,b,c$ very small or very large numbers would not
be represented. The real main factor is actually the addition of $1$ in
formulas (1), which always makes that only very few small numbers are
represented. The following theorem proves this fact for numbers less than any
given threshold.

######  Theorem 1.

Let $M$ be a fixed integer. Than, $\mathcal{R}_{H}(a,b,c)\cap(-\infty,M]$ is
finite for any $a,b,c\in\mathbb{Z}$.

Let us note that the sets of represented numbers by $H$ are actually closely
related to one another by the equality

$\mathcal{R}_{H}(a+h,b+h,c+h)=\mathcal{R}_{H}(a,b,c)+h$ (4)

for any $a,b,c,h\in\mathbb{Z}$. Indeed, this follows since if the triple
$(A,B,C)$ is obtained from $(a,b,c)$ through the sequence of operations
$H^{[n]}$, then through exactly the same sequence of operations we obtain (see
Lemma 2.1)

$(a+h,b+h,c+h)\xrightarrow{\text{ $H^{[n]}$ }}(A+h,B+h,C+h)\,.$

In Section 2 we show that essentially there are only two distinct sets of
representatives, namely $\mathcal{R}_{H}(0,0,0)$ and $\mathcal{R}_{H}(0,1,1)$,
and all the others are obtained by translations (4). A combined graphical
representation of the fundamental sets $\mathcal{R}_{H}(0,0,0)$ and
$\mathcal{R}_{H}(0,1,1)$ and the set of triples $\mathcal{T}_{H}(0,0,0)$ and
$\mathcal{T}_{H}(0,1,1)$ is shown in Figure 1.

These tessellations of the plain with integers have the particular
characteristic that in any basic adjacent triangles that together form a
lozenge (diamond consisting of any four close circles positioned in such a way
that each circle is adjacent to at least two of the remaining three) the sum
of the numbers in the nodes on the longer diagonal is with $1$ larger than the
sum of the numbers on the shorter diagonal.

Figure 1. The geometrical representation of the core of
$\mathcal{R}_{H}(0,0,0)$ and $\mathcal{R}_{H}(0,1,1)$ as the sets of nodes in
the tiling of the plane with triples in $\mathcal{T}_{H}(0,0,0)$ and
$\mathcal{T}_{H}(0,1,1)$.

Lozenge tilings of the plane or of some distinguished domains with special
characteristics have been intensively studied [10, 12, 13, 8, 5], in part due
to the connection with practical applications of a random tiling model for two
dimensional electrostatics [9, 11]. While the main focus in these mentioned
works is the evaluation of the number of distinct tilings, our objectives
include characterizing the weights in the nodes of a network generated by a
certain triple, classifying the networks of numbers and highlighting certain
unique paths that connect the network nodes.

Once we observe that any integer $a$ is represented in
$\mathcal{R}_{H}(*,*,*)$, with $a$ being one of the stars, the natural problem
is to find an efficient algorithm to decide whether or not a positive $a$
belongs to $\mathcal{R}_{H}(0,0,0)$, $\mathcal{R}_{H}(0,1,1)$,
$\mathcal{R}_{H}(1,0,1)$ or $\mathcal{R}_{H}(1,1,0)$. For example, one finds
that $2024\not\in\mathcal{R}_{H}(0,1,1)$, but $2023\in\mathcal{R}_{H}(0,1,1)$
(see the graphical representation in Figure 9). For $2023$, there are
infinitely many nontrivial paths to get from $(0,1,1)$ to a triple that
represents $2023$, but one of the shortest takes $99$ steps. Such a path
passes by the intermediate triple $(1089,1156,1123)$ through the sequence of
operations:

$(0,1,1)\xrightarrow{\text{ $\big{(}H^{\prime\prime}\circ
H^{\prime\prime\prime}\circ H^{\prime}\big{)}^{[22]}$ }}(1089,1151,1123),$

and then, changing the direction, reaches the target through:

$(0,1,1)\xrightarrow{\text{ $\big{(}H^{\prime\prime\prime}\circ
H^{\prime\prime}\circ H^{\prime}\big{)}^{[11]}$ }\circ\text{
$\big{(}H^{\prime\prime}\circ H^{\prime\prime\prime}\circ
H^{\prime}\big{)}^{[22]}$ }}(1956,1939,2023).$

Searching for common patterns among the sets of triples
$\mathcal{T}_{H}(a,b,c)$, one can sometimes notice that starting from two
different triples $(a_{1},b_{1},c_{1})$ and $(a_{1},b_{1},c_{1})$, and
performing an identical sequence of operations, it can happen that one of the
components of the resulting triples ends up being the same. For example,
$G=H^{\prime\prime\prime}\circ H^{\prime\prime}\circ
H^{\prime\prime\prime}\circ H^{\prime}$ transforms $(1,3,6)$ and $(6,7,9)$
into two triples that have the third component equal to $17$:

$\begin{split}(1,3,6)&\xrightarrow{\text{ $H^{\prime}$
}}(9,3,6)\xrightarrow{\text{ $H^{\prime\prime\prime}$
}}(9,3,7)\xrightarrow{\text{ $H^{\prime\prime}$ }}(9,14,7)\xrightarrow{\text{
$H^{\prime\prime\prime}$ }}(9,14,17)\\\ (6,7,9)&\xrightarrow{\text{
$H^{\prime}$ }}(11,7,9)\xrightarrow{\text{ $H^{\prime\prime\prime}$
}}(11,7,10)\xrightarrow{\text{ $H^{\prime\prime}$ }}(11,15,10)\\\
&\xrightarrow{\text{ $H^{\prime\prime\prime}$ }}(11,15,17)\,.\end{split}$ (5)

Continuing from this point, the search reveals that even more triples obtained
in a similar way satisfy the same property. Thus, in the example above, one
even finds an infinite sequence of such triples obtained through a periodic
series of transformations. Indeed, if $F=H^{\prime\prime\prime}\circ
H^{\prime}\circ H^{\prime\prime\prime}\circ H^{\prime\prime}$, then the third
component of the sequences $\\{F^{(k)}(9,14,17)\\}_{k\geq 0}$ and
$\\{F^{(k)}(11,15,17)\\}_{k\geq 0}$ are equal, and the sequence of these equal
components starts with $17,19,27,41,$ $61,$ $87,119,157,201,251,307,\dots$,
the general formula for the general term being $\\{3k^{2}-k+17\\}_{k\geq 0}$.
In the following remark we note that this observation is universally valid.

######  Remark 1.1.

Let $(a_{1},b_{1},c_{1})$ and $(a_{2},b_{2},c_{2})$ be triples of integers.
Suppose $G$ is a composition of a finite sequence of operators from
$\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$ and
$G(a_{1},b_{1},c_{1})=(A,*,*)$ and $G(a_{2},b_{2},c_{2})=(A,*,*)$, where
$A\in\mathbb{Z}$, and the $*$’s may represent any integer. Then, there exists
an infinite sequence of distinct integers $\\{A_{n}\\}_{n\geq 0}$ and a
sequence $\\{G_{n}\\}_{n\geq 0}$, where each $G_{n}$ is a sequence of
compositions of $H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}$, such
that

$G_{n}(a_{1},b_{1},c_{1})=(A_{n},*,*)=G_{n}(a_{2},b_{2},c_{2})\ \ \text{ for
$n\geq 0$.}$

We will revisit this remark in Section 5.2 after we will obtain the
characterization of the set of triples $\mathcal{T}_{H}(a,b,c)$.

The next result shows that the represented numbers in both fundamental
tessellations $\mathcal{T}_{H}(0,0,0)$ and $\mathcal{T}_{H}(0,1,1)$ cover all
the residue classes modulo any prime $p\geq 5$. The graphical representations
$\mod p$ reveal intricate patterns (see Figures 2 and 4). Note that while the
density of the represented weights in residue classes is uniform, there are
two particular residue classes where the density is either very small or very
large compared to the others.

######  Theorem 2.

Let $p$ be prime and let $d_{p}(u)$ denote the limit density of the residue
class $u=R\pmod{p}$ of the represented integers in
$R\in\mathcal{R}_{H}(a,b,c)$, for $u=0,1,\dots,p-1$. Then:

(1) If the germ is $(0,0,0)$, then the values of the density are:

  * $\circ$

If $p=2$, then $d_{2}(0)=3/4$ and $d_{2}(1)=1/4$.

  * $\circ$

If $p=3$, then $d_{3}(0)=3/9$ and $d_{3}(1)=3/9$ and $d_{3}(2)=3/9$.

  * $\circ$

If $p\geq 5$ and $p\equiv 1\pmod{6}$, then
$d_{p}\big{(}(p-1)/3\big{)}=(2p-1)/p^{2}$ and $d_{p}(u)=(p-1)/p^{2}$ for
$u=1,\dots,p-1$ and $\neq(p-1)/3$.

  * $\circ$

If $p\geq 5$ and $p\equiv 5\pmod{6}$, then
$d_{p}\big{(}(2p-1)/3\big{)}=1/p^{2}$ and $d_{p}(u)=(p+1)/p^{2}$ for
$u=0,\dots,p-1$ and $\neq(2p-1)/3$.

(2) If the germ is $(0,1,1)$, then the values of the density are:

  * $\circ$

If $p=2$, then $d_{2}(0)=1/4$ and $d_{2}(1)=3/4$.

  * $\circ$

If $p=3$, then $d_{3}(0)=3/9$ and $d_{3}(1)=6/9$ and $d_{3}(2)=0$.

  * $\circ$

If $p\geq 5$ and $p\equiv 1\pmod{6}$, then $d_{p}(0)=(2p-1)/p^{2}$ and
$d_{p}(u)=(p-1)/p^{2}$ for $u=1,\dots,p-1$.

  * $\circ$

If $p\geq 5$ and $p\equiv 5\pmod{6}$, then $d_{p}(0)=1/p^{2}$ and
$d_{p}(u)=(p+1)/p^{2}$ for $u=1,\dots,p-1$.

Figure 2. A cut-off representation of the triangular networks generated by
$(9,2,6)$ (left) and $(1,8,3)$ (right). The weights are taken modulo $23$ on
the left and modulo $37$ on the right. In order to distinguish them, the
residue classes are represented in distinct colors in each of the two cases.

Note that Theorem 2 allows us to distinguish or even precisely determine which
tower a given triple $(a,b,c)$ belongs to. For example, knowing additionally
that $(a,b,c)$ appears in one of the basic tessellations
$\mathcal{T}_{H}(0,0,0)$, $\mathcal{T}_{H}(0,1,1)$, $\mathcal{T}_{H}(1,0,1)$,
$\mathcal{T}_{H}(1,0,1)$, and one of $a,b$ or $c$ is congruent to $2$ mod $3$,
then it would follow that $(a,b,c)\in\mathcal{T}(0,0,0)$. With less
information, one needs to combine Theorem 2 with finding a path to the center
(the triangle with the minimum weights) of the tessellation and then travel by
translation through the tower to find the basic germ (see the discussion in
Section 5.3.1).

The work is structured as follows. In Section 2 we present the first
implications of employing the iterative composition of any combination of the
operators in $\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$, and
then prove the existence of the four towers of tesselations corresponding to
the fundamental germs. Next, in Section 3, we describe the geometric
representation of the image of the composed operators $H$ as lozenge tilings
of the plane with the represented integers placed as weights in the nodes.
Continuing on, in Section 4, we prove Theorem 2, which provides explicitely
the densities of the weights in residue classes modulo any prime number $p$.
Our work concludes in Section 5 by achieving the parameterization of all
weights in a tessellation, which consequently supports the fact stated in
Theorem 1 that only a finite number of weights can be found under any given
threshold. Additionally, we also present the procedure to be followed on the
shortest route taken on the network nodes from a specific triangle towards its
origin.

##  2\. Properties of $H$

######  Lemma 2.1.

For any integer $a,b,c,h$ we have:

$\begin{split}\mathcal{T}_{H}(a+h,b+h,c+h)&=\mathcal{T}(a,b,c)+h\,,\\\
\mathcal{R}_{H}(a+h,b+h,c+h)&=\mathcal{R}(a,b,c)+h\,.\end{split}$ (6)

###### Proof.

We use the usual notations for the translations of a tuple and of a set, by
which $(a,b,c)+h:=(a+h,b+h,c+h)$, and $\mathcal{S}+h$ represents the set
obtained by adding $h$ to each element in the set $\mathcal{S}$.

The equality of the sets of represented integers in (6) follows from the
equality of the sets of triples above, which in turn is implied by the
definitions of the three analogous relations for $H^{\prime},H^{\prime\prime}$
and $H^{\prime\prime\prime}$, of which, for exemplification, the first one is

$\begin{split}H^{\prime}(a+h,b+h,c+h)&=\big{(}-(a+h)+1+(b+h)+(c+h),b+h,c+h\big{)}\\\
&=(-a+1+b+c,b,c)+h\\\ &=H^{\prime}(a,b,c)+h\,.\end{split}$

This completes the proof of the lemma. ∎

######  Proposition 2.1.

We have:

  1. (1)

The operators $H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}$ are
involutions.

  2. (2)

Any two distinct operators
$H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}$ do not commute.

  3. (3)

$H^{\prime\prime}\circ H^{\prime\prime\prime}\circ
H^{\prime\prime}=H^{\prime\prime\prime}\circ H^{\prime\prime}\circ
H^{\prime\prime\prime}$.

  4. (4)

$\big{(}H^{\prime}\circ H^{\prime\prime}\big{)}^{[3]}=Id$.

###### Proof.

It is enough to check that $H^{\prime}\circ H^{\prime}=Id$ and
$H^{\prime}\circ H^{\prime\prime}\neq H^{\prime}\circ H^{\prime\prime}$, as
for the other operators the analoguos relations follow by a rotation of the
variables. Also, let us note that the other permutations of the components as
in part (3) or part (4) can be obtained in the same way through the
corresponding composition of the analogous permuted operators.

Let $x,y,z$ be fixed. (1) For the fist part, we have:

$\begin{split}H^{\prime}\big{(}H^{\prime}(x,y,z)\big{)}&=H^{\prime}(-x+1+y+z,y,z)\\\
&=\big{(}-(-x+1+y+z)+1+y+z,y,z\big{)}\\\ &=(x,y,z).\end{split}$

(2) The necommutativity follows since

$\begin{split}H^{\prime}\big{(}H^{\prime\prime}(x,y,z)\big{)}&=H^{\prime}(x,-y+1+z+x,z)\\\
&=\big{(}-x+1+(-y+1+z+x)+z,-y+1+z+x,z\big{)}\\\
&=(2-y+2z,1-y+z+x,z),\end{split}$

and

$\begin{split}H^{\prime\prime}\big{(}H^{\prime}(x,y,z)\big{)}&=H^{\prime\prime}(-x+1+y+z,y,z)\\\
&=\big{(}-x+1+y+z,-y+1+z+(-x+1+y+z),z\big{)}\\\
&=(1-x+y+z,2-x+2z,z),\end{split}$

so that $\big{(}H^{\prime}\circ
H^{\prime\prime}\big{)}(x,y,z)=\big{(}H^{\prime\prime}\circ
H^{\prime}\big{)}(x,y,z)$ if and only if $y=x$ and $z=x-1$.

(4) As before, we have

$\begin{split}H^{\prime\prime}\big{(}H^{\prime\prime\prime}(x,y,z)\big{)}&=(x,2+2x-z,1+x+y-z),\end{split}$

which, applied twice, implies

$\begin{split}\big{(}H^{\prime\prime}\circ
H^{\prime\prime\prime}\big{)}^{[2]}(x,y,z)&=\big{(}x,2+2x-(1+x+y-z),\\\
&\phantom{(x,xX}1+x+(2+2x-z)-(1+x+y-z)\big{)}\\\
&=(x,1+x-y+z,2+2x-y),\end{split}$

and finally, for the third time,

$\begin{split}\big{(}H^{\prime\prime}\circ
H^{\prime\prime\prime}\big{)}^{[3]}(x,y,z)&=\big{(}x,2+2x-(2+2x-y),\\\
&\phantom{(x,xX}1+x+(1+x-y+z)-(2+2x-y)\big{)}\\\ &=(x,y,z).\end{split}$

The same calculation with components and operators interchanged proves formula
(4), and also the fact that $\big{(}H^{\prime}\circ
H^{\prime\prime}\big{)}^{[3]}=\big{(}H^{\prime}\circ
H^{\prime\prime\prime}\big{)}^{[3]}=Id$.

(3) By part (4) we know that $H^{\prime\prime}\circ
H^{\prime\prime\prime}\circ H^{\prime\prime}\circ H^{\prime\prime\prime}\circ
H^{\prime\prime}\circ H^{\prime\prime\prime}=Id$, from which, composing on the
right with $H^{\prime\prime\prime},H^{\prime\prime}$ and
$H^{\prime\prime\prime}$, in that order, and using (1), we obtain
$H^{\prime\prime}\circ H^{\prime\prime\prime}\circ
H^{\prime\prime}=H^{\prime\prime\prime}\circ H^{\prime\prime}\circ
H^{\prime\prime\prime}$. These completes the proof of the proposition. ∎

######  Theorem 3.

Let $a,b,c\in\mathbb{Z}$. Then, there exists $h\in\mathbb{Z}$ such that either

$\begin{split}\mathcal{R}_{H}(a,b,c)&=\mathcal{R}_{H}(0,0,0)+h\quad\text{ or
}\\\
\mathcal{R}_{H}(a,b,c)&=\mathcal{R}_{H}(0,1,1)+h=\mathcal{R}_{H}(1,0,1)+h=\mathcal{R}_{H}(1,1,0)+h.\end{split}$

###### Proof.

From Theorem 1 we know that the set $\mathcal{R}_{H}(a,b,c)$ has a first
element. Then, let $m=\min\mathcal{R}_{H}(a,b,c)$, and let
$(m,u,v)\in\mathcal{T}_{H}(a,b,c)$ be a triple in which $m$ is represented. We
do not know in advance on which position $m$ appears in the triple, but the
other two situations in which $m$ would be in the second or the third position
are treated the same.

Then there are two possibilities: $m\leq v\leq u$ or $m\leq u\leq v$. Again,
it is enough to discuss only one of the cases, the other being treated
similarly by rotating the variables and the corresponding operators involved.
Moreover, we will see that the cases blend together, because we will find that
in fact $u=v$.

Thus, without restricting the generality, we may also suppose that $m\leq
u\leq v$. Since $(m,u,v)\in\mathcal{T}_{H}(a,b,c)$, it follows that the third
component of $H^{\prime\prime\prime}(m,u,v)$, which equals $-v+1+m+u$ belongs
to $\mathcal{R}_{H}(a,b,c)$. Then, since $m$ is the minimum, $m\leq-v+1+m+u$,
which implies $v\leq u+1$. Therefore, either $v=u$ or $v=u+1$.

Case $v=u+1$. Since $(m,u,u+1)\in\mathcal{T}(a,b,c)$ it follows that
$H^{\prime\prime\prime}(m,u,u+1)$ $=(m,u,u)$ is also in $\mathcal{T}(a,b,c)$,
so that it is enough to consider the case $u=v$.

Case $v=u$. Starting from $(m,u,u)$, the following neighbor triples are also
in $\mathcal{T}_{H}(a,b,c)$:

$(m,u,u)\xrightarrow{\text{ $H^{\prime\prime\prime}$
}}(m,u,m+1)\xrightarrow{\text{ $H^{\prime\prime}$ }}(m,-u+2m+2,m+1).$

The condition that $m$ is the minimum of $\mathcal{R}_{H}(a,b,c)$ implies
$m\leq-u+2m+2$, that is $u\leq m+2$. As a consequence, for $u$ are only three
possible values, $m,m+1$ or $m+2$, which we will analyze next.

  1. (1)

If $u=m$, then we have $(m,m,m)\in\mathcal{R}_{H}(a,b,c)$.

  2. (2)

If $u=m+1$, since $v=u$, we ha that $(m,m+1,m+1)\in\mathcal{R}_{H}(a,b,c)$.

  3. (3)

If $u=m+2$ then $(m,u,u)=(m,m+2,m+2)$ and the following three-steps path also
arrives at $(m,m,m)$:

$(m,m+2,m+2)\xrightarrow{\text{ $H^{\prime\prime\prime}$
}}(m,m+2,m+1)\xrightarrow{\text{ $H^{\prime\prime}$
}}(m,m,m+1)\xrightarrow{\text{ $H^{\prime\prime\prime}$ }}(m,m,m).$

In conclusion, we have shown that, provided $m=\min\mathcal{R}(a,b,c)$, then
either $(m,m,m)\in\mathcal{T}_{H}(a,b,c)$ or
$(m,m+1,m+1)\in\mathcal{T}_{H}(a,b,c)$. By Lemma 2.1 it then follows that
$(0,0,0)\in\mathcal{T}_{H}(a-m,b-m,c-m)$ in the first case, and
$(0,1,1)\in\mathcal{T}_{H}(a-m,b-m,c-m)$ in the second. ∎

The two fundamental sequences of the represented integers, the only ones that
exist according to Theorem 3, arranged in increasing order, are:

$\begin{split}\mathcal{R}(0,0,0)&=\\{0,1,2,4,6,8,9,12,14,16,20,21,22,25,30,32,36,\dots\\},\\\
\mathcal{R}(0,1,1)&=\\{0,1,3,4,7,9,12,13,16,19,21,25,27,28,31,36,37,\dots\\}.\end{split}$
(7)

The numbers can be obtained starting with the germs $(0,0,0)$ and $(0,1,1)$
and then step by step using the definition, keeping or changing the operator
from $\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$, that is,
moving around in all directions. Another more efficient approach is utilizing
the parametrization of the elements of the sequences presented in Section 5.

The second of the two sequences in (7) are the Löschian numbers [28, A003136].
Named after August Lösch, the sequence is a bi-product of a study in the field
of economics, regarding market development, population distribution, and the
size of regions approximated using a honeycomb network [26, Chapter 10]. The
ordered sequence $\mathcal{R}(0,1,1)$ is abundant in properties (see [25] and
the references therein) and besides algebra and number theory [18, 21, 22, 19]
it arises in very diverse contexts, such as a counter of the protein coats in
a virus shell model [29, Chapter 3], at the confluences between art and
mathematics [32], a fractal generator in the theory of place geometry [1, 2],
or in the geographical Central Place theory [3, 27, 2].

##  3\. The lozenge representation of the operators $H$

Given three integers $a,b,c$, we place them at the vertices of an equilateral
triangle, called the base triangle. Then, by fixing any two of the numbers
$a,b,c$, we place in the fourth vertex of the lozenge111A lozenge is the union
of two triangles, which are identical to the base triangle and share a common
side. generated by the base triangle (the endpoints of its smaller diagonal
being the fixed numbers) the value of the corresponding operator from
$\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$ (see Figure 5 and
the further in-depth description in Section 5).

By applying this procedure to every pair of two numbers from $\\{a,b,c\\}$, we
obtain three lozenges with a common base triangle, a star with three
equilateral triangles built adjacent to the sides of the base triangle.
Continuing in the same way, we proceed with the newly obtained equilateral
triangles, and then repeat the process endlessly, resulting in a triangular
tiling of the plane with integers.

######  Remark 3.1.

In essence, the geometric interpretation of relations (1) is a $2$-dimensional
tiling with integers in nodes, which has the property that in any lozenge of
it, the sum of the numbers on the long diagonal is always $1$ more than the
sum of the numbers on the short diagonal.

It is fundamental, and in Section 5 we will ascertain, that this construction
is indeed consistent, meaning that any triangle, which is obtained in exactly
three ways as the intersection of three lozenges is uniquely and unequivocally
determined. One can check the outlined procedure starting with any triple of
numbers from the adjacent circles in Figures 1 and 9.

With this geometric representation it becomes clear that for any two triangles
or triples of represented integers, there are infinitely many paths and
corresponding sequencers of operators taken from
$\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$ that composed
successively into a new operator $H$ connects one triple to the other. In
order to mark the association, on the set of all triples $\mathbb{Z}^{3}$, let
us define a relation according to which two triples are equivalent if they
belong to the same tiling. This is the same as saying that the triples in any
set $\mathcal{T}_{H}(a,b,c)$ defined by (3) are pairwise equivalent. The
equivalence classes of this relation are these parallel triangular tilings,
each of them being generated by any base triangle it contains.

It should be noted that formally, any triangle of adjacent numbers obtained by
iterating the operators from
$\\{H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}\\}$ is associated with
a specific order. The classification of triples in the equivalence classes
$\mathcal{T}_{H}$ has its unique features, which causes nearby triples in
$\mathbb{Z}^{3}$ to not always be equivalent. As a result, in general, the
permutations of the triple $(a,b,c)$ do not necessarily belong to the same
tiling $\mathcal{T}_{H}(a,b,c)$.

Figure 3. The triples and how they appear oriented in the triangular networks
generated by $(0,0,0)$ (left) and $(0,1,1)$ (right). Once can check that every
triple that occurs appears only once in the left image and exactly six times
in the right image.

######  Remark 3.2.

There is a difference between the rotation symmetries of the two triangular
networks in Figure 3. In the network from the left-side, a spiral of lozenges
is formed around a center consisting of the triangle with vertices $(0,0,0)$,
while in the network on the right-side, although the generation also starts
around a triangle, namely $(0,1,1)$, the network ends up having a single
symmetry center, the node with weight $0$, instead of a triangle as in the
left figure.

Additionally, we define an equivalence relation on the set of triples
$\mathcal{T}_{H}$ from (3), according to which
$\mathcal{T}_{H}(a_{1},b_{1},c_{1})$ is equivalent to
$\mathcal{T}_{H}(a_{2},b_{2},c_{2})$ if there exists an integer $h$ such that
$\mathcal{T}_{H}(a_{1},b_{1},c_{1})=\mathcal{T}_{H}(a_{2},b_{2},c_{2})+h$. In
this way, $\mathbb{Z}^{3}$ is partitioned into equivalence classes formed by
towers of tilings, in which the elements are obtained from each other by
translations, according to Lemma 2.1. Furthermore, Theorem 3 states that every
triple $(a,b,c)$ can be found in a unique tiling, and there are exactly four
distinct towers. One of these, to which $\mathcal{T}_{H}(0,0,0)$ belongs, is
distinguished, while the other three have one of the three permutations of
$(0,1,1)$ as the generating germ for their classes of equivalence.

######  Remark 3.3.

The consequence of the difference between the two types of symmetry centers,
as pointed out in Remark 3.2, is the fundamental distinction between the two
types of networks in the four towers.

Thus, any ordered triple $(a,b,c)$ appears exactly once in each triangular
tiling $\mathcal{T}_{H}(a,b,c)$ that is equivalent to $\mathcal{T}_{H}(0,0,0)$
and exactly six times if it belongs to a tiling belonging to any of the other
three towers generated by $(0,1,1)$, $(1,0,1)$ or $(1,1,0)$.

##  4\. The densities $\pmod{p}$ – Proof of Theorem 2

In the following, we use the parametrization of the represented integers from
Section 5, Theorem 4.

###  4.1. The germ $(0,1,1)$

Let $F(x,y)=x^{2}+xy+y^{2}$. We find the distribution of the residue classes
of $F(x,y)$ modulo $p$ for $(x,y)$ in the modulo $p$ box $[0,p-1]^{2}$. For
this, we need to calculate the number of solutions
$N_{p}(l)=\\#\mathcal{M}_{p}(l)$, where

$\mathcal{M}_{p}(l)=\big{\\{}(x,y):x^{2}+xy+y^{2}\equiv l\pmod{p}\big{\\}},$
(8)

for any $l$, $0\leq l\leq p-1$. Then the limit density is
$d_{p}(l)=N_{p}(l)/p^{2}$.

Figure 4. Comparison between local symmetries of the two types of triangular
networks. On the top row is the network generated by $(0,0,0)$, and on the
bottom row is the one generated by $(0,1,1)$. In each of these, the weights
are taken modulo $2,3,5$, and $7$, and the residue classes are represented in
distinct colors, so that the cases from Theorem 2 can be verified.

Case $p=2$. Modulo $2$, the values of $F(x,y)$ for $x,y\in\\{0,1\\}$ are
$\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}$, so that $d_{2}(0)=1/4$ and
$d_{2}(1)=3/4$.

Case $p=3$. The values of $F(x,y)\pmod{3}$, for $x,y=0,1,2$, are:
$\begin{pmatrix}0&1&1\\\ 1&0&1\\\ 1&1&0\\\ \end{pmatrix}$. Therefore, the
value of the densities are: $d_{3}(0)=3/3^{2}=1/3$, $d_{3}(1)=6/3^{2}=2/3$ and
$d_{3}(2)=0$.

Case $p\geq 5$. The handling of the congruence in (8) differs depending on
whether $l$ is zero or not modulo $p$. First, let us make a variable
substitution to remove the term $xy$ from the left-hand side. To do this, we
multiply the congruence by $4$ and complete a square. Then, we replace the
variables $x,y$ with $u,v$, where $u=x+2^{-1}y$ and $v=2^{-1}y$. With the
simplified congruence, the problem reduces to counting
$N^{*}_{p}(l)=\\#\mathcal{M}_{p}^{*}(l)$, where

$\mathcal{M}^{*}_{p}(l)=\big{\\{}(u,v):u^{2}+3v^{2}\equiv l\pmod{p}\big{\\}},$
(9)

since there is a one-to-one correspondence between the solutions of the two
congruences, implying $N^{*}_{p}(l)=N_{p}(l)$.

A. Suppose $l\equiv 0\pmod{p}$. Then the nature of the congruence in (9)
depends on the fact that $-3$ is a quadratic residue or not. Firstly, if
$\left(\frac{-3}{p}\right)=-1$222The Legendre symbol denoted by
$\left(\frac{a}{p}\right)$ evaluates to $1$ if $a$ is a quadratic residue and
to $-1$ if $a$ is a non-residue modulo $p$, for any $a$ relatively prime to
$p$., the congruence has only a single solution, $(u,v)=(0,0)$.

Secondly, if $\left(\frac{-3}{p}\right)=1$, let $a$ be a square root of $-3$.
Then, for each $v\in\\{1,\dots,p-1\\}$, the congruence in (9) has exactly two
distinct solutions $(av,v)$ and $(p-av,v)$ because $2av\not\equiv 0\pmod{p}$.
Then, we have a total of $2p-1$ solutions in this case, including solution
$(u,v)=(0,0)$.

######  Remark 4.1.

We have $\left(\frac{-3}{p}\right)=1$ if $p\equiv 1\pmod{6}$ and
$\left(\frac{-3}{p}\right)=-1$ if $p\equiv 5\pmod{6}$. This result is obtained
by combining Euler’s criterion with the quadratic reciprocity law as follows:

$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}\cdot(-1)^{\frac{p-1}{2}\cdot\frac{3-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right).$

In conclusion, we have proved that

$N^{*}_{p}(0)=\begin{cases}1&\text{ if $p\equiv 5\pmod{6}$}\\\\[2.0pt]
2p-1&\text{ if $p\equiv 1\pmod{6}$}.\end{cases}$ (10)

B. Now suppose $l\not\equiv 0\pmod{p}$. First let us note that if $a\not\equiv
0\pmod{p}$, then $N^{*}_{p}(l)=N^{*}_{p}(a^{2}l)$ because the correspondence
$(u,v)\leftrightsquigarrow(au,av)$ is a a one-to-one mapping between the
solutions of the congruences counted in the two sets $\mathcal{M}^{*}_{p}(l)$
and $\mathcal{M}^{*}_{p}(a^{2}l)$. It follows that $N^{*}_{p}(l)$ remains
constant for all quadratic residues $l$ and still, $N^{*}_{p}(l)$ also is
constant for all non-residues $l$. We denote by $R$ and $N$ the cardinalities
in the two cases:

$\begin{split}R:=N^{*}_{p}(l)\quad\text{ if
$\left(\frac{l}{p}\right)=1$}\,,\quad\text{ and }\quad
N:=N^{*}_{p}(l)\quad\text{ if $\left(\frac{l}{p}\right)=-1$}\,.\end{split}$
(11)

Next, to find $R$, we may assume that $l=1$. Let us introduce a
parametrization of the curve that defines the congruence in
$\mathcal{M}_{p}^{*}(1)$, intersecting it with all possible lines that pass
through the base solution $(1,0)$. This will determine a second solution,
which satisfies the system

$\begin{cases}u^{2}+3v^{2}\equiv 1\pmod{p}\\\\[4.0pt] r(u-1)\equiv
v\pmod{p}\end{cases}$ (12)

for $r=0,1,\dots,p-1$. In addition to the lines in (12), the vertical line
$u=1$ gives only the base solution $(1,0)$. Note that all solutions in
$\mathcal{M}_{p}^{*}(1)$ are obtained in this way, and to find their exact
number we need to check for any repetitions.

If $r=0$, then $v=0$, which gives two solutions $(1,0)$ and $(p-1,0)$.

If $r\in\\{1,\dots,p-1\\}$, then (12) implies $u^{2}+3r^{2}(u-1)^{2}\equiv
1\pmod{p}$, that is,

$(3r^{2}+1)u^{2}-6r^{2}u+(3r^{2}-1)\equiv 0\pmod{p}\,.$ (13)

Besides the base solution $u=1$, congruence (13) has the solution
$u=(3r^{2}-1)(3r^{2}+1)^{-1}$, provided that $3r^{2}+1\not\equiv 0\pmod{p}$.

Let us check if these solutions $u=u(r)$ are distinct for different $r$’s. We
see that the equality
$(3r_{1}^{2}-1)(3r_{1}^{2}+1)^{-1}=(3r_{2}^{2}-1)(3r_{2}^{2}+1)^{-1}$, for
distinct $r_{1},r_{2}\in\\{0,1,\dots,p-1\\}$, is equivalent with
$(r_{1}-r_{2})(r_{1}+r_{2})\equiv 0\pmod{p}$. Then, in the only case with
uncertainty regarding the possible coincidence of solutions, that is when
$r_{2}=p-r_{1}$, we see that the corresponding two $v$’s, that is,
$r_{1}(u-1)$ and $r_{2}(u-1)$ are distinct because $p\neq 2$, while the case
$u=1$ was settled before.

If $3r^{2}+1\equiv 0\pmod{p}$, then congruence (13) reduces to $2u-2\equiv
0\pmod{p}$, so that system (12) also has a unique solution, the same $(1,0)$.
This counts only for just two values of $r$ that are square roots of
$(-3)^{-1}$, and it occurs, according to Remark 4.1, only for $p\equiv
1\pmod{6}$. Thus, checking all $r\in\\{0,1,\dots,p-1\\}$, we find that if
$p\equiv 1\pmod{6}$, the base solution $(1,0)$ appears repeated three times,
when $r=0,r_{1},r_{2}$, where and $r_{1}$ and $r_{2}$ are the square roots mod
$p$ of $(-3)^{-1}$, and the base solution $(1,0)$ is never repeated otherwise,
when $(-3)^{-1}$ is not a quadratic residue.

In conclusion, we have shown that system (12) has a single solution for
$r\in\\{1,2,\dots,p-1\\}$ and two solutions if $r\equiv 0\pmod{p}$, all
together, being in a total of $2+(p-1)=p+1$ distinct solutions, provided that
$p\not\equiv 1\pmod{6}$, and two less, because of the noted repetitions,
otherwise. Therefore:

$R=N^{*}_{p}(1)=\begin{cases}p-1&\text{ if $p\equiv 1\pmod{6}$}\\\\[4.0pt]
p+1&\text{ if $p\equiv 5\pmod{6}$}.\end{cases}$ (14)

To complete the analysis, we still need to treat the case when $l$ is a
quadratic non-residue, that is, to find $N$ defined by (11). This can be done
by reckoning that $\mathbb{F}_{p}^{2}$ is partitioned into subsets
$\mathcal{M}_{p}^{*}(l)$ grouped by their equal cardinalities, as follows:

$\mathbb{F}_{p}^{2}=\mathcal{M}_{p}^{*}(0)\cup\bigcup_{\begin{subarray}{c}l=1\\\
\left(\frac{l}{p}\right)=1\end{subarray}}^{p-1}\mathcal{M}_{p}^{*}(l)\cup\bigcup_{\begin{subarray}{c}l=1\\\
\left(\frac{l}{p}\right)=-1\end{subarray}}^{p-1}\mathcal{M}_{p}^{*}(l).$ (15)

Then, on combining (10), (11), and the fact that there are an equal number of
$(p-1)/2$ of quadratic residues and non-residues, from (15) we find that

$p^{2}=\begin{cases}2p-1+\frac{p-1}{2}R+\frac{p-1}{2}N&\text{ if $p\equiv
1\pmod{6}$}\\\\[6.0pt] 1+\frac{p-1}{2}R+\frac{p-1}{2}N&\text{ if $p\equiv
5\pmod{6}$}.\end{cases}$

On employing (14), it follows that

$N=\begin{cases}p-1&\text{ if $p\equiv 1\pmod{6}$}\\\ p+1&\text{ if $p\equiv
5\pmod{6}$},\end{cases}$ (16)

which, compared to (14), means that $N=R$ for all $p$.

To complete the proof of Theorem 2 for the germs $(0,1,1)$, $(1,0,1)$ and
$(1,1,0)$, we only have to replace the values of $N^{*}_{p}(l)$, explicitly
obtained in relations (10), (11), (14) and (16), in the definition
$d_{p}(l)=N_{p}^{*}(l)/p^{2}$, for $p\geq 5$ and $0\leq l\leq p-1$, to obtain
the formulas of the densities from the statement of the theorem.

###  4.2. The germ $(0,0,0)$

Let

$N_{p}(l)=\\#\big{\\{}(x,y):x^{2}+y^{2}+xy-x-y\equiv l\pmod{p}\big{\\}}.$ (17)

For any $l$, $0\leq l\leq p-1$, as seen above, the limit density is
$d_{p}(l)=N_{p}(l)/p^{2}$.

Case $p=2$. With $F(x,y)=x^{2}+y^{2}+xy-x-y$, and $x,y\in\\{0,1\\},$ the
values modulo $2$, are $0,0,0,1$, so that $d_{2}(0)=3/4$ and $d_{2}(1)=1/4$.

Case $p=3$. The values of $F(x,y)\pmod{3}$, for $y=0,1,2$, are: $0,0,2$ if
$y=0$; $0,1,1$ if $y=1$; and $2,1,2$ if $y=2$. Then
$d_{3}(0)=d_{3}(1)=d_{3}(2)=3/3^{2}=1/3$.

Case $p\geq 5$. As before, we change the variables to transform the expression
from the left side of the congruence into a canonical quadratic form. First,
we multiply the congruence by $4$ and complete the square to get rid of the
$xy$ term. Then replacing $x$ by $u$, where $u=2x+y$, the congruence inside
(17) becomes

$u^{2}+3y^{2}-2u-2y\equiv 4l\pmod{p}\,.$

Next we complete the square to get rid of the linear term $-2u$. For this, we
replace $u$ by $w$, where $w=u-1$, and the above congruence becomes

$w^{2}+3y^{2}-2y\equiv 4l+1\pmod{p}\,.$

Now we eliminate the remaining linear term by multiplying the congruence by
$3$ and completing the square. We replace $y$ by the new variable $v$, where
$v=3y-1$, and arrive at the new form of the congruence

$3w^{2}+v^{2}\equiv 4(3l+1)\pmod{p}\,.$

In the end, we can go back to the original variables by putting $x=2^{-1}v$
and $y=2^{-1}w$. Considering the fact that $p\neq 2,3$, the solutions of the
several congruences above are in a one-to-one correspondence with each other,
the number of solutions counted by $N_{p}(l)$ in (17) equals
$N^{*}_{p}(3l+1)$, the number of solutions of the last new congruence, where

$N^{*}_{p}(3l+1)=\\#\big{\\{}(x,y):x^{2}+3y^{2}\equiv 3l+1\pmod{p}\big{\\}}.$
(18)

This is the same as in the case of germ $(0,1,1)$, except that $l$ is replaced
by $3l+1$. Adapting the analysis from there, the cases are distinguished
depending on whether $3l+1\equiv 0\pmod{p}$ or $3l+1\not\equiv 0\pmod{p}$.

Note that $(-3)^{-1}=\frac{p-1}{3}$ if $p\equiv 1\pmod{6}$ and
$(-3)^{-1}=\frac{2p-1}{3}$ if $p\equiv 5\pmod{6}$.

If $3l+1\equiv 0\pmod{p}$, then the number of solution of the congruence in
(18), which becomes $x^{2}\equiv-3y^{2}\pmod{p}$, depends on whether $-3$ is a
quadratic residue or not modulo $p$.

If $\left(\frac{-3}{p}\right)=-1$, then there is only one solution at
$(x,y)=(0,0)$, so that $d_{p}\big{(}\frac{p-1}{3}\big{)}=\frac{1}{p^{2}}$ or
$d_{p}\big{(}\frac{2p-1}{3}\big{)}=\frac{1}{p^{2}}$, depending on the case
where $p\equiv 1\pmod{6}$ or $p\equiv 5\pmod{6}$.

If $\left(\frac{-3}{p}\right)=1$, then for each $x\in\\{1,\dots,p-1\\}$, there
are two distinct solutions $(x,\pm y)$ for $y\in\\{1,\dots,p-1\\}$, plus one
more $(0,0)$, resulting in a total of $2p-1$ solutions. Therefore,
$d_{p}\big{(}\frac{p-1}{3}\big{)}=\frac{2p-1}{p^{2}}$ or
$d_{p}\big{(}\frac{2p-1}{3}\big{)}=\frac{2p-1}{p^{2}}$, depending on the case
where $p\equiv 1\pmod{6}$ or $p\equiv 5\pmod{6}$.

It remains to be seen if this is consistent with the statement of the theorem,
and this can be checked because condition $\left(\frac{-3}{p}\right)=1$ holds
if and only of $3$ divides $p-1$ (which reduces in our case to $p\equiv
1\pmod{6}$). The last equivalence follow by combining Euler’s criterion with
the quadratic reciprocity because law, because

$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}\cdot(-1)^{\frac{p-1}{2}\cdot\frac{3-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right).$

This conclude the proof of Theorem 2.

##  5\. The number tiling of the plane and the proof of Theorem 1

The aim of this section is to prove Theorem 1. To do so, we will begin by
offering a more detailed account of the tiling coverage of the plane with
integers, which will lead to a parameterization of all nodes within the
triangular network $\mathcal{T}_{H}(a,b,c)$.

###  5.1. A parametrization of the represented integers

Let $a,b,c$ be fixed integers. In order to determine which elements belong to
the sequence of integers represented in $\mathcal{R}_{H}(a,b,c)$, we need to
describe the structure of the set of triples $\mathcal{T}_{H}(a,b,c)$. In this
set, although the components of the triples appear multiple times in different
triples, one can track the step-by-step generation of triples starting from
the basic germ $(a,b,c)$.

We start building an interlinked network of numbers based at the nodes of a
$2$-dimensional triangular network by placing $a,b,c$ in the nodes of one of
the smallest equilateral triangles of the grid. This is the the base triangle
that stores the germ $(a,b,c)$. We will identify and then refer to the nodes
by the integers they contain. The integers in the nodes will be called
weights. Then, we place the weight $d=H^{\prime}(a,b,c)=-a+b+c+1$ in the
fourth corner of the lozenge $abdc$ with the shorter diagonal $bc$.

Figure 5. The lozenge expansion of $d$ from $a$ across the short diagonal
$bc$.

The process of expanding towards a future node from a previous one, by placing
the child-integer symmetrically across a side of a triangle, as shown in
Figure 5, is then iterated to generate new integers across any side of the
triangles in the current generation. The expansion can be done in any order
using either of the corresponding operators
$H^{\prime},H^{\prime\prime},H^{\prime\prime\prime}$, with some rule, such as
to produce a certain linear development in a given direction, randomly, or in
a circular fashion, as shown by the individual steps from Figure 6. (The
circular spiral construction is also employed in generating the first two
circular annuli around the base germs $(0,0,0)$ and $(0,1,1)$ in Figure 3.)

Figure 6. Consistency check on the immediately adjacent hexagonal path around
the initial triangle $abc$. One finds the same value of $g$ regardless of
whether one goes around in trigonometric direction in four steps or vice versa
in just one step.

We need to ensure that this procedure can be repeated indefinitely to fill all
the nodes in the triangular network covering the plane. The construction must
be consistent, meaning that the same weights are obtained regardless of the
path taken to reach the nodes. To verify this, we will check the correctness
on the minimal paths around a node and use the fact that
$H^{\prime},H^{\prime\prime}$ and $H^{\prime\prime\prime}$ are involutions (as
we know from Proposition 2.1), which allows to go back and forth on the paths.
This will guarantee a unique and consistent filling of the nodes of the entire
triangular tiling, as we will then have established step by step the
uniqueness of the weights on any path between two fixed nodes.

In counterclockwise direction, the weights in Figure 6 are:

$\begin{split}d&=H^{\prime}(a,b,c)=-a+b+c+1\\\
e&=H^{\prime\prime}(d,b,c)=-b+d+c+1=-a+2c+2\\\
f&=H^{\prime}(d,e,c)=-d+e+c+1=-b+2c+2\\\
g&=H^{\prime\prime}(f,e,c)=-e+f+c+1=a-b+c+1,\\\ \end{split}$ (19)

and then clockwise:

$\begin{split}g&=H^{\prime\prime}(a,b,c)=-b+a+c+1\,.\end{split}$ (20)

Since the same weight $g$ is produced on both paths (19) and (20), we have
proven the consistency of the construction.

######  Remark 5.1.

The symmetries in relations (1) and (19) imply:

  1. (1)

The symmetric correspondence of the formulas of $e$ and $f$ in (19) reflect
the position of nodes $a$ and $b$ in Figure 6.

  2. (2)

Let $x,y,z$ be the weights of three consecutive adjacent nodes in the
triangular network. Then $z=-x+2y+2$.

Iterating the second part of Remark 5.1, it follows that the weights that
follow $x$ and $y$ in straight line are: $-x+2y+2$, $-2x+3y+6$, $-3x+4y+12$,
$-4x+5y+20$, and so on. Then, by induction, we obtain the following result.

######  Lemma 5.1.

The weights of the nodes on the straight line determined by the adjacent nodes
$x$ and $y$ are

$L(x,y\mid k)=-(k-1)x+ky+k(k-1)\ \ \text{ for $k\geq 0$.}$ (21)

Now, just using the one-dimensional parametrization (21), one can prove
Theorem 1. For this it suffices to check the weights on the sets of parallel
half-lines on the six directions rotated by $60$ degrees around any fixed
origin, and note that the size of the quadratic term $k^{2}-k$ allows for at
most a finite number of negative weights around the origin.

A complete two-dimensional parametrization of the weights is given in the next
section.

###  5.2. Two-dimensional parametrization of the weights

Let $\varrho=1/2+\sqrt{3}i/2$ be a root of order $6$ of $1$. Any node of the
triangular lattice can then be uniquely expressed using the base
$\\{1,\varrho\\}$ as the linear combination $m+n\varrho$ with
$m,n\in\mathbb{Z}$. Thus, the nodes of the network are the elements of the set

$\begin{split}\mathcal{H}:=\\{m+n\varrho:m,n\in\mathbb{Z}\\}\,.\end{split}$

Let $G(a,b,c\mid m,n)$ denote the value of the weight associated to the node
$m+n\varrho$ in the triangular network generated by the base triangle of
weights $a$,$b$, and $c$.

To compare, it should be noted that $G$ must be two-dimensional and dependent
on $c$, while $L$ is one-dimensional and independent of $c$, and they are
related by

$L(a,b\mid m)=G(a,b,c\mid m,0)\ \ \text{ for any $a,b,c,m\in\mathbb{Z}$.}$

The next theorem provides the closed-form expression of $G(a,b,c\mid m,n)$.

######  Theorem 4.

For any base triple $(a,b,c)$ and any integers $m,n$, we have

$G(a,b,c\mid m,n)=-(m+n-1)a+mb+nc+(m^{2}+n^{2}+mn-m-n)\,.$ (22)

###### Proof.

Let $m,n\in\mathbb{Z}$ and let $a,b,c$ be fixed real numbers. While the linear
expansion in Lemma 5.1 uses the weight $c$ just as a support to generate the
weights on the line $ab$, here we generate and also record all the weights on
the line that starts at $c$ and is adjacent to $ab$. Then, iterating the
process, the general formula will follow.

Figure 7. The _parallelogram rule_ used to obtain the closed formula for
$G(a,b,c\mid m,n)$. In order to find $z$ from the opposite corner of $a$ and
having the coordinates $m,n$ in base $\\{1,\varrho\\}$, one applies Lemma 5.1
a few times to find $x$, the $m$-th element on the line $ab$, then $d$, with
$d=H^{\prime}(a,b,c)$, next $y$, the $m$-th element on the line $cd$, and
finally one arrives at $z$, the $n$-th element on line $xy$.

Let $z=G(a,b,c\mid m,n)$ be a short notation of the unknown variable for the
fixed variables $a,b,c,m,n$. Next we use Figure 7 as a support in the
following successive steps:

  1. (1)

By (21) we find $x=L(a,b\mid m)$.

  2. (2)

Then $d=H^{\prime}(a,b,c)=-a+b+c+1$.

  3. (3)

Next, by (21) we find $y=L(c,d\mid m)$.

  4. (4)

Finally, by (21) again we find $z=L(x,y\mid n)$.

Then writing $x,y$, and $z$ in terms of $a,b,c$, we find

$\begin{split}x=&-(m-1)a+mb+m(m-1),\\\ y=&-(m-1)c+md+m(m-1)=-ma+mb+c+m^{2},\\\
\end{split}$

so that

$\begin{split}z=&-(n-1)x+ny+n(n-1)\\\ =&-(m+n-1)a+mb+nc+(m^{2}+n^{2}+mn-
m-n)\,.\end{split}$

The last line gives us the needed formula for $z=G(a,b,c\mid m,n)$, which
completes the proof of the theorem. ∎

The operators $H^{\prime},H^{\prime\prime}$, and $H^{\prime\prime\prime}$ are
not linear because of the constant $1$ that is added in their definition.
However, by subtracting component-wise the triples obtained as images of
triples with one equal component on the same position obtained through a
composition of a finite sequence of $H^{\prime},H^{\prime\prime}$ and
$H^{\prime\prime\prime}$, the constant term cancels. As a result we obtain a
linear function with integer coefficients that has a zero at a lattice point.
Consequently, it will have infinitely many zeros at other lattice points
equally spaced on a straight line, as noted in the example before Remark 1.1.
Alternatively, the same result can be achieved by using the complete
characterization of the weights from Theorem 4.

Figure 8.  The spectrum of colors corresponding to the weights of $5801$ nodes
generated by (1) in a rectangular neighborhood of the initial triangle with
weights $2023,2109,2092$.

###  5.3. Final notes

Except for a few special cases, regardless of the initial triangle, the size
of the numbers generated by (1), shows a wide spectrum of nuances. A colored
representation of the weights along and across the plane and a few explicit
values of the weights are shown in Figure 8 and Figure 9, respectively. Two
examples of an extract of the network of weights generated by the initial
triangles $4,7,5$ and $2023,2109,2092$, respectively, are shown in Figure 9.
It should be noted that although the numbers that generate the spectrum are
discrete, it has a continuum appearance without a distinct break separating
the colors. Also, the ‘visual spectrum’ is similar all over the plane,
regardless of the area where one makes an analogous representation.

Figure 9. Triangular lattices filled recursively in the nodes by weights given
by relations (1) across any short diagonal of a lozenge. The image on the left
shows $167$ nodes whose maximum value is $67$, all of which are generated by
the initial triangle with weights of $4,5,7$. In the image on the right there
are shown only $77$ nodes with weights between $1699$ and $2509$. Color
intensity shades indicate the size of the numbers in the nodes to distinguish
them from each other.

####  5.3.1. Evaluation of the number of negative weights

Let us find the path from $(0,0,c)$, where $c\geq 0$, to the minimum triple in
the covering that contains $(0,0,c)$. Along the way we will meet three
remarkable sequences and also find the tower to which $(0,0,c)$ belongs to.

Let $H$ be the composition operator defined by $H:=H^{\prime}\circ
H^{\prime\prime}\circ H^{\prime}\circ H^{\prime\prime\prime}$. A direct
calculation shows that, for any triple $(a,b,c)$ with $a=b$, we have

$H(a,a,c)=\big{(}3a-2c+4,3a-2c+4,2a-c+1\big{)}\,.$

This is a zigzag operator whose iterations define the shortest path in the
triangular network from $(0,0,c)$ towards the center. (The center is the
triple in the network whose components are the smallest and closest to each
other.) Iterating the application of $H$, by induction we find that

$\begin{split}H^{[n]}(a,a,c)=\big{(}&(2n+1)a-2nc+n(3n+1),\\\
&(2n+1)a-2nc+n(3n+1),\ \\\ &2na-(2n-1)c+n(3n-2)\big{)}\end{split}$ (23)

for $n\geq 1$.

######  Remark 5.2.

Formula(23) employs two remarkable sequences if $a=c=0$.

The first one is $4,14,30$, $52,80$, $114$, $154$, $200$, $252$, $310,\dots$
and the formula for the $n$-th element is $\mathtt{NE}(n):=n(3n+1)$ for $n\geq
1$. It is the so-called sequence of the twice second pentagonal numbers [28,
A049451].

The second sequence is that of octagonal or the star numbers [28, A000567],
with elements $1,8,21,40,65,96,133,176,225,280,\dots$, and the formula for the
$n$-th element given by $\mathtt{SW}(n):=n(3n-2)$ for $n\geq 1$ (see [7] for
the occurrence of this sequence in the context of some polytopal conjectures
for Coxeter groups).

We note that creating a spiral pattern by arranging the natural numbers,
starting from the center and moving towards the bottom left, and then in a
thread-like manner around an ever-expanding hexagon, $\mathtt{NE}(n)$ is the
sequence of numbers on the radius formed from the center towards the North-
East corner, while $\mathtt{SW}(n)$ represents the sequence formed on the
opposite radius starting from the center towards the South-West corner.

Case $c\equiv 0\pmod{3}$. Starting with $(0,0,c)$, the iterations (23) lead to
a triple of the form $(s,s,s)$. For this to happen, the necessary number of
steps can be deduced from the equality

$-2nc+n(3n+1)=-(2n-1)c+n(3n-2),$

which implies $n=\frac{c}{3}$. Then the triple with all components equal is

$\left(-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c),-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c),-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c)\right).$
(24)

Translating up this triple by
$\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c)$, we arrive at the germ
$(0,0,0)$. Employing Theorem 3, it follows that if $c\equiv 0\pmod{3}$, then
$-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c)$ is the minimum weight in a
node of the triangular network that contains the triple $(0,0,c)$.

Case $c\equiv 1\pmod{3}$. Starting from $(0,0,c)$ and applying the iterations
of $H$, we arrive next to the center of the triangular network at a triple of
the form $(s,s,s+1)$. Employing formula (23), we see that this occurs if

$-2nc+n(3n+1)+1=-(2n-1)c+n(3n-2),$

that is, $n=\frac{c-1}{3}$. Then, applying the operator
$H^{\prime\prime\prime}$ to this last triple, we arrive at the triple with all
components equal, triple given by the same formula in (24). Then, as in the
previous case, translating the triple up by
$\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c)$, we arrive at the germ
$(0,0,0)$, so that, by Theorem 3, it follows that if $c\equiv 1\pmod{3}$, then
$-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c)$ is the minimum weight in a
node of the triangular network that contains the triple $(0,0,c)$.

Case $c\equiv 2\pmod{3}$. Proceeding as in the previous cases, one finds that
the first triple that is close to the center and is in the image of the
iterations of $H$ has the form $(s,s,s+2)$. In order for this to happen, we
need to take $n=\frac{c-2}{3}$. Then, applying $H^{\prime\prime\prime}$ on the
triple in (23) with $n=\frac{c-2}{3}$, we obtain

$\left(-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c-2),-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c-2),-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c+1)\right),$
(25)

triple which translated up by
$\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c+1)$ is seen to belong to the
tower of germ $(1,1,0)$.

In conclusion, by Theorem 3 it follows that if $c\equiv 2\pmod{3}$, then
$-\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c+1)$ is the minimum weight in
a node of the triangular network that contains the triple $(0,0,c)$.

Putting together the results in the tree cases, we find that the minimum
weight of a node in the triangular network that contains the triple $(0,0,c)$
is
$-\left\lfloor\mbox{\small$\displaystyle\frac{1}{3}$}(c^{2}-c+1)\right\rfloor$
for any $c\geq 1$. The sequence of the absolute values of the minima is
$0,1,2,4,7,10,14,19,24,30,37,\dots$, the remarkable sequence [28, A007980].
Among the many properties it has (see [33]), we mention that it equals the
number of partitions of $2n$ into at most three parts, and it is the number of
linearly-independent terms at $2n$-th order in the power series expansion of a
trigonal rotational-energy-surface that describes the potential energy of a
molecule as it undergoes rotational motion around a trigonal axis.

######  Remark 5.3.

Based on the analysis above, and in accordance with Theorem 1, it is possible
to make an accurate estimation and even calculate algorithmically the number
of nodes with negative weights. Thus, since the considered triples $(0,0,c)$,
with $c\geq 0$, are located in the triangular network they generate on the
boundary that separates negative weights from positive ones, and since the
minimum triple is in the center, it follows that the number of negative
weights is asymptotically equal to $\frac{2\pi}{3\sqrt{3}}c^{2}$ as
$c\to\infty$, at any level, in both towers that contain the considered
triples.

In order to give a sense of scale, we mention that the minimum weight in a
node of the triangular network that contains $(0,0,c)$, with $c=100$, is
$-3300$, and there are precisely $11946$ negative weights in the nodes. Then,
the ratio between the number of negative weights and the just mentioned
approximation is, in this case, only $11946\cdot\frac{3\sqrt{3}}{2\pi
c^{2}}\approx 0.98793$ instead of the limit $1$, which will be attained as $c$
tends to infinity.

### Conclusion

We have introduced a three folded operator (1), which, starting with a triple
of integers and moving back and forth from any point in any direction,
generates a triangular tessellation of the plane with integers in the nodes,
called weights. This operator further leads to a natural relation that defines
as equivalent tessellations obtained from each other through a translation. As
a consequence, we obtain exactly four classes of equivalence of tessellations
organized in four towers, generated in their very beginnings by $(0,0,0)$, on
one hand, and the more closely related the other three built around $(0,1,1)$,
$(1,0,1)$ and $(1,1,0)$, on the other hand. We have obtained a
characterization of the weights (see Theorems 1, 3, 4) and the densities of
their distribution in residue classes modulo a prime number (Theorem 2), and
along the way, we have encountered remarkable sequences that reveal intricate
patterns concealed within the networks. This has the potential to offer a
fresh perspective or open up an alternative approach to the practical
applications in which they appear, including the geographical model known as
the Central Place Theory [3, 27], recently enhanced with a fractal insertion
[1, 2], in a molecule or a virus model [29, Chapter 3], at the intersection
between arts and mathematics [32], or in the modern communication systems (see
[25] and the references therein). Additionally, it could be worthwhile to
continue studying the potential implications of the tesselations with integers
studied above in relation to the special lozenge tilings [11, 9] and their
implications in two dimensional electrostatics.

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