# Topological densities in Einstein-scalar-Gauss-Bonnet gravity

M. Bousder1 and Z Sakhi2
1Faculty of Science, Mohammed V University in Rabat, Rabat, Morocco
2Quantum Physics and Applications Team, LPMC, Faculty of Science Ben M’sik,
Casablanca Hassan II University, Morocco<EMAIL_ADDRESS>
###### Abstract

The present work is devoted to studying the background dynamical evolution of
a scalar field in Einstein-Gauss-Bonnet gravity in maximally symmetric space-
time. This study is useful for giving meaning to the presence of two Gauss-
Bonnet vacua, instead of using the spherically symmetric bubbles of the ”true”
vacuum expand in the ”false” vacuum. The theory admits two possible effective
cosmological constants, which lead to two maximally symmetric vacuum
solutions. The first solution corresponds to the dynamics of dark energy. When
there is matter, the second solution describes dark matter. In Einstein-Gauss-
Bonnet gravity, we establish the expression of the topological mass spectrum
which depends on the golden ratio and its inverse. In the Schwarzschild limit,
the topological density corresponds to the standard model radiation energy
density. We find the mass loss rate which gives the evolution of mass over
time.

##  1  Introduction

The observations of the rotation of galaxies and gravitational lenses indicate
the presence of a new mass, called dark matter (DM) hiding in galaxies, which
does not interact with radiation and matter, but can be detected by its
gravitational effect. The $\Lambda$CDM model is a cosmological model,
parametrized by a cosmological constant $\Lambda$ associated with cold dark
matter. It is often called the standard Big Bang model because it is the
simplest model that accounts for the properties of the cosmos: the large-scale
structure of the observable universe and the distribution of galaxies, the
abundance of light elements (hydrogen, helium and lithium) and the expansion
of the universe. $\Lambda$CDM model assumes that general relativity theory
correctly describes gravity on a cosmological scale. However, the $\Lambda$CDM
model presents several problems, such as the cosmological constant, fine-
tuning problem, and the problem of cosmic coincidence [1]. Recent developments
of non-trivial extension of Lovelock theory, namely Einstein-Gauss-Bonnet
(EGB) theory have been proposed [2], providing new insights into the
4-dimensional theory of gravity [3]. Their idea is that before taking the
limit $D\longrightarrow 4$, they multiplied the Gauss-Bonnet (GB) term by the
factor $1/(D-4)$. The divergent factor $1/(D-4)$ is canceled by the vanishing
GB contributions, which leads to a theory of gravity with only two dynamical
degrees of freedom, which is in contradiction with Lovelock theorem [4] which
describes the gravity at $D\geq 5$. However, it was shown in several papers
that the idea of the limit $D\longrightarrow 4$ is not clearly defined, as
well as the absence of proper action [5, 6, 7, 8]. It was explicitly confirmed
by a direct product $D$-dimensional spacetime or by adding a counter term,
before taking the limit $D\longrightarrow 4$, which can be seen as a class of
Horndeski theory [9] but with $2+1$-dofs. Although the EGB gravity is
currently debatable, the spherically symmetric black hole solution is still
meaningful and worthy of study [10].
The 4D symmetrical static and spherical black hole solution in EGB gravity
were obtained [11], also, solutions of static and spherically symmetric
compact stars [12, 13, 14]. Many researchers have studied the mass-radius
profile and the maximum mass in EGB gravity [15, 16]. Consequently, it is
possible to describe the matter inside compact objects (COs) and the dynamical
evolution of the matter at high density and the behavior of violent events.
The Einstein-scalar-Gauss-Bonnet (EsGB) gravity [17], are the class of
classical scalar-tensor theories that have second-order EOMs, as a special
case the Horndeski gravity [18]. The EGB gravity admits two maximally
symmetric vacuum solutions as the Einstein vacuum in $\alpha\rightarrow 0$,
and the Gauss-Bonnet vacuum in $\alpha\neq 0$ [19, 20]. Previous studies [21]
including the effect of varying the cosmological constant, showed that the
correspondence between ordinary thermodynamic systems and black hole mechanics
would be completed to include a variable cosmological constant. The bubble
nucleation probability depends on the curvature coupling of the Higgs fields,
which is a renormalizable parameter of the Standard Model (SM) in curved
spacetime [22]. In the effective field theory, the thermal and quantum
fluctuations to overcome the barrier are characterized by the decrease of the
vacuum. The bubble nucleation in thermal fluctuations can be described in
terms of Euclidean time coordinate by instantons [23]. The action of the
Coleman-de Luccia instanton determines the rate of vacuum decay [24]. The
classic solutions for switching from a false vacuum to a true vacuum are
called instantons. We suppose that the Boltzmann constant $k_{B}$, the reduced
Planck constant $\hbar$ and light speed $c,$ are such that: $k_{B}=\hbar=c=1.$
We begin in section 2 with a discussion of the equation of motion for a
coupling between a scalar field and the model. the chameleon mechanism and the
scalaron mass are investigated in section 3. Section 4 is devoted to the two
branches of solutions for the effective cosmological constant in a maximally
symmetric vacuum. In Section 5, we study the scalar dark matter in the
exterior region of CO by the EsGB and the regions of validity of the GB
functional coupling. In Section 6, we explore the applications for the 4D
Einstein-Gauss-Bonnet black hole. Finally, the paper ends in Section 7 by
summarizing our main results.

##  2 Einstein-scalar-Gauss-Bonnet gravity

We start by the action of Einstein-scalar-Gauss-Bonnet (EsGB) gravity [25, 26]
in $4$-dimensions as

$\mathcal{S}=\int
d^{4}x\sqrt{-g}\left(\frac{M_{p}^{2}}{2}R+f\left(\phi\right)\mathcal{G}+\mathcal{L}_{DM}\right)+\mathcal{S}_{m},$
(2.1)

where $S_{m}$ is the matter action, $M_{p}=\frac{c^{4}}{8\pi G}\approx 2\times
10^{18}\left[GeV\right]$, $R$ is the Ricci scalar, $f\left(\phi\right)$ is a
Gauss-Bonnet coupling function, which is ultraviolet (UV) corrections to
Einstein theory. We define the Gauss-Bonnet invariant [27] as

$\mathcal{G}=R^{2}-4R_{\mu\nu}R^{\mu\nu}+R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}.$
(2.2)

The variation with respect to the field $\phi$ gives us the equation of motion
for the scalaron field

$\square\phi-\partial_{\phi}V\left(\phi\right)+\mathcal{G}\partial_{\phi}f\left(\phi\right)=0.$
(2.3)

The variation of the action over the metric $g_{\mu\nu}$ simplified by the
Bianchi identity gives the equations of motion in [25]. In the Jordan frame,
the scalar DM Lagrangian reads [28]

$\mathcal{L}_{DM}=-\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-V\left(\phi\right)$
(2.4)

the kinetic term is invariant under the transformation $\phi\rightarrow-\phi$
of the $Z_{2}$ symmetry. The metric of a spatially flat homogeneous and
isotropic universe in FLRW model is given by:

$ds^{2}=-dt^{2}+a^{2}(t)\sum_{i=1}^{3}\left(dx^{i}\right)^{2},$ (2.5)

where $a(t)$ is a dimensionless scale factor, from which we define the Ricci
scalar $R$ and the GB invariant $\mathcal{G}$ in FLRW geometry as

$R=6\left(2H^{2}+\dot{H}\right)\text{ \ \ \ \
}\mathcal{G}=24H^{2}\left(\dot{H}+H^{2}\right).$ (2.6)

We start by considering $\phi=\phi\left(t\right)$. where
$\dot{\phi}=\partial_{t}\phi$,
$f^{\prime}\left(\phi\right)=\partial_{\phi}f\left(\phi\right)$ and
$H=\dot{a}/a$ is the Hubble parameter. Eq.(2.3) can be written as the Klein
Gordon equation in a simple form

$\ddot{\phi}+3H\left[\dot{\phi}-8f^{\prime}\left(\phi\right)H\left(\dot{H}+H^{2}\right)\right]+V^{\prime}\left(\phi\right)=0.$
(2.7)

##  3 Scalaron mass in EsGB gravity

Recently, there has been a renewed interest in the relationship between dark
matter and the scalaron mass (i.e. the mass of fields $\phi$) [31]. We will
study later the scalaron mass $m_{\phi}$ which will describe the dark matter.
To describe the mass of any scalar field, we use the Klein-Gordon equation
$\square\phi=\partial_{\phi}V_{eff}(\phi)$. Note that Eq.(2.3) gives the same
form as in the last equation. Since $\mathcal{G}$ in Eq.(2.3), does not depend
on $\phi$, in this case, the effective potential is evaluated as follows

$V_{eff}(\phi)=V\left(\phi\right)-f\left(\phi\right)\mathcal{G}.$ (3.1)

We notice that the effective potential of the scalaron includes the Gauss-
Bonnet coupling and the Gauss-Bonnet invariant. In other words, the Gauss-
Bonnet term affects the potential structure of the scalaron; thus, the
scalaron mass depends on the matter’s contribution. The particles of the field
$\phi$ come from the fluctuation around the minimum of the effective potential
$V_{eff}(\phi)$. The mass of small fluctuations around $\phi_{\min}$
Eq.(LABEL:acc) give the scalaron mass [55] is determined as

$m_{\phi}^{2}=\frac{\partial^{2}}{\partial\phi^{2}}V_{eff}(\phi_{\min}),$
(3.2)

where $V_{eff}(\phi_{\min})$ as a minimum value of the scalaron effective
potential $V_{eff}$. Also we have
$\mathcal{G}=\frac{\partial_{\phi}^{2}V\left(\phi\right)}{\partial_{\phi}^{2}f\left(\phi\right)}-\frac{\partial_{\phi}^{2}V_{eff}(\phi)}{\partial_{\phi}^{2}f\left(\phi\right)}$.
To make progress, let us express the $\mathcal{G}$ in a more convenient form

$\mathcal{G}=\frac{\partial_{\phi}^{2}V\left(\phi_{\min}\right)}{\partial_{\phi}^{2}f\left(\phi_{\min}\right)}-\frac{m_{\phi}^{2}}{\partial_{\phi}^{2}f\left(\phi_{\min}\right)}.$
(3.3)

For $f\left(\phi\right)=f_{0}e^{k\phi}$ and
$V\left(\phi\right)=V_{0}e^{-k\phi}$[25, 3] we obtain

$f\left(\phi_{\min}\right)\mathcal{G}=V\left(\phi_{\min}\right)-\frac{m_{\phi}^{2}}{k^{2}},$
(3.4)

From Eqs. (3.1, 3.3), we get

$V_{eff}(\phi_{\min})=\frac{m_{\phi}^{2}}{k^{2}}.$ (3.5)

where $\partial_{\phi}^{2}=\partial^{2}/\partial\phi^{2}$ and $V_{eff,\min}$
as a minimum value of the scalaron effective potential $V_{eff}$. The minimum
of the potential at $\phi=\phi_{\min}$ should satisfy
$\partial_{\phi}V_{eff}(\phi_{\min})=0$. It is shown that scalar fields can
explain the abundance of dark matter. The scalaron mass change according to
the trace of the energy-momentum tensor [33]. The minimum of the potential at
$\phi=\phi_{\min}$ should satisfy $\partial_{\phi}V_{eff}(\phi_{\min})=0$,
which give a special value of the GB invariant as

$\mathcal{G}=-\frac{V_{0}}{f_{0}}e^{-4R_{0}\phi_{\min}}$ (3.6)

also we have

$V_{eff}(\phi)=V\left(\phi\right)\left[1+e^{4R_{0}\left(\phi-\phi_{\min}\right)}\right]$
(3.7)

If $\phi_{\min}=0$, we get $V_{eff}(\phi)=2V_{0}\cosh\left(2R_{0}\phi\right)$.
Substituting Eqs.(LABEL:va,3.6) into Eq.(3.1) then into Eq.(3.2), the scalaron
mass can then be expressed as

$m_{\phi}=2\sqrt{2V_{0}}R_{0}e^{-R_{0}\phi_{\min}}.$ (3.8)

It is shown that scalar fields can explain the abundance of dark matter. The
scalaron mass change according to the trace of the energy-momentum tensor
[33]. In the large curvature limit [34] we have $R_{0}\phi_{\min}=1$, one
finds

$R\longrightarrow R_{0}=1/\phi_{\min}=M_{p}^{-2}\rho,$ (3.9)

where $\rho$ is the matter matter-energy. We can then express the scalaron
mass $m_{\phi}$ as a function of the matter-energy density

$m_{\phi}=\frac{2\sqrt{2V_{0}}}{eM_{p}^{2}}\rho.$ (3.10)

The scalaron field $\phi$ becomes dynamical in the low energy density
environment on the cosmological scale. Since the mass $m_{\phi}$ depends on
the matter density $\rho$, thus, the scalaron becomes heavy in the high-
density region of matter. This feature is called the chameleon mechanism which
is one of the screening mechanism in the modified gravity [35]. The chameleon
mechanisms is defined when the scalaron mass depends on the environment
surrounding the scalaron field. The scalaron $\phi$ is regarded as a dynamical
dark matter and can be a dark matter candidate. Similar topics had been
researched in many literatures [35, 36].

##  4 Maximally symmetric vacuum solutions

Recently, there are several papers which study the Particle production induced
by vacuum decay [45]. Following these concepts, here, we will interpret the
Gauss-Bonnet vacua by the production of dark matter particles. When gravity is
taken into consideration, the vacua are those with maximally symmetric spaces
[22]. In the maximally symmetric space, the scalar curvature of de Sitter
space is given by

$R=\frac{2D}{D-2}\Lambda,$ (4.1)

where $\Lambda$ is the cosmological constant. In the case of the positive
$\Lambda$, we have the de Sitter solution. In the large limit, we obtain$\
1/\phi_{\min}=\frac{2D}{D-2}\Lambda_{0}$. In $4$-dimensional vacuum, the
equation of motion has a solution that $R_{\mu\nu}=\Lambda g_{\mu\nu}$,
implies that $R=4\Lambda$. The Minkowski spacetime as the vacuum of the
$\Lambda_{0}=0$. The singularity problem [37] corresponds to
$R_{0}\rightarrow\infty$ (the curvature singularity) or $\phi_{\min}$ $=0$.
From Eq.(3.8), Eq.(3.9) and Eq.(4.1), for $e^{-2R_{0}\phi_{\min}}\approx
1-2R_{0}\phi_{\min}$ and $R_{0}\longrightarrow R$ one obtain

$2\phi_{\min}R^{2}-R+\frac{m_{\phi}}{2\sqrt{2V_{0}}}=0,$ (4.2)

which can reduce to

$4\phi_{\min}\left(\frac{D}{D-2}\right)^{2}\Lambda^{2}-\frac{D}{D-2}\Lambda+\frac{m_{\phi}}{4\sqrt{2V_{0}}}=0.$
(4.3)

In maximally symmetric vacuum solutions, there are two branches of solutions
for the effective cosmological constant,

$\Lambda_{\pm}=\frac{D-2}{8D\phi_{\min}}\left(1\pm\sqrt{1-\frac{4m_{\phi}}{\sqrt{2V_{0}}}\phi_{\min}}\right),$
(4.4)

In the limit where $2m_{\phi}\phi_{\min}\ll\sqrt{2V_{0}}$, the two branches
are given by

$\displaystyle\Lambda_{+}$ $\displaystyle=$
$\displaystyle\frac{D-2}{4D\phi_{\min}}\left(1-\frac{m_{\phi}\phi_{\min}}{\sqrt{2V_{0}}}\right),$
(4.5) $\displaystyle\Lambda_{-}$ $\displaystyle=$
$\displaystyle\frac{D-2}{4D}\frac{m_{\phi}}{\sqrt{2V_{0}}}.$ (4.6)

Using Eq. (3.10) and $1/\phi_{\min}=\frac{2D}{D-2}\Lambda_{0}$, the above
equations can be further rewritten as

$\Lambda_{+}=\frac{\Lambda_{0}}{2}\left(1-\frac{2\phi_{\min}\rho}{eM_{p}^{2}}\right)\text{;
\ }\Lambda_{-}=\Lambda_{0}\frac{\phi_{\min}\rho}{eM_{p}^{2}}.$ (4.7)

We notice that $\Lambda_{+}=\frac{\Lambda_{0}}{2}-\Lambda_{-}$. In particular,
the large curvature limit $1/4=\Lambda_{0}\phi_{\min}$, we have

$\frac{\Lambda_{0}}{2}=\Lambda_{+}+\Lambda_{-}\text{;
}\Lambda_{-}=\frac{\rho}{4eM_{p}^{2}};\text{ \ \ \
}\Lambda_{0}=\frac{1}{4\phi_{\min}}=\frac{\rho}{4M_{p}^{2}},$ (4.8)

or equivalently

$\Lambda_{+}=\frac{\left(e-2\right)}{2e}\Lambda_{0}\text{ ;\
}\Lambda_{-}=\frac{\Lambda_{0}}{e};\text{ \ \ \
}\Lambda_{0}=\frac{\rho}{4M_{p}^{2}}$ (4.9)

Let us now comment on the two solutions above. The cosmological constant
$\Lambda_{+}$ is equivalent to that found by [26] which is proportional to the
mass of the scalar field (chameleonic dark matter [33]). If $\rho$ decreases
over time, the value of $\Lambda_{+}$ increases to reach $\Lambda_{0}$. On the
other hand, the second branch $\Lambda_{-}$ depends on the matter density. We
notice that $\Lambda_{+}>0\ $and $\Lambda_{-}>0$, instead of studying the
false vacuum forming inside the true vacuum in the bubble geometry [39]. The
two roots $\left(\Lambda_{-},\Lambda_{+}\right)$ can represent two faces of
the same true vacuum. In a vacuum (without matter), the chameleon mechanism
will be zero, which corresponds to dark energy. The chameleon mechanism
appears when there is the matter Eq.(3.10), the vacua are those with
$\Lambda_{-}$. The vacuum $\Lambda_{-}$ will be spontaneously produced with
ordinary matter fields. The Gauss-Bonnet vacuum becomes a chameleon if there
is matter, which solves the problem of the Gauss-Bonnet vacuum suffering from
perturbative ghost instability [19]. This justifies that the matter has an
impact on the vacuum, is that the two vacua are separated by a domain wall,
composed of ordinary matter in the thin wall approximation [39]. In
particular, the large curvature limit $1/\phi_{\min}=R_{0}=4\Lambda_{0}$
Eq.(4.9), must satisfy $\Lambda_{0}=\rho/4M_{p}^{2}$ and
$\Lambda_{-}=\rho/4eM_{p}^{2}$, we can show that
$\Lambda_{+}+\Lambda_{-}=\Lambda_{0}$, which corresponds to
$\Lambda_{-}/\Lambda_{0}\approx 0,368\equiv 36,8\%$ and
$\Lambda_{+}/\Lambda_{0}\approx 0,132\equiv 13,2\%$. The percentage
$\Lambda_{-}/\Lambda_{0}\equiv 36,8\%$ is close to the density of matter and
dark matter in the universe [43]. It is interesting to note that
$50\%=\left(36,8\%+13,2\%\right)$ of space-time gets a mass in a region
occupied by $\left(\Lambda_{-},\Lambda_{+}\right)$. In this case, we introduce
the parameter $X$ which explains the lack of $50\%$ of $\Lambda_{0}$:
$\Lambda_{0}=\Lambda_{+}+\Lambda_{-}+X$, i.e. $X/\Lambda_{0}=0,5\equiv 50\%$.
So, we can associate the parameter $X$ to dark energy.

Energy content | parameter | pourcentage
---|---|---
Matter+Dark matter | $\Lambda_{-}$ | $\Lambda_{-}/\Lambda_{0}\equiv 36,8\%$
Dark energy | $\Lambda_{+}+X$ | $\left(\Lambda_{+}+X\right)/\Lambda_{0}=63,2\%$
Universe | $\Lambda_{0}$ | $100\%$
Table 1: Numerical estimate of the percentages of ordinary matter, dark matter
and dark energy in the universe [43].

We notice that the dark energy is described by the parameter $\Lambda_{+}$ and
an unknown parameter $X$ table (2).

##  5 Functional coupling and Barrow entropy

In this section, we explain in detail how to construct the equation of motion
of the Einstein-Gauss-Bonnet gravity. We begin by reviewing the the 4D EsGB
action

$\mathcal{S}=\frac{M_{p}^{2}}{2}\int
d^{4}x\sqrt{-g}\left(R+f\left(\phi\right)\mathcal{G}-\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-V\left(\phi\right)\right)+\mathcal{S}_{m},$

where $M_{p}=1/\left(8\pi G_{N}\right)=1.221\times 10^{19}GeV$ is the reduced
Planck mass, $R$ is the Ricci scalar, $S_{m}$ is the matter action and
$f\left(\phi\right)$ is a functional coupling of the scalar field $\phi$. In
the above equation $\left(\mu,\nu\right)=\left(0,1,2,3\right)$. We define the
GB term as

$\mathcal{G}\equiv
R^{2}-4R_{\mu\nu}R^{\mu\nu}+R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}.$ (5.1)

The Kretschmann scalar is $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$. The
variation with respect to the field $\phi$ gives us the equation of motion for
the scalar field

$\square\phi=\partial_{\phi}V_{eff}\left(\phi\right),$ (5.2)

where $\square\equiv\nabla_{\mu}\nabla^{\mu}$ and the effective potential is

$V_{eff}\left(\phi\right)=V\left(\phi\right)-f\left(\phi\right)\mathcal{G}.$
(5.3)

Varying the action (LABEL:F1) over the metric $g_{\mu\nu}$, we obtain the
following equations of motion:

$G^{\mu\nu}+\mathcal{K}^{\mu\nu}+f\left(\phi\right)\mathcal{H}^{\mu\nu}+\frac{1}{2}\left[\mathcal{T}_{\phi}^{\mu\nu}-g^{\mu\nu}V_{eff}\left(\phi\right)\right]=\frac{1}{2}\kappa^{2}T^{\mu\nu},$

where the Einstein tensor is $G^{\mu\nu}=R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R$,
the matter stress tensor is
$T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta\mathcal{S}_{m}}{\delta
g_{\mu\nu}}$. On the other hand the $\mathcal{K}^{\mu\nu}$ and
$\mathcal{H}^{\mu\nu}$ are given by

$\displaystyle\mathcal{K}^{\mu\nu}$ $\displaystyle=$ $\displaystyle
4[G^{\mu\nu}\square+\frac{1}{2}R\nabla^{\mu}\nabla^{\nu}+\left(g^{\mu\nu}R^{\rho\sigma}-R^{\mu\rho\nu\sigma}\right)\nabla_{\rho}\nabla_{\sigma},$
(5.4)
$\displaystyle-R^{\nu\rho}\nabla_{\rho}\nabla^{\mu}+R^{\mu\rho}\nabla_{\rho}\nabla^{\nu}]f\left(\phi\right)$
$\displaystyle\mathcal{H}^{\mu\nu}$ $\displaystyle=$ $\displaystyle
2R^{\mu\rho\sigma\tau}R_{\text{ \ }\rho\sigma\tau}^{\nu}-RR^{\mu\nu}$ (5.5)
$\displaystyle+\frac{1}{2}R_{\text{ \
}\rho}^{\mu}R^{\nu\rho}-R^{\mu\rho\sigma\tau}R_{\text{ \
}\rho\sigma\tau}^{\nu}.$

The tensor $\mathcal{K}^{\mu\nu}$ represents an operator which acts on
$f\left(\phi\right)$. The energy-momentum tensor for the scalar field is

$\mathcal{T}_{\phi}^{\mu\nu}=\nabla^{\mu}\phi\nabla^{\nu}\phi-\frac{1}{2}g^{\mu\nu}\nabla_{\rho}\phi\nabla^{\rho}\phi.$
(5.6)

The stress tensor for anisotropic compact object is given as

$T^{\mu\nu}=\left(\rho+P_{t}\right)u^{\mu}u^{\nu}+P_{t}g^{\mu\nu}+\left(P-P_{t}\right)\chi^{\mu}\chi^{\nu},$
(5.7)

with energy density $\rho_{CO}=\rho(r)\equiv c^{2}\rho(r)$, transverse
pressure $P_{t}(r)$ and radial pressure $P(r)$ of the homogeneously
distributed matter in the compact object (CO), where $u^{\mu}$ is the four-
velocity of the fluid and $\chi^{\mu}$ is the unit space-like vector in the
radial direction.
In limit $\alpha\longrightarrow 0$, one can see that this is equivalent to the
original form given by the Bekenstein-Hawking entropy. One can check that for
$D=2$, the topological term characterized by a vanishes Lovelock coupling
$\alpha$. The Barrow entropy [59] is a new black hole entropy which is given
by

$S=\pi\left(\frac{A}{A_{0}}\right)^{1+\delta/2},$ (5.8)

where $0\leq\delta\leq 1$, $A$ is the black hole horizon area and $A_{0}$ is
the Planck area [56]. When $\delta=0$, the area law is restored, i.e.
$S=\frac{A}{4G}$ $\left(\text{where }A_{0}=4G\right)$ while $\delta=1$
represents the most intricate and fractal structure of the horizon.
The form the functional $f\left(\phi\right)$ can take
$f\left(\phi\right)=e^{-\gamma\phi}$ [60], where $\gamma$ is a constant, which
corresponds to EGB gravity coupled with dilaton that arises as a low-energy
limit of the string theory [61]. Using the same principle as this entropy, we
assume that for the compact object (or black hole) exterior, there is a
presence of scalar fields $\phi$, while inside the compact object is replaced
by the Gauss-Bonnet (GB) coupling $\alpha$ as

$\begin{array}[]{c}\text{coupling constant=}f\left(\phi\right),\text{ \ \ CO\
exterior }\left(D=4\right)\\\ \text{coupling
constant=}\frac{\alpha}{D-4},\text{ \ \ CO\ interior }\left(D\rightarrow
4\right)\end{array}.$ (5.9)

The GB coupling $\alpha$ is measured in $km^{2}$. In the CO exterior interior,
we have rescaled the coupling constant
$\alpha\rightarrow\alpha/\left(D-4\right)$. The negative (positive) $\alpha$
leads to a decrease (increase) of the CO radius and the maximum mass [63]. If
$\alpha<0$ the solution is still the anti-de Sitter (AdS) space, if $\alpha>0$
the solution is the de Sitter (dS) space [62]. We investigate in detail the
impact of the Gauss-Bonnet coupling on the properties of an anisotropic
compact object, such as mass, radius and the factor of compactness.
Considering the limit $D\rightarrow 4$, and it has an effect on gravitational
dynamics in 4D. Additionally, at the CO boundary $(r=R)$, the GB coupling must
be continuous, i.e. $f\left(\phi\right)\equiv\frac{\alpha}{D-4}$. On the other
hand, the function $f\left(\phi\right)$ describes the star’s exterior region.
To study the equations of motion inside and outside CO, we differentiate
between two cases:
In CO interior ($D\rightarrow 4$) we have:

$G^{\mu\nu}+\alpha\left(\mathcal{H}^{\mu\nu}+\frac{1}{2}g^{\mu\nu}\mathcal{G}\right)=\frac{\kappa^{2}}{2}T^{\mu\nu}.$
(5.10)

In CO exterior ($D=4$) we have:

$G^{\mu\nu}+\mathcal{K}^{\mu\nu}+f\left(\phi\right)\left[\mathcal{H}^{\mu\nu}-\frac{1}{2f\left(\phi\right)}g^{\mu\nu}\left(\nabla_{\lambda}\phi\nabla^{\lambda}\phi+4V_{eff}\left(\phi\right)\right)\right]=\frac{1}{2}\kappa^{2}T^{\mu\nu},$
(5.11)

with
$g_{\mu\nu}\mathcal{T}_{\phi}^{\mu\nu}=-\nabla_{\lambda}\phi\nabla^{\lambda}\phi$.
In CO interior ($D\rightarrow 4$) we have: The GB invariant can be greatly
simplified to the matter density [64] and using Eq.(5.3) we obtain

$\begin{array}[]{c}\rho_{f(\phi)}=4\mathcal{G}-\frac{4}{f\left(\phi\right)}\left(\frac{1}{4}\nabla_{\lambda}\phi\nabla^{\lambda}\phi+V\left(\phi\right)\right),\text{
\ \ CO\ exterior }\left(D=4\right)\\\ \rho_{CO}=\mathcal{G},\text{ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ CO\ interior }\left(D\rightarrow 4\right)\end{array}.$ (5.12)

In this case, the term $\rho_{CO}$ represents the density of matter in compact
objects, and $\rho_{f(\phi)}$ is the density of dark matter surrounding these
objects. Note that the relation between $\rho_{f(\phi)}$ and $\rho_{CO}$
highlight the chameleon dark matter [65, 66]. Formally, at the points where
the dark matter density $\rho_{DM}$ equal to $\rho_{f(\phi)}$. For
$V\left(\phi\right)\approx-\frac{1}{4}\nabla_{\lambda}\phi\nabla^{\lambda}\phi$,
we obtain $\rho_{DM}\approx 4\rho_{CO}$, which is in good agreement with the
observation data of the percentages of dark matter and the matter in the
Universe [69, 57]: $\rho_{DM}\equiv 80\%$ and $\rho_{matter}(\rho_{CO})\equiv
20\%$.
Next, we assume that $\phi=\phi(t)$. In cosmological and quintessence behavior
[67, 68], the energy density $\rho_{\phi}$ and pressure $P_{\phi}$ of the
scalar field are given by

$\displaystyle\rho_{\phi}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\dot{\phi}^{2}+V\left(\phi\right),$ (5.13)
$\displaystyle P_{\phi}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\dot{\phi}^{2}-V\left(\phi\right).$

The quintessence models describe dark energy with a scalar field $\phi$. In
this case, $\rho_{\phi}$ and $P_{\phi}$ are respectively, the density and the
pressure of the dark energy (DE). The Planck Collaboration [69] provides
constraints on the equation of state
$\omega_{\phi}=P_{\phi}/\rho_{\phi}\approx-1.028\pm 0.032$. Starting from Eqs.
(5.13), we obtain

$\rho_{DM}=4\rho_{CO}+\frac{1}{f\left(\phi\right)}\left(P_{\phi}-3\rho_{\phi}\right).$
(5.14)

For small $f\left(\phi\right)$, the DM chameleon effect vanishes and the DM
density depends only on
$\left(P_{\phi}-3\rho_{\phi}\right)/f\left(\phi\right)$.
The scalar field always sits at the minimum of its effective potential. We
assume that a massive scalar field begins oscillating about a minimum. The
mass of small fluctuations around $\phi_{\min}$ gives a new scalar field mass
as effective mass by
$m_{eff}^{2}=\left.\frac{\partial^{2}}{\partial\phi^{2}}V_{eff}(\phi)\right|_{\phi=\phi_{\min}}$.
From Eqs. (5.3,5.13,5.14) we obtain

$m_{eff}^{2}=\frac{1}{2}\frac{\partial^{2}}{\partial\phi^{2}}\left(\rho_{\phi}-P_{\phi}+\frac{\rho_{CO}}{\rho_{DM}-4\rho_{CO}}\left(3\rho_{\phi}-P_{\phi}\right)\right)_{\phi=\phi_{\min}}.$
(5.15)

From Eq. (5.14), the functional coupling is given by

$f\left(\phi\right)=\frac{P_{\phi}-3\rho_{\phi}}{\rho_{DM}-4\rho_{CO}}.$
(5.16)

In the void ($\rho_{CO}=0,\omega_{\phi}\sim-1$), we have
$f\left(\phi\right)=-4\rho_{\phi}/\rho_{DM}$. In the CO surface, we assume
that $\rho_{DM}\approx 0$ and $P_{\phi}\approx 0$, so we get
$f_{surface}\left(\phi\right)\approx 3\rho_{\phi}/4\rho_{CO}$. Since
$\rho_{\phi}$ represents the density of DE according to quintessence, the
effect of DE is weak on the CO surface, which shows that
$f_{surface}\left(\phi\right)\left(\propto\rho_{\phi}\right)\rightarrow 0$.
Inside CO, we have ($P_{\phi}=\rho_{\phi}=0$), i.e. $f\left(\phi\right)=0$,
which exactly corresponds with the assumption Eq. (5.9). For this reason, we
exclude $f\left(\phi\right)$ inside matter, and we replace it with the GB
coupling $\alpha$ (see the next section).

##  6 EGB primordial black holes

We start by the action of Einstein-Gauss-Bonnet (EGB) gravity [48] in
$4$-dimensions as

$\mathcal{I}=\int
d^{D}x\sqrt{-g}\left(\frac{M_{p}^{2}}{2}R-2\Lambda+f\left(\phi\right)\left(R^{2}-4R_{\mu\nu}R^{\mu\nu}+R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\right)\right),$
(6.1)

where $S_{m}$ is the matter action, $M_{p}^{2}=\frac{c^{4}}{8\pi G}\approx
2\times 10^{18}\left[GeV\right]$, $R$ is the Ricci scalar. We define the
Gauss-Bonnet invariant as
$\mathcal{G}=\mathcal{L}_{2}=R^{2}-4R_{\mu\nu}R^{\mu\nu}+R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma},$
where $\alpha_{2}=\alpha/\left(D-4\right)$ is the Gauss-Bonnet coupling have
dimensions of $\left[length\right]^{2}$, that represent ultraviolet (UV)
corrections to Einstein theory. To solving the field equation we obtain the
black hole solution
$ds^{2}=-f(r)dt^{2}+\frac{1}{f(r)}dr^{2}+r^{2}\left(d\theta^{2}+\sin^{2}\theta
d\phi^{2}\right)$. Taking the limit $D\longrightarrow 4$, we obtain the exact
solution in closed form

$f(r)\approx
1+\frac{r^{2}}{2\alpha}\left(1-\sqrt{1+4\alpha\left(\frac{2M}{r^{3}}\pm\frac{1}{l^{2}}\right)}\right).$
(6.2)

This last solution could be obtained directly from the derivation done in
[49]. In the limit $r\longrightarrow\infty$ with vanishing black hole charge,
we asymptotically obtain the GR Schwarzschild solution. In the limit
$\alpha\longrightarrow 0$, we can recover the Reissner-Nordström-AdS solution.
If $\alpha<0$ the solution is still an AdS space, if $\alpha>0$ the solution
is a de Sitter (dS) space, [50]. The solutions show that the event horizon is
located at $R_{\pm}=M\pm\sqrt{M^{2}-\alpha},$ where $R_{H}=R_{+}$ and $R_{-}$
are the event horizon and the Cauchy horizon radius of the EGB black hole
[47]. We can express the ADM mass $M$ of the black hole in terms of $R_{H}$ by
solving $f(r)=0$ for $r=R_{H}$ resulting in

$M=\frac{l^{-2}R_{H}^{4}+R_{H}^{2}+\alpha}{2R_{H}}.$ (6.3)

The Hawking temperature of the EBG black hole can be calculated as

$T=\frac{3l^{-2}R_{H}^{4}+R_{H}^{2}-\alpha}{8\pi\alpha R_{H}+4\pi R_{H}^{3}}.$
(6.4)

The thermodynamic volume $V=\frac{4\pi R_{H}^{3}}{3}$ is the conjugate
variable to the pressure. The parameters $V$ and $A$ are the conjugate
quantities of the pressure $P$ and GB coupling parameter $\alpha$. The event
horizon in spacetime can be located by solving the metric equation: $f(r)=0$,
and from Eq .(6.2) we obtain

$\rho_{BH}+P=\frac{3}{8\pi R_{H}^{2}}+\frac{3\alpha}{8\pi R_{H}^{4}},$ (6.5)

where $M=\rho_{BH}V$, $V=4\pi r^{3}/3$ and $P=3/8\pi l^{2}$. For the limit
$\alpha\longrightarrow 0$, we can recover the density $\rho_{\Lambda}=3/16\pi
R_{H}^{2}$ of the holographic dark energy (DE) [70]. In the spatially flat
homogeneous and isotropic universe in the FLRW, the (modified) Friedmann
equations can be obtained [48] $H^{2}+\alpha H^{4}=\frac{8\pi
G}{3}\rho+\frac{\Lambda}{3}$ and $\left(H^{2}+\alpha H^{4}\right)\dot{H}=-4\pi
G\left(\rho+P\right)$. This equation looks like the Eq .(6.5), i.e. can
describe the dynamics in space-time associated to a black hole. For
Gauss–Bonnet branch, we introduce the topological density as

$\rho_{\alpha}=\rho_{\alpha}^{\left(2\right)}=\frac{\alpha}{16\pi R_{H}^{4}}.$
(6.6)

From Eq. (6.5) we obtain the Van der Waals equation:

$\left(P+\rho_{BH}\right)\left(V-\frac{4\pi\alpha
R_{H}}{3}\right)=\frac{R_{H}}{2}.$ (6.7)

In the limit $\alpha\longrightarrow 0$ and $P\gg\rho$, we can recover the
ideal gas law. The critical point occurs when $P(V)$. We note that the Gauss-
Bonnet coupling $\alpha$ represente measure of the average attraction between
particles. Before proceeding further, we note that the mass $M$ can be
interpreted as a chemical enthalpy [51], which is the total energy of the
black hole [52] including both the energy $PV$ and the internal energy $E$ and
required to displace the vacuum energy of its environment. The Hubble horizon
mass is connected with the Smarr formula as

$M_{H}+PV=\frac{R_{H}}{2}+\frac{\alpha}{2R_{H}},$ (6.8)

where $M_{H}=\frac{4\pi R_{H}^{3}}{3}\rho$ is the Hubble horizon mass. We have
also

$R_{H}\approx 2M-\frac{\alpha}{2M}.$ (6.9)

We can recover the Schwarzschild radius for $\alpha\approx 0$. This expression
gives an interpretation of the difference between Schwarzschild and EGB black
holes. From Eqs. (6.8,6.9), we obtain the the Smarr formula:

$M=M_{H}+PV+\frac{\alpha}{4M}\left(1-\frac{1}{1-\frac{\alpha}{4M^{2}}}\right).$
(6.10)

In the limit $\alpha\longrightarrow 0,$ we can recover the standard Smarr
formula. Let us mention that for the AdS-Schwarzschild limit and using
$M\approx M_{H}$, we get $PV\propto=-\alpha/4M\propto T$. This expression
gives an interpretation to the term that depends on $\alpha$, as a solution of
AdS-GB gravity in the presence of a perfect fluid [53]. This is not valid for
the dS-GB black holes.

$PV=\left(M-M_{H}\right)-\frac{\alpha}{4M}\left(1-\frac{1}{1-\frac{\alpha}{4M^{2}}}\right).$
(6.11)

For Gauss-Bonnet branch we obtain
$\left(1-\frac{4M^{2}}{\alpha}\right)\left(1-\frac{\alpha}{4M^{2}}\right)=1$.
One of two solutions to this equation is the Golden ratio ($\alpha>0$) in de
Sitter (dS) space: $\frac{\alpha}{4M^{2}}=\frac{1+\sqrt{5}}{2}$ or
$\frac{4M^{2}}{\alpha}=\frac{1+\sqrt{5}}{2}$. If $\alpha<0$ the solution is
still an AdS space: $\frac{\alpha}{4M^{2}}=\frac{1-\sqrt{5}}{2}$ or
$\frac{4M^{2}}{\alpha}=\frac{1-\sqrt{5}}{2}.$ We obtain the spectrum of mass
in the Gauss-Bonnet branch is

$M_{\alpha}=\frac{\sqrt{\gamma\alpha}}{2},$ (6.12)

with
$\gamma=\left\\{\frac{2}{1-\sqrt{5}},\frac{1-\sqrt{5}}{2},\frac{2}{1+\sqrt{5}},\frac{1+\sqrt{5}}{2}\right\\}$.
It can be written as
$\gamma\approx\left\\{-1.618,-0.618,0.618,1.618\right\\}$. This shows that the
Gauss-Bonnet coupling represents a topological black hole mass.

$\alpha$ | $\gamma$ | $M_{\alpha}$
---|---|---
$-0.5$ | $-1.618$ | $0.449$
$-0.5$ | $-0.618$ | $0.277$
$0.5$ | $0.618$ | $0.277$
$0.5$ | $1.618$ | $0.449$
Table 2: Numerical estimate of the values of the PBH mass, according to the
frequency $f$ and the interval of $N$ between $10^{-47}$ and $60$. Such as
$f=2.561\times 10^{33}\frac{N}{M}.$

Taking the derivative with respect to time $t$ in both sides of Eq. (6.9),
thus the mass loss rate of a black hole is obtained as

$\frac{dM}{dt}\approx\frac{\dot{R}_{H}}{2}\left(1-\frac{\alpha}{2M^{2}}\right),$
(6.13)

with $\dot{R}_{H}=\frac{dR_{H}}{dt}$. Using the 4D Stefan-Boltzmann law [54,
58] $\frac{dM}{dt}=-\pi^{2}AT^{4}/60$, with $T$ and $A$ are temperature and
area of black hole, respectively. Evidently, we have $\alpha/M^{2}\propto
T^{4}$. For the Gauss-Bonnet branch and from Eq. (6.12), we have
$\frac{dM}{dt}\approx\frac{\dot{R}_{H}}{2}\left(1-2\gamma\right)$. Taking the
Friedmann equation $HR_{H}=1$ [70] and from Eq .(6.9) we have

$M_{\pm}(t)=\frac{1}{4H\left(t\right)}\pm\frac{1}{4}\left(\frac{1}{H\left(t\right)}+4\alpha\right)^{1/2}.$
(6.14)

Focusing on $M_{-}(t)$, one can find the condition
$\frac{1}{H}\left(\frac{1}{H}-1\right)\geq\frac{4\alpha}{\theta^{2}}$.

Figure 1: Left, curves of $M_{\pm}$ vs $H$ from Eq. (6.14)with $\alpha=0.5$.
Right, $M_{+}$ vs $t$ (dashed line) and $M_{-}$ vs $t$ (solid line) for the
values $q=1/2$ and $\alpha=-0.5$ .

For $\alpha=0.5$ we plot $M_{\pm}$ vs $H$ Fig. (1, left). Using $H=q/t$ in the
radiation-dominated era ($q=1/2$) for $\alpha=-0.5$, we plot $M_{\pm}$ vs $t$
Fig. (1, right). In Fig. (1, right), the domain in red, represents the
interval where the evolution of the mass with respect to time begins to
increase. While, before this interval, the mass decreases over time which is
due to Hawking evaporation. According to this figure, two types of PBHs have
opposite properties such as the evolution of their masses over time.

##  7 Conclusion

The coupling between the scalar field $\phi$ and the 4D Einstein-Gauss-Bonnet
gravity action is studied. We have studied the model describing the compact
stars in 4D Einstein-Gauss-Bonnet gravity surrounded by scalar dark matter. We
have made a comparison between the star interior and the exterior region. In
this case, the Gauss-Bonnet coupling describes the interior structure of the
black hole, while the coupling function $f\left(\phi\right)$ describes the
black hole exterior region. We have shown that $f\left(\phi\right)=0$ inside
the black hole. By solving the equations of motion based on the chameleon
mechanism. We refer to $\phi$ as a chameleon field, since its physical
properties, such as its mass, depend sensitively on the environment. It can be
used to demonstrate the relation between the cosmological constant and the
matter density. The Gauss-Bonnet vacuum have a chameleon structure, and the
mass $m_{\phi}$ appears with a small fluctuations. Moreover, we have studied
the vacua in maximally symmetric solutions, and we obtain two branches of
solutions, i.e. there are two such vacua. The first solution corresponds to
the dark energy and the second represents the chameleonic vacua or dark
matter. We have found strong mixing between the two vacua, which is presented
by the chameleon mechanism. We also discussed the equation of state parameter
in the model of the scalar field $\phi$ minimally coupled to EGB gravity, also
in the maximally symmetric space. The percentages of the effective
cosmological constants are in good agreement with those of dark matter and
dark energy in the Universe.
We have considered 4D Einstein-Gauss-Bonnet gravity. We establish the
relationship between a topological mass spectrum and the golden ratio. We have
obtained the mass loss rate, which gives by taking the Friedmann equation of
type $HR_{H}=1$ [70], the evolution of the mass over time, and the evaporation
Hawking time. For such a choice and fixed values of Lovelock coupling, we have
plotted the mass of PBH versus the Hubble parameter and time.

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# Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers

Tomislav Došlić1,2, Luka Podrug${}^{1},$111Corresponding author.

1 University of Zagreb, Faculty of Civil Engineering
Kačićeva 26, 10000 Zagreb, Croatia
2 Faculty of Information Studies, Novo mesto, Slovenia

<EMAIL_ADDRESS>
###### Abstract

In this paper we explore two types of tilings of a honeycomb strip and derive
some closed form formulas for the number of tilings. Furthermore, we obtain
some new identities involving tribonacci numbers, Padovan numbers and
Narayana’s cow sequence and provide combinatorial proofs for several known
identities about those numbers.

Keywords: tilings; hexagonal lattice; tribonacci numbers; tetranacci numbers;
Padovan numbers; Fibonacci numbers; Narayana’s cow sequence

##  1 Introduction

Tilings or tesselations have been attracting human attention since the time
immemorial. They appear as natural solutions of many practical problems and
their aesthetic appeal motivates the interest that goes way beyond the limits
of their practical relevance. In mathematics, tiling-related problems appear
in almost all areas, ranging from purely recreational settings of plane
geometry all the way to the deep questions of eigenvalue count asymptotics for
boundary-value problems in higher-dimensional spaces [11, 12]. Many of those
problems, formulated in simple and intuitive terms and seemingly innocuous,
quickly turn out to be quite intractable in their generality. That motivates
interest in their restricted versions that might be more accessible. In this
paper we look at several such restricted problems when the area being tiled
has a given structure and the allowed tiles belong to a small set of given
shapes. In particular, we consider problems of tiling a narrow strip of the
hexagonal lattice in the plane by several types of tiles made of regular
hexagons. Similar problems for strips in square and triangular lattices have
been considered in several recent papers [1, 4, 5, 8].

The substrate (i.e., the area to be tiled) is a honeycomb strip $H_{n}$
composed of $n$ regular hexagons arranged in two rows in which the hexagons
are numbered starting from the bottom left corner, as shown in Figure 1 for
$n=12$. The number of hexagons in the strip will be called its length. The
choice of the substrate might seem arbitrary, but it provides a neat visual
model for a linear array of locally interacting units with additional longer
range connections: The inner dual of a strip of length $n$ is, in fact,
$P_{n}^{2}$, the path on $n$ vertices with edges between all vertices at
distance 2 in $P_{n}$. Another way to look at it is as the ladder graph with
descending diagonals, another familiar structure. Clearly, tilings with
monomers and dimers in the strip correspond to matchings in its inner dual,
thus enabling us to directly transfer known results about matchings into our
context. We refer the reader to the classical monograph by Lovász and Plummer
[9] for all necessary details on matchings.

246810121357911 Figure 1: A tiling of a honeycomb strip of length 12 using 4
dimers.

We start by examining the tilings of such strips by monomers (i.e., single
hexagons) and dimers made of two hexagons joined along an edge. Such tilings
have been considered recently by Dresden and Ziqian [4], who found that the
total number of such tilings is given by the tetranacci numbers. We refine
their results in several ways. First, in Section 2, we obtain formula for the
number of such tilings with a specified number of dimers. Then we consider
tilings with colored monomers and dimers in Section 3. Along the way we obtain
combinatorial proofs for generalizations of several identities involving
tetranacci numbers from the paper by Dresden and Ziqian; we present them in
Section 4. Section 5 is devoted to another type of restricted tilings of the
honeycomb strip. There we prohibit horizontal dimers but allow trimers of
consecutive hexagons. The total number of such tilings is given in terms of
tribonacci numbers, and Padovan and Narayana’s cow numbers appear as special
cases. Combinatorial proofs of some related identities are presented in
Section 6. The paper is closed by some remarks listing some open problems and
indicating several possible directions for future work.

##  2 Tiling a honeycomb strip with exactly $k$ dimers

In this section we consider a honeycomb strip of length $n$ and its tilings by
hexagonal monomers and dimers shown in Figure 2. We are interested in the
number of such tilings with a given number of dimers. The dimers can be in any
position; in Figure 2 we see a descending, a horizontal and an ascending
dimer, from left to right, respectively. Ascending and descending dimers will
be both called slanted when their exact orientation is not important. We
denote the number of all possible tilings of a honeycomb strip of length $n$
using exactly $k$ dimers by $c_{n,k}$.

Figure 2: Monomer and three possible positions of a dimer tile.

Dresden and Ziqian [4] proved that the total number of all possible ways to
tile a strip with monomers and dimers $h_{n}$ satisfies recursion

$h_{n}=h_{n-1}+h_{n-2}+h_{n-3}+h_{n-4}$ (2.1)

with initial values $h_{1}=1$, $h_{2}=2$, and $h_{3}=4$. It makes sense to
define $h_{0}=1$, accounting for the only possible tiling (the empty one) of
the empty honeycomb strip. Their recurrence is the same as the recurrence for
the tetranacci numbers $Q_{n}$ (A000078 in [10]) with shifted initial values.
Hence, $h_{n}=Q_{n+3}$. We wish to determine $c_{n,k}$, the number of such
tilings using exactly $k$ dimers, and hence $n-2k$ monomers. It is easy to see
that $c_{n,k}=0$ for $k>\left\lfloor\frac{n}{2}\right\rfloor$, since the strip
with $n$ hexagons can contain at most $\left\lfloor\frac{n}{2}\right\rfloor$
dimers. On the lower end, there is only one tiling without dimers, so
$c_{n,0}=1$ for all $n$. By stacking $k$ dimers at the beginning of the strip,
it is always possible to tile the remainder by monomers, so it follows that
all $c_{n,k}$ for $k$ between $1$ and $\left\lfloor\frac{n}{2}\right\rfloor$
will be strictly positive. Hence the numbers $c_{n,k}$ will be arranged in a
triangular array without internal zeros. In table 1 we give the list of
initial values that can be easily verified.

$\begin{array}[]{cc}c_{0,0}=1\\\ c_{1,0}=1\\\ c_{2,0}=1&c_{2,1}=1\\\
c_{3,0}=1&c_{3,1}=3\end{array}$

Table 1: Initial values of $c_{n,k}$.

In the next theorem we give a recurrence relation for $c_{n,k}$.

######  Theorem 1.

Let $n\geq 4$ be an integer and $c_{n,k}$ be the number of ways to tile a
honeycomb strip of a length $n$ by using exactly $k$ dimers and $n-2k$
monomers. Then the numbers $c_{n,k}$ satisfy the recurrence relation

$c_{n,k}=c_{n-1,k}+c_{n-2,k-1}+c_{n-3,k-1}+c_{n-4,k-2}$ (2.2)

with the initial conditions given in Table 1.

###### Proof.

We consider an arbitrary tiling which uses $k$ dimers and note that the $n$-th
hexagon can be tiled either by a dimer or by a monomer. The number off all
such tilings with the last hexagon tiled by a monomer is $c_{n-1,k}$, since
the number of dimers $k$ remains the same. If the last hexagon is a part of a
dimer, then we distinguish two possible situations: either the dimer is
slanted or it is horizontal. The number of tilings ending in a slanted dimer
is $c_{n-2,k-1}$, since the last dimer increases the length of a strip by two
and number of dimers by one. If the dimer is horizontal, it means that it must
cover the $(n-2)$-nd and the $n$-th hexagon. In that case, we have two
subcases: either the $(n-1)$-st hexagon is tiled by monomer, and the rest of
the strip can be tiled in $c_{n-3,k-1}$ ways, or $(n-1)$-st hexagon forms a
dimer with $n-3$-rd hexagon, and rest of the strip can be tiled in
$c_{n-4,k-2}$ ways. Described cases are illustrated in Figure 3 from left to
right, respectively.

$c_{n-1,k}$$n$

$c_{n-2,k-1}$$n-1$$n$

$c_{n-3,k-1}$$n-1$$n-2$$n$

$c_{n-4,k-2}$$n-3$$n-1$$n$$n-2$

Figure 3: All possible endings of a tiled honeycomb strip of length $n$.

Since the listed cases and subcases are disjoint and describe all possible
situations, the total number of tilings is the sum of the respective counting
numbers i.e., $c_{n,k}=c_{n-1,k}+c_{n-2,k-1}+c_{n-3,k-1}+c_{n-4,k-2}$, which
proves our theorem. ∎

We are now able to list the initial rows of the triangle of $c_{n,k}$, which
we do in Table 2 below.

$\begin{array}[]{c|ccccccccccc}n/k&0&1&2&3&4&5&6&\cdots&Q_{n}\\\ \hline\cr
0&1&&&&&&&&1\\\ 1&1&&&&&&&&1\\\ 2&1&1&&&&&&&2\\\ 3&1&3&&&&&&&4\\\
4&1&5&2&&&&&&8\\\ 5&1&7&7&&&&&&15\\\ 6&1&9&16&3&&&&&29\\\
7&1&11&29&15&&&&&56\\\ 8&1&13&46&43&5&&&&108\\\ \end{array}$

Table 2: The initial values of $c_{n,k}$.

The triangle of table 2 appears as the sequence A101350 in [10]. Its leftmost
column consists of all 1’s, counting the unique tilings without dimers. The
second column seems to be given by $c_{n,1}=2n-3$. Indeed, the only dimer in
the tiling can cover either hexagons $(i,i+1)$ for $1\leq i\leq n-1$ or
hexagons $(i,i+2)$ for $1\leq i\leq n-2$, resulting in $2n-3$ possible
tilings. As expected, the rows of the triangle sum to the (shifted) tetranacci
numbers,
$\sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}c_{n,k}=Q_{n+3}$,
since by disregarding values $k$, recurrence 2.2 becomes the defining
recurrence for the tetranacci numbers. The appearance of the Fibonacci numbers
as the rightmost diagonal, $c_{2n,n}=F_{n+1}$, can be readily explained by
looking at the inner dual of the strip. As mentioned before, it is the ladder
graph with the descending diagonal in each square, as shown in Figure 4 .
Clearly, tilings with $n$ dimers correspond to perfect

$L_{n}$$H_{2n}$$\longrightarrow$ Figure 4: Hexagonal strip of a length $2n$
and its inner dual

matchings in the inner dual. A simple parity argument dictates that no
diagonal can participate in such a perfect matching. By omitting the diagonals
we are left with a ladder graph and it is a well known folklore result that
perfect matchings in ladder graphs are counted by Fibonacci numbers. Somewhat
less obvious is the appearance of the convolution of Fibonacci numbers and
shifted Fibonacci numbers as the first descending subdiagonal,
$c_{2n+1,n}=\frac{1}{5}((n+2)F_{n+4}+(n-1)F_{n+2})=A023610(n)$, but it follows
by observing that the only monomer breaks the strip into two pieces each of
which can be tiled by dimers only, and the number of such tilings is obtained
by summing the corresponding products, hence leading to convolution. There are
no formulas in the OEIS for other columns or diagonals. In the rest of this
section we determine formulas for all elements of the triangle $c_{n,k}$.

It is well known that for the Fibonacci numbers one has
$c_{2n,n}=\sum\limits_{m=0}^{n}\binom{n-m}{m}$. By writing this as
$c_{2n,n}=\sum\limits_{m=0}^{n}\binom{n-m}{m}\binom{n-m}{0}$ and by noting
that a similar formula
$c_{2n+1,n}=\sum\limits_{m=0}^{n-1}\binom{n-m}{m}\binom{n-m}{1}$ can be
readily verified by induction, it becomes natural to consider
$\sum\limits_{m=0}^{n-k}\binom{n-m}{m}\binom{n-m}{k}$ as the formula for the
elements on descending diagonals. By shifting the indices $n\to n-k$ and $k\to
n-2k$ one arrives at expression for $c_{n,k}$.

######  Theorem 2.

The number of ways to tile a honeycomb strip of length $n$ using $k$ dimers
and $n-2k$ monomers is equal to

$c_{n,k}=\sum\limits_{m=0}^{k}\binom{n-k-m}{m}\binom{n-k-m}{n-2k}.$ (2.3)

Theorem 2 can be proven by induction, but we prefer to present a combinatorial
proof. To do that we need to some new terms and one lemma.

We say that a tiling of a honeycomb strip is breakable at the position $k$ if
a given tiling can be divided into two tiled strips, first strip of length $k$
and second of length $n-k$. Note that breaking the strip is only allowed along
the edge of the tile. If no such $k$ exist, we say that tiling is unbreakable.

For example, if first two hexagons form a dimer, the tiling is unbreakable at
position $1$, since it is not allowed to break a tiling through the dimer. As
an example, Figure 5 illustrates all breakable positions of a given tiling.

24681357 Figure 5: Tiling of a honeycomb strip that is breakable at positions
$2$, $3$ and $7$.

######  Lemma 1.

For $n>4$, every tiled strip of length $n$ is breakable into four types of
unbreakable tiled strips: length-one strip tiled with a single monomer,
length-two strip tiled with a single dimer, length-three strip tiled with a
horizontal dimer and a monomer, and length-four strip tiled with two
horizontal dimers.

###### Proof.

Every left or right slanted dimer forms a strip of length two. When removed,
we are left with smaller strips, each of them tiled with hexagons and
horizontal dimers. Every horizontal dimer occupies positions in the form
$\left\\{i,i+2\right\\}$. If position $i+1$ is occupied by a monomer, hexagons
in position $i,i+1$ and $i+2$ form a length-three tiled strip. If position
$i+1$ is occupied by another horizontal dimer, that dimer can occupy positions
$i-1,i+1$ or $i+1,i+3$. Either way, those two horizontal dimers form a length-
four tiled strip. After they are removed, we are left with only monomers,
where each monomer forms a simple tiled strip of length one. Those are only
for types of unbreakable tilings. They are illustrated in Figure 6. ∎

Figure 6: All unbreakable types of tiled strip. The second and the fourth can
be left or right slanted, and the third can be upside down, depending on the
parity of the position.

###### Proof of the Theorem 2.

We denote types of tiled strip from Figure 6 by $M$, $D$ ,$T$ and $V$, from
left to right, respectively. By Lemma 1, an arbitrary tiling of a strip
$H_{n}$ of length $n>4$ can be broken into those four types of unbreakable
tilings. Breaking of a given tiled strip into unbreakable strips produces
unique number of tiled strips of each type. So, let $k_{1}$ denotes the number
of strips of type $D$, $k_{2}$ the number of strips of type $T$, $k_{3}$ the
number of strips of type $V$, and since the strip has length $n$, what remains
are $n-2k_{1}-3k_{2}-4k_{3}$ strips of type $M$. Now we establish 1-1
correspondence between two sets: the first set, that contains all tilings of a
strip $H_{n}$ which by braking produce $k_{1}$ strips of type $D$, $k_{2}$
strips of type $T$, $k_{3}$ strips of type $V$ and $n-2k_{1}-3k_{2}-4k_{3}$
strips of type $M$, and the second set that contains all permutations with
repetition of a set with $n-k_{1}-2k_{2}-3k_{3}$ elements, where there are
$k_{1}$ elements of type $d$, $k_{2}$ elements of type $t$, $k_{3}$ elements
of type $v$ and $n-2k_{1}-3k_{2}-4k_{3}$ elements of type $m$. From an
arbitrary permutation we obtain the corresponding tiling as follows: we
replace elements $v$, $t$, $d$ and $m$ with a tiled strips of type $V$, $T$,
$D$ and $M$, respectively. For example, permutation $dmdvt$ yields the tiling
shown in Figure 7. The way to obtain a permutation from a given tiling is
obvious.

$d$$m$$d$$v$$t$ Figure 7: Tiling that corresponds to permutation $dmdvt$.

The number of all permutation in the second set is
$\frac{(n-k_{1}-2k_{2}-3k_{3})!}{k_{1}!k_{2}!k_{3}!(n-2k_{1}-3k_{2}-4k_{3})!}$.
One can easily verify that this expression can be written as
$\binom{n-k_{1}-2k_{2}-3k_{3}}{k_{1}}\binom{n-2k_{1}-2k_{2}-3k_{3}}{k_{3}}\binom{n-2k_{1}-2k_{2}-4k_{3}}{k_{2}}$.
Since we are interested in the number $c_{n,k}$ which denotes the number of
ways to tile a length-$n$ strip that contains exactly $k$ dimers, note that
$k_{1}+k_{2}+2k_{3}$ must be equal to $k$. Hence, we have

$\displaystyle c_{n,k}$
$\displaystyle=\sum\limits_{k_{1}+k_{2}+2k_{3}=k}\binom{n-k_{1}-2k_{2}-3k_{3}}{k_{1}}\binom{n-2k_{1}-2k_{2}-3k_{3}}{k_{3}}\binom{n-2k_{1}-2k_{2}-4k_{3}}{k_{2}}.$

By introducing new index of summation $m=k_{2}+k_{3}$ and by substitutions
$k_{2}=m-k_{3}$ and $k_{1}=k-k_{2}-2k_{3}=k-m-k_{3}$ we obtain:

$\displaystyle c_{n,k}$
$\displaystyle=\sum\limits_{m=0}^{k}\sum\limits_{k_{3}=0}^{m}\binom{n-k-m}{k-m-
k_{3}}\binom{n-2k+k_{3}}{k_{3}}\binom{n-2k}{m-k_{3}}$
$\displaystyle=\sum\limits_{m=0}^{k}\sum\limits_{k_{3}=0}^{m}\binom{n-k-m}{n-2k+k_{3}}\binom{n-2k+k_{3}}{n-2k}\binom{n-2k}{m-k_{3}}.$

Finally, by using identity
$\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m}$ on the first two
binomial coefficients and Vandermonde’s convolution
$\sum\limits_{k}\binom{n}{k}\binom{r}{m-k}=\binom{n+r}{m}$ (see [7]), we
arrive to:

$\displaystyle c_{n,k}$
$\displaystyle=\sum\limits_{m=0}^{k}\binom{n-k-m}{n-2k}\left(\sum\limits_{k_{3}=0}^{m}\binom{k-m}{k_{3}}\binom{n-2k}{m-k_{3}}\right)$
$\displaystyle=\sum\limits_{m=0}^{k}\binom{n-k-m}{n-2k}\binom{n-k-m}{m},$

which concludes our proof. ∎

Since the row sums in the Table 2 are tetranacci numbers, the Theorem 2 gives
us identity

$Q_{n+3}=\sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum\limits_{m=0}^{k}\binom{n-k-m}{m}\binom{n-k-m}{n-2k}.$

##  3 Tilings of honeycomb strip with colored dimers and monomers

Katz and Stenson [8] used colored squares and dominos to tile ($2\times
n$)-rectangular board and obtained a recursive relation for the number of all
ways to tile a board. They also proved some combinatorial identities involving
the number of such tilings. In this section we do a honeycomb strip analogue.
We continue to count tilings of a hexagon strip with dimers and monomers, but
we allow $a$ different colors for monomers and $b$ different colors for
dimers. Let $h^{a,b}_{n}$ denotes the number of all different tilings of a
strip with $n$ hexagons. It is convenient to define $h^{a,b}_{0}=1$. We start
with initial values illustrated in Figure 8. One can easily see that
$h^{a,b}_{1}=a$ since we have $a$ colors to choose from for a monomer.
Similarly, $h^{a,b}_{2}=a^{2}+b$, since we can tile a strip with two monomers
in $a^{2}$ ways or with one dimer in $b$ ways. For $n=3$, note that if we use
only monomers, we can choose colors in $a^{3}$ ways, and if we use one dimer
and one monomer, we can put dimer in $3$ different positions and for each of
that positions we can choose colors for tiles in $ab$ ways. Hence,
$h^{a,b}_{3}=a^{3}+3ab$.

$a$ colors

$a^{2}$ colors

$b$ colors

$a^{3}$ colors

$ab$ colors

$ab$ colors

$ab$ colors

Figure 8: All possible tilings for $n=1,2,3$.

In the next theorem we give a recursive relation for $h^{a,b}_{n}$.

######  Theorem 3.

For $n\geq 4$, the number of all possible tilings of the honeycomb strip
containing $n$ hexagons with $a$ different kinds of monomer and $b$ different
kinds of dimer satisfies the recursive relation

$h^{a,b}_{n}=a\cdot h^{a,b}_{n-1}+b\cdot h^{a,b}_{n-2}+ab\cdot
h^{a,b}_{n-3}+b^{2}\cdot h^{a,b}_{n-4}$

with the initial conditions $h^{a,b}_{0}=1$, $h^{a,b}_{1}=a$,
$h^{a,b}_{2}=a^{2}+b$, and $h^{a,b}_{3}=a^{3}+3ab$.

###### Proof.

The proof is similar to the proof of Theorem 1, but here we must also pay
attention to the colors. We consider an arbitrary tiling and note that $n$-th
hexagon can either be tiled by monomer or dimer. In the case when $n$-th
hexagon is tiled by monomer, the rest of the strip can be tiled in
$h^{a,b}_{n-1}$ ways, but the monomer can be colored in $a$ different ways,
which gives us the total of $a\cdot h^{a,b}_{n}$ possible ways. If the last
hexagon is a part of a dimer, then we distinguish two possible situations:
either the dimer is slanted, or the dimer is horizontal. The number of tilings
ending in a slanted dimer is $h^{a,b}_{n-2}$, and the last dimer can be
colored in $b$ ways. So there are $b\cdot h^{a,b}_{n-2}$ such tilings. As in
the proof of Theorem 1, if the dimer is horizontal, it means that it covers
the $(n-2)$-nd and the $n$-th hexagon. In that case, the $(n-1)$-st hexagon
can be tiled by monomer, we can choose colors in $ab$ ways, and the rest of
the strip can be tiled in $h^{a,b}_{n-3}$ ways. This gives us the $ab\cdot
h^{a,b}_{n-3}$ possible tiling in this case. The last case is if the
$(n-1)$-st hexagon forms a dimer with $(n-3)$-rd hexagon. There are
$b^{2}\cdot h^{a,b}_{n-4}$ such tilings. All cases are illustrated in Figure
9.

$h^{a,b}_{n-1}$$a$ colors$n$

$h^{a,b}_{n-2}$$b$ colors$n-1$$n$

$h^{a,b}_{n-3}$$ab$ colors$n-1$$n-2$$n$

$h^{a,b}_{n-4}$$b^{2}$ colors$n-3$$n-1$$n$$n-2$

Figure 9: All possible endings of a colored tiling of a strip with $n$
hexagons.

This gives us relation $h^{a,b}_{n}=a\cdot h^{a,b}_{n-1}+b\cdot
h^{a,b}_{n-2}+ab\cdot h^{a,b}_{n-3}+b^{2}\cdot h^{a,b}_{n-4}$, which proves
our theorem. ∎

We can now list some first values of $h^{a,b}_{n}$.

$\begin{array}[]{r|l}n&h^{a,b}_{n}\\\ \hline\cr 0&1\\\ 1&a\\\ 2&a^{2}+b\\\
3&a^{3}+3ab\\\ 4&a^{4}+5a^{2}b+2b^{2}\\\ 5&a^{5}+7a^{3}b+7ab^{2}\\\
6&a^{6}+9a^{4}b+16a^{2}b^{2}+3b^{3}\end{array}$

Table 3: Some first values of $h^{a,b}_{n}$.

We can notice that the values $c_{n,k}$ from the last section appear in every
row as coefficients of a bivariate polynomial. Connection between these values
is given in the next theorem:

######  Theorem 4.

The number $h^{a,b}_{n}$ of all possible tilings of the honeycomb strip of
length $n$ with monomers of $a$ different colors and dimers of $b$ different
colors is given by

$h^{a,b}_{n}=\sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}c_{n,k}a^{n-2k}b^{k}=\sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum\limits_{m=0}^{k}\binom{n-k-m}{m}\binom{n-k-m}{k-m}a^{n-2k}b^{k}.$

###### Proof.

We could prove the theorem by induction, but again we present a simple
combinatorial proof. The number $h^{a,b}_{n}$ denotes the number of all
possible tilings of the strip of a length $n$. For a fixed $0\leq
k\leq\lfloor\frac{n}{2}\rfloor$, there are $c_{n,k}$ possible ways to tile a
strip with exactly $k$ dimers, and since this tiling have $k$ dimers and
$n-2k$ monomers, the colors can be selected in $a^{n-2k}b^{k}$ ways which
gives a total of $c_{n,k}a^{n-2k}b^{k}$ possible tilings. Since every tiling
of the strip can contain $0$, $1$, … ,
$\left\lfloor\frac{n}{2}\right\rfloor-1$ or
$\left\lfloor\frac{n}{2}\right\rfloor$ dimers, the overall number of tilings
is the sum of these cases, that is
$h^{a,b}_{n}=\sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}c_{n,k}a^{n-2k}b^{k}$.
∎

##  4 Some (generalized) combinatorial identities involving tetranacci numbers

In this section we generalize several of the identities obtained by Dresden
and Ziqian [4] to the case of colored tilings of a honeycomb strip. All of the
following identities reduce to the mentioned identities of Dresden and Ziqian
by setting $a=b=1$.

######  Theorem 5.

For every $m,n\geq 0$

$h^{a,b}_{m+n}=h^{a,b}_{m}h^{a,b}_{n}+h^{a,b}_{m-1}\left(bh^{a,b}_{n-1}+abh^{a,b}_{n-2}+b^{2}h^{a,b}_{n-3}\right)+h^{a,b}_{m-2}\left(abh^{a,b}_{n-1}+b^{2}h^{a,b}_{n-2}\right)+b^{2}h^{a,b}_{n-1}h^{a,b}_{m-3}.$

###### Proof.

We consider a tiling of a honeycomb strip containing $m+n$ hexagons. We have
$h^{a,b}_{m+n}$ such tilings. On the other hand, there are $h^{a,b}_{m}\cdot
h^{a,b}_{n}$ tilings that are breakable at position $m$, as shown in the
Figure 10.

$h^{a,b}_{m}$$h^{a,b}_{n}$ Figure 10: Breakable tiling at position $m$.

All other tilings are unbreakable at position $m$. If that is the case,
unbreakability can occur because of the right-inclined, left-inclined or
horizontal dimer crossing the line of the break. Figure 11 shows all possible
situations that can occur if tiling is not breakable at position $m$. Note
that any tiling of a honeycomb strip is breakable if $n>4$.

$h^{a,b}_{m-1}$$h^{a,b}_{n-1}$$b$ colors

$h^{a,b}_{m-2}$$h^{a,b}_{n-1}$$ab$ colors

$h^{a,b}_{m-1}$$h^{a,b}_{n-2}$$ab$ colors

$h^{a,b}_{m-3}$$h^{a,b}_{n-1}$$b^{2}$ colors

$h^{a,b}_{m-2}$$h^{a,b}_{n-2}$$b^{2}$ colors

$h^{a,b}_{m-1}$$h^{a,b}_{n-3}$$b^{2}$ colors

Figure 11: Layouts that can occur if tiling is not breakable at position $m$.

Summing all these cases gives us the proof of the theorem. ∎

Our second identity counts tilings of the strip containing at least one dimer.

######  Theorem 6.

For every integer $n\geq 1$,

$h^{a,b}_{n}-a^{n}=bh^{a,b}_{n-2}+2b\sum\limits_{k=3}^{n}a^{k-2}h^{a,b}_{n-k}+b^{2}\sum\limits_{k=3}^{n}a^{k-3}h^{a,b}_{n-k-1}.$
(4.1)

###### Proof.

We prove the result by double counting of all ways of to tile a strip in which
there is at least one dimer. On one hand, there are $h^{a,b}_{n}-a^{n}$ such
tilings, since the only tiling without dimers uses only monomers, and we can
choose colors in $a^{n}$ ways. The other way to count such tilings keeps trace
of the position where the first dimer occurs. Since the dimer covers two
positions in the strip, we use the larger number to determine it position. For
example, dimer occupying hexagons $1$ (or $2$) and $3$ has a position $3$.
First we start with slanted dimers. If the position of the first dimer is $k$
for $k\geq 2$, the first part of the strip consists of $k-2$ monomers and the
rest of the strip can be tiled in $h^{a,b}_{n-k}$ ways, which gives us total
of $b\sum\limits_{k=2}^{n}a^{k-2}h^{a,b}_{n-k}$ ways. We must now consider the
horizontal dimer case. Note that the horizontal dimer cannot have positions
$1$ and $2$. If the position of dimer is $k$ for $k\geq 3$, then the dimer
occupies hexagons $k-2$ and $k$. We have two subcases, depending on whether
the $(k-1)$-st hexagon is tiled by a monomer or by a dimer. In the second
case, it must be paired with $(k+1)$-st hexagon, since position $k$ is first
to occur. In the first subcase, dimer and monomer can be colored in $ab$ ways,
the first part of the strip consisting of $k-3$ monomers can be colored in
$a^{k-3}$ ways, and the rest of the strip can be tiled in $h^{a,b}_{n-k}$
ways, which gives us the total of $b\sum\limits_{k=3}^{n}a^{k-2}h^{a,b}_{n-k}$
ways. The latter subcase involves two dimers, the first occupying hexagons
$k-2$ and $k$, and the second covering $k-1$ and $k+1$. These dimers can be
colored in $b^{2}$ ways, the first part of the strip consisting of $k-3$
monomers can be colored in $a^{k-3}$ ways, and the rest can be tiled in
$h^{a,b}_{n-k-1}$ ways. Since all the cases are disjoint, the overall number
is the sum of the respective counting numbers, which proves our theorem. ∎

We conclude this section with a pair of identities counting tilings of the
strip containing at least one monomer.

######  Theorem 7.

For every integer $n\geq 0$ we have

$h^{a,b}_{2n}-b^{n}F_{n+1}=a\sum\limits_{k=0}^{n}b^{k}h^{a,b}_{2n-2k-1}F_{k+2}$
(4.2)

and

$h^{a,b}_{2n-1}=a\sum\limits_{k=0}^{n}b^{k}h^{a,b}_{n-2k-1}F_{k+2}.$ (4.3)

###### Proof.

The number of tilings of the $2n$-strip containing only dimers is
$b^{n}F_{n+1}$. Hence, the number of tilings containing at least one monomer
is $h^{a,b}_{n}-b^{n}F_{n+1}$. On the other hand, we can count such tilings
based on the position of the first monomer. First we consider the odd
positions in the strip. If the first monomer occurs at position $2k+1$, for
some $0\leq k\leq n-1$, the first part of the strip is tiled only by dimers,
and that can be done in $b^{k}F_{k+1}$ ways, the monomer can be colored in $a$
ways, and the rest of the strip can be tiled in $h^{a,b}_{2n-2k-1}$ ways.
Figure 12 illustrates this case.

$b^{k}F_{k+1}$$h^{a,b}_{2n-2k-1}$$a$ colors2k+1 Figure 12: The hexagon occurs
at position $2k+1$.

Since the monomer can occur at any position $2k+1$ for $0\leq k\leq n-1$, the
total number of ways that monomer occurs at odd position is
$a\sum\limits_{k=0}^{n-1}b^{k}h^{a,b}_{2n-2k-1}F_{k+1}$.

Now we consider the even positions. The case is similar, but there are some
different details. If the first monomer occurs at position $2k$ for $1\leq
k\leq n$, then all $2k-1$ hexagons must be tiled with dimers. For this to be
possible, the $(2k+1)$-st hexagon must be tiled by the same dimer as
$(2k-1)$-st. This dimer and monomer can be colored in $ab$ ways. The first
part of the strip containing $2k-2$ hexagons can be tiled only by dimers and
in $b^{k-1}F_{k}$ ways, and the rest of the strip in $h^{a,b}_{2n-2k-1}$ ways.
This case is illustrated in Figure 13.

$b^{k-1}F_{k}$$h^{a,b}_{2n-2k-1}$$a$ colors$b$ colors2k2k+12k-1 Figure 13: The
hexagon occurs at position $2k$.

The number of tilings where monomer occurs at even position is
$ab\sum\limits_{k=0}^{n-1}b^{k-1}h^{a,b}_{2n-2k-1}F_{k}$, and the total number
of tilings is the sum of these two cases. Since $h^{a,b}_{-1}=0$ and
$F_{0}=0$, the first sum can be extended to $k=n$ and the second to $k=0$.

$\displaystyle h^{a,b}_{2n}-b^{n}F_{n+1}$
$\displaystyle=a\sum\limits_{k=0}^{n-1}b^{k}h^{a,b}_{2n-2k-1}F_{k+1}+ab\sum\limits_{k=1}^{n}b^{k-1}h^{a,b}_{2n-2k-1}F_{k}$
$\displaystyle=a\sum\limits_{k=0}^{n}b^{k}h^{a,b}_{2n-2k-1}F_{k+1}+a\sum\limits_{k=0}^{n}b^{k}h^{a,b}_{2n-2k-1}F_{k}$
$\displaystyle=a\sum\limits_{k=0}^{n}b^{k}h^{a,b}_{2n-2k-1}F_{k+2}$

The proof of second identity is similar. When the length of the strip is odd,
i.e. $2n-1$, the left hand side is $h^{a,b}_{n}$, since it cannot be tiled
only by dimers, and the proof for the right hand side is the same, hence the
theorem follows.

∎

##  5 Tiling of a honeycomb strip and tribonacci numbers

The tribonacci numbers (sequence A000073 in OEIS [10]) are the sequence of
integers starting with $T_{0}=0$, $T_{1}=0$ and $T_{2}=1$ and defined by
recursive relation

$T_{n}=T_{n-1}+T_{n-2}+T_{n-3},\text{ for }n\geq 3.$ (5.1)

For the reader’s convenience we list a first few values of the sequence in
Table 4.

$\begin{array}[]{c|ccccccccccc}n&0&1&2&3&4&5&6&7&8&9&10\\\ \hline\cr
T_{n}&0&0&1&1&2&4&7&13&27&44&81\end{array}$

Table 4: The first few values of tribonacci numbers.

In this section we are still interested in counting all tilings of a honeycomb
strip of a given length, but now by using different types of tiles. We still
allow monomers and slanted dimers, but we prohibit horizontal dimers. In
addition, we allow trimers of consecutively numbered hexagons. By prohibiting
horizontal dimers we effectively suppress longer-range connections represented
by horizontal edges in the inner dual. Also, by allowing trimers of the form
$\left\\{i-1,i,i+1\right\\}$ we abandon the context of matchings and instead
work with packings in the inner dual. The allowed tiles are illustrated in
Figure 14.

Figure 14: The allowed types of tiles.

Let $g_{n}$ denotes the number of ways to tile a hexagonal strip of length $n$
by using only the allowed tiles. It is convenient to define $g_{0}=1$, and it
is immediately clear that $g_{1}=1$, $g_{2}=2$.

######  Theorem 8.

Let $g_{n}$ denote the number of all ways to tile a honeycomb strip of length
by using only the allowed tiles. Then

$g_{n}=T_{n+2},$

where $T_{n}$ denotes $n$-th tribonacci number.

###### Proof.

We start with an arbitrary tiling of a strip. There are three disjoint cases
involving the $n$-th hexagon. If the hexagon is tiled by a monomer, then the
rest of the strip can be tiled in $g_{n-1}$ ways. If it is covered by a dimer,
there are $g_{n-2}$ such tilings, and finally, if the rightmost hexagon is
covered by a trimer, the are $g_{n-3}$ such tilings. By summing the respective
numbers we obtain a recurrence that is the same as the defining recurrence for
the tribonacci numbers, and the initial values determine the value of the
shift. ∎

In the next part, we refine our results by counting the number of tilings with
a fixed number of trimers, dimers or monomers. We denote these numbers by
$t_{n,k}$, $u_{n,k}$ and $v_{n,k}$, respectively, where $n$, as usual, denotes
the length of a strip, and $k$ the number of tiles of a certain kind. We can
also fix the number of all types of tiles. Let $g_{n}^{k,l}$ denotes the
number of all ways to tile a strip of a length $n$ using exactly $k$ trimers,
$l$ dimers and $n-3k-2l$ monomers. We list some first values in the Table 5.
From the definition it is clear that $t_{n,k}=0$ for
$k>\left\lfloor\frac{n}{3}\right\rfloor$, $u_{n,k}=0$ for
$k>\left\lfloor\frac{n}{2}\right\rfloor$ and $v_{n,k}=0$ for $k>n$. It is also
convenient to define $t_{0,0}=u_{0,0}=v_{0,0}=1$. For these sequences we can
obtain recursive relations in the obvious way, by considering the state of the
last hexagon to see whether it is covered by a trimer, by a dimer, or by a
monomer. The recursive relations are:

$t_{n,k}=t_{n-1,k}+t_{n-2,k}+t_{n-3,k-1},$ (5.2)
$u_{n,k}=u_{n-1,k}+u_{n-2,k-1}+u_{n-3,k}$ (5.3)

and

$v_{n,k}=v_{n-1,k-1}+v_{n-2,k}+v_{n-3,k}.$ (5.4)

We can now list some first values of the corresponding triangles:

$\begin{array}[]{ccc}\begin{array}[]{c|ccc}n/k&0&1\\\ \hline\cr 0&1\\\ 1&1\\\
2&2\\\ 3&3&1\\\ 4&5&2\\\ 5&8&5\\\ 6&13&10&1\\\ 7&21&20&3\\\
8&34&38&9\end{array}&\begin{array}[]{c|ccccc}n/k&0&1\\\ \hline\cr 0&1\\\
1&1\\\ 2&1&1\\\ 3&2&2\\\ 4&3&3&1\\\ 5&4&6&3\\\ 6&6&11&6&1\\\ 7&9&18&13&4\\\
8&13&30&27&10&1\\\
\end{array}&\begin{array}[]{c|ccccccccc}n/k&0&1&2&3&4&5&6&7&8\\\ \hline\cr
0&1\\\ 1&0&1\\\ 2&1&0&1\\\ 3&1&2&0&1\\\ 4&1&2&3&0&1\\\ 5&2&3&3&4&0&1\\\
6&2&6&6&4&5&0&1\\\ 7&3&7&12&10&5&6&0&1\\\ 8&4&12&16&20&15&6&7&0&1\\\
\end{array}\\\ t_{n,k}&u_{n,k}&v_{n,k}\\\ \end{array}$

Table 5: Initial values for $t_{n,k}$, $u_{n,k}$ and $v_{n,k}$.

The first and the second triangle of Table 5 are not in the OEIS, while the
third one appears as A104578 [10], the Padovan convolution triangle. The same
arguments as the ones used on $c_{n,k}$ shows that the rows of those triangles
do not have internal zeros, with the obvious exception of the zeros appearing
in the first descending subdiagonal of $v_{n,k}$.

Before we go any further, we introduce two closely related sequences defined
by Fibonacci-like recurrences of length three, namely the Narayana’s cows
sequence (A000930) and the Padovan sequence (A000931). We denote the $n$-th
element of these sequences by $N_{n}$ and $P_{n}$, respectively. The initial
values are $N_{0}=N_{1}=N_{2}=1$ and $P_{0}=1$, $P_{1}=P_{2}=0$, and for
$n\geq 3$ we have recursive relation $N_{n}=N_{n-1}+N_{n-3}$ and
$P_{n}=P_{n-2}+P_{n-3}$. We refer the reader to [10] for more details about
those sequences. In particular, we draw the reader’s attention to the fact
that there are several other sequences referred to as the Narayana numbers,
for example A001263, a very important triangle of numbers refining the Catalan
numbers and appearing in many different contexts. In the rest of this paper,
when we refer to Narayana’s numbers, we always mean A000930.

We now take a closer look at sequences $t_{n,0}$, $u_{n,0}$ and $v_{n,0}$,
i.e., at the number of tilings where one type of tile is omitted. The sequence
$t_{n,0}=F_{n+1}$, since such tilings contain only slanted dimers and
monomers; since such tilings correspond to matchings in the path on $n$
vertices, they are counted by Fibonacci numbers.

The sequence $u_{n,0}$ counts the number of all ways to tile a length-$n$
strip by using only monomers and trimers, hence its elements satisfy the
defining recurrence for the Narayana’s cow sequence. Similarly, since the
elements of the sequence $v_{n,0}$ are the numbers of all different tiling
where monomers are omitted, they satisfy Padovan’s recursion. We have
$u_{n,0}=N_{n}$ and $v_{n,0}=P_{n+3}$. In the next three theorems we present
connection between the number of tilings and above listed sequences. It turns
out that the elements of the three triangles of Table 5 can be expressed by
convolution-like formulas involving the Fibonacci, the Narayana’s and the
Padovan numbers. Such formulas could have been anticipated from the second
column of triangle $t_{n,k}$ which seems to be the (shifted) self-convolution
of Fibonacci numbers and also from the name of the entry A104578 in OEIS.

######  Theorem 9.

For $n\geq 0$, the number of ways to tile a strip with $n$ hexagons using
exactly $k$ trimers is

$t_{n,k}=\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-2k+1\end{subarray}}F_{i_{0}}\cdots F_{i_{k}}.$ (5.5)

###### Proof.

If there are no trimers in the tiling, one can only use dimers or monomers to
tile a strip and the number ways to do that is $t_{n,0}=F_{n+1}$. If we use
exactly $k$ trimers, those trimers divide our strip into $k+1$ smaller strips.
In this sense we allow the strip to be of a length $0$ if two trimers are
adjacent; the sub-strips of length 0 can also appear at the beginning or at
the end of a strip. We have a strip with $n$ hexagons which is tiled with $k$
trimers, so there are $n-3k$ hexagons left to tile. Since the position of each
trimer is arbitrary, the lengths of strips between and around them can vary
from $0$ to $n-3k$, but the sum of the lengths must be constant, that is
$i_{0}+i_{1}+\dots+i_{k}=n-3k$. Each of those smaller strips can be tiled only
by dimers or monomers, hence in $t_{i_{j},0}$ ways, where $0\leq j\leq k$.
Summing this over all positions of the trimers we have:

$\displaystyle t_{n,k}=$
$\displaystyle\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-3k\end{subarray}}t_{i_{0},0}\cdots t_{i_{k},0}$
$\displaystyle=$
$\displaystyle\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-3k\end{subarray}}F_{i_{0}+1}\cdots F_{i_{k}+1}$
$\displaystyle=$
$\displaystyle\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-2k+1\end{subarray}}F_{i_{0}}\cdots F_{i_{k}}.$

∎

Note that Theorem 9 allow us to express the tribonacci numbers as a double sum

$T_{n+2}=\sum_{k=0}^{n}t_{n,k}=\sum_{k=0}^{n}\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq
0\\\ i_{0}+\cdots+i_{k}=n-2k+1\end{subarray}}F_{i_{0}}\cdots F_{i_{k}}.$ (5.6)

######  Theorem 10.

For $n\geq 0$, the number of ways to tile a strip with $n$ hexagons using
exactly $k$ dimers is

$u_{n,k}=\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-2k\end{subarray}}N_{i_{0}}\cdots N_{i_{k}}.$ (5.7)

###### Proof.

We already know that the number of tilings with no dimers is $u_{n,0}=N_{n}$.
Now we look at the tilings of the strip with $n$ hexagons that have exactly
$k$ dimers. That leaves us with $n-2k$ hexagons to be tiled by monomers and
trimers. As in the proof of Theorem 9, we note that $k$ dimers divide the
strip into $k+1$ smaller strips, each of the length $0\leq i_{j}\leq n-2k$.
Each smaller strip can be tiled in $N_{i_{j}}$ ways, and after summing over
all possible positions of $k$ dimers we have

$u_{n,k}=\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-2k\end{subarray}}N_{i_{0}}\cdots N_{i_{k}}.$

∎

The next result gives a new combinatorial interpretation of sequence A104578
of [10].

######  Theorem 11.

For $n\geq 0$, the number of ways to tile a strip with $n$ hexagons using
exactly $k$ monomers is

$v_{n,k}=\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}>0\\\
i_{0}+\cdots+i_{k}=n+2k+3\end{subarray}}P_{i_{0}}\cdots P_{i_{k}}.$ (5.8)

###### Proof.

The proof will be analogous to the two previous proofs. The number of tilings
with no monomers is $v_{n,0}=P_{n+3}$. A monomer does not divide our strip,
but if it first appears in position $i$, we will consider strips left and
right from it. We count the number of tilings of the strip $H_{n}$ that have
exactly $k$ monomers. That leaves us $n-k$ untiled hexagons. Omitting $k$
hexagons leaves us with with $k+1$ smaller strips, each of the length $0\leq
i_{j}\leq n-k$. Each smaller strip can be tiled in $P_{i_{j}+3}$ ways, and
after summing over all possible positions of $k$ monomers we have

$\displaystyle v_{n,k}=$
$\displaystyle\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}\geq 0\\\
i_{0}+\cdots+i_{k}=n-k\end{subarray}}P_{i_{0}+3}\cdots P_{i_{k}+3}$
$\displaystyle=$
$\displaystyle\sum\limits_{\begin{subarray}{c}i_{0},\dots,i_{k}>0\\\
i_{0}+\cdots+i_{k}=n+2k+3\end{subarray}}P_{i_{0}}\cdots P_{i_{k}}.$

∎

Now we turn our attention to the number of tilings of a strip of length $n$
with numbers of all types of tiles fixed. Recall that number of tiling
consisting of $k$ trimers, $l$ dimers and $n-3k-2l$ monomers is denoted by
$g^{k,l}_{n}$. In the next theorem we give a closed form formula for
$g^{k,l}_{n}$.

######  Theorem 12.

For $n\geq 0$, the number of ways to tile a strip with $n$ hexagons using
exactly $k$ trimers, $l$ dimers and $n-2k-l$ monomers is

$g^{k,l}_{n}=\binom{n-3k-l}{l}\binom{n-2k-l}{k}.$ (5.9)

###### Proof.

Consider a set consisting of all arbitrary tilings of a length-$n$ strip that
have exactly $k$ trimers, $l$ dimers and $n-3k-2l$ monomers. To prove this
theorem, we establish a 1-1 correspondence between that set and the set of all
permutations of $n-2k-l$ elements where we have $k$ elements $t$, $l$ elements
$d$ and $n-3k-2l$ elements $m$. From an arbitrary permutation we obtain the
corresponding tiling as follows: we replace each element $t$ with a trimer,
each element $d$ with a dimer, and each element $m$ with a monomer. In this
manner we obtained a tiling of a strip of length $n$ with prescribed number of
tiles of each type. For example, the permutation $tmdmt$ corresponds to the
tiling shown in the Figure 15.

Figure 15: Tiling corresponding to the permutation $tmdmt$.

In an obvious way we can also obtain a permutation from a given tiling. Since
the total number of permutations of this set is
$\frac{(n-2k-l)!}{k!l!(n-3k-2l)!}$, we arrive to:

$\displaystyle g^{k,l}_{n}$
$\displaystyle=\frac{(n-2k-l)!}{k!l!(n-3k-2l)!}\cdot\frac{(n-3k-l)!}{(n-3k-l)!}$
$\displaystyle=\frac{(n-3k-l)!}{l!(n-3k-2l)!}\cdot\frac{(n-2k-l)!}{k!(n-3k-l)!}$
$\displaystyle=\binom{n-3k-l}{l}\binom{n-2k-l}{k}.$

∎

From Theorem 12 we arrive to yet another identity for tribonacci numbers:

$T_{n+2}=\sum\limits_{k=0}^{\left\lfloor\frac{n}{3}\right\rfloor}\sum\limits_{l=0}^{\left\lfloor\frac{n-3k}{2}\right\rfloor}\binom{n-3k-l}{l}\binom{n-2k-l}{k}$
(5.10)

Specially, if we set $k=0$ in the first equation we have

$\displaystyle t_{n,0}$
$\displaystyle=\sum\limits_{l=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n-l}{l}\binom{n-l}{0}$
$\displaystyle=\sum\limits_{l=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n-l}{l}$
$\displaystyle=F_{n+1}.$

Since Theorem 12 gives us the number of all tilings using the prescribed
number of tiles of each type, we can express values $t_{n,k}$ and $u_{n,k}$ in
a new way by summing over $l$ and $k$, respectively.

######  Corollary 1.

For $n\geq 0$,

$t_{n,k}=\sum\limits_{l=0}^{\left\lfloor\frac{n-3k}{2}\right\rfloor}\binom{n-3k-l}{l}\binom{n-2k-l}{k},$
(5.11)

and

$u_{n,l}=\sum\limits_{k=0}^{\left\lfloor\frac{n-2l}{3}\right\rfloor}\binom{n-3k-l}{l}\binom{n-2k-l}{k}.$
(5.12)

##  6 Some identities involving tribonacci numbers

In this section we prove, in a combinatorial way, several identities involving
the tribonacci, Narayana’s, Padovan and Fibonacci numbers. We begin with a
well-known identity for tribonacci numbers and we give it a new combinatorial
interpretation:

######  Theorem 13.

For $n\geq 4$,

$T_{n}+T_{n-4}=2T_{n-1}.$

###### Proof.

Let $\mathcal{G}_{n}$ denotes the set of all tilings of a length-$n$ strip,
$\mathcal{M}_{n}$, $\mathcal{D}_{n}$ and $\mathcal{T}_{n}$ the tilings ending
with a monomer, dimer or trimer, respectively. As before, the cardinal number
of the set $\mathcal{G}_{n}$ is $g_{n}$. It is clear that
$\mathcal{T}_{n}=\mathcal{M}_{n}\dot{\cup}\mathcal{D}_{n}\dot{\cup}\mathcal{T}_{n}$.
To prove the theorem we have to establish 1-1 correspondence between sets
$\mathcal{G}_{n-2}\cup\mathcal{G}_{n-6}$ and
$\mathcal{G}_{n-3}\times\left\\{0,1\right\\}$.

To each tiling from the set $\mathcal{G}_{n-3}$ we add a monomer at the end to
obtain an element of $\mathcal{M}_{n-2}$. Thus, we obtained bijection between
the sets $\mathcal{G}_{n-3}$ and $\mathcal{M}_{n-2}$. In this way, we have
used all the tiling of the set $\mathcal{G}_{n-3}$ once. Now we take the
tilings from the set $\mathcal{G}_{n-3}$ again, and if it ends with a trimer,
i.e. if it is an element of $\mathcal{T}_{n-3}$, we remove it to obtain a
tiling of length $n-6$, i.e., element of a set $\mathcal{G}_{n-6}$. If it ends
with a dimer (an element of $\mathcal{D}_{n-3}$), we remove it and replace it
with a trimer to obtain element from $\mathcal{T}_{n-2}$. Finally, if the
tiling is an element of $\mathcal{M}_{n-3}$, we replace the last monomer with
a dimer, to obtain an element of $\mathcal{D}_{n-2}$. In this way we have used
every tiling of a length $n-3$ twice and obtained all tilings of a length
$n-2$ and $n-6$ exactly once. Diagram that visualize 1-1 correspondence
between the two sets is shown in the Figure 16.

$\mathcal{G}_{n-2}$$\mathcal{M}_{n-2}$$\mathcal{D}_{n-2}$$\mathcal{T}_{n-2}$$\mathcal{G}_{n-6}$$\mathcal{G}_{n-3}$$\mathcal{G}_{n-3}$$\mathcal{M}_{n-3}$$\mathcal{D}_{n-3}$$\mathcal{T}_{n-3}$
Figure 16: 1-1 correspondence between sets
$\mathcal{G}_{n-2}\cup\mathcal{G}_{n-6}$ and
$\mathcal{G}_{n-3}\times\left\\{0,1\right\\}$.

It follows that $g_{n-2}+g_{n-6}=2g_{n-3}$, and since $g_{n}=T_{n+2}$, the
theorem follows.∎

For the next few identities is is useful to recall the definition of
breakability. We say that a tiling of a honeycomb strip is breakable at the
position $k$ if given tiling can be divided into two tiled strips, the first
one containing the leftmost $k$ hexagons and the second one containing the
rest.

Our next identity differentiates tilings based on the breakability.

######  Theorem 14.

For any integers $m,n\geq 1$ we have the identity

$T_{m+n}=T_{m}T_{n}+T_{m+1}T_{n+1}+T_{m-1}T_{n}+T_{m}T_{n-1}.$

###### Proof.

We consider an arbitrary tiling of a strip of length $m+n-2$. If the tiling is
breakable at position $m-1$, we divide it into two strips of a length $m-1$
and $n-1$. Hence, the total number of tiling in this case is $g_{m-1}g_{n-1}$.
If the tiling is not breakable at position $m-1$, that means that either a
dimer or a trimer is blocking it. If the dimer is preventing the tiling from
breaking, there are strips of lengths $m-2$ and $n-2$ on each side, so the
total number of tilings in this case is $g_{m-2}g_{n-2}$. If the trimer is
blocking it, it can reduce the length of the left or of the right strip by
two. So the total number of tilings in this case is
$g_{m-3}g_{n-2}+g_{m-2}g_{n-3}$. By summing the contributions of all these
cases we obtain
$g_{m+n-2}=g_{m-1}g_{n-1}+g_{m-2}g_{n-2}+g_{m-3}g_{n-2}+g_{m-2}g_{n-3}$, and
by using the equality $g_{n}=T_{n+2}$ we have
$T_{m+n}=T_{m}T_{n}+T_{m+1}T_{n+1}+T_{m-1}T_{n}+T_{m}T_{n-1}$. ∎

The next identity was proved by Frontczak [6] by using generating functions.
Here we provide a combinatorial interpretation.

######  Theorem 15.

For any integer $n\geq 0$ we have the identity

$T_{n+2}=\sum\limits_{k=0}^{n+1}F_{k}T_{n-k}.$

###### Proof.

We prove this theorem by counting all ways to tile a strip by using at least
one trimer. The total number of ways to tile a length-$n$ strip without
trimers is $t_{n,0}=F_{n+1}$, hence the number of tilings having at least one
trimer is $T_{n+2}-F_{n+1}$. On the other hand, we can count the same tilings
by observing where the first trimer appears. If the leftmost trimer occupies
hexagons $\left\\{i,i+1,i+2\right\\}$, we say that the position of trimer is
$i$. So, all possible positions range from $1$ to $n-2$. If a trimer first
appears at position $k$, the leftmost $k-1$ hexagons are tiled only by
monomers and dimers, and the number of all ways to do that is $F_{k}$. The
rest of the strip, of length $n-k-2$, can be tiled in $T_{n-k}$ ways. By
summing over all possible positions of the leftmost trimer, we have
$T_{n+2}-F_{n+1}=\sum\limits_{k=1}^{n-2}F_{n}T_{n-k}$. By using
$T_{n-3}=T_{n}-T_{n-1}-T_{n-2}$, one can extend the tribonacci numbers to
negative integers and obtain $T_{-1}=1$. Since $T_{1}=T_{0}=0$, the sum above
can be extended to obtain $T_{n+2}=\sum\limits_{k=1}^{n+1}F_{k}T_{n-k}$, which
concludes our proof. ∎

######  Theorem 16.

For any integer $n\geq 0$ we have the identity

$T_{n+2}=\sum\limits_{k=0}^{n}N_{k}T_{n-k}+N_{n}.$

###### Proof.

We prove this theorem by counting all ways to tile a strip by using at least
one dimer. The proof is analogous to the previous one. The total number of
ways to tile a strip of length $n$ without dimers is $u_{n,0}=N_{n}$, hence
the number of tilings having at least one dimer is $T_{n+2}-N_{n}$. Similarly
as before, we can count the same thing by observing where the leftmost dimer
appears. If the leftmost dimer occupies hexagons $\left\\{i,i+1\right\\}$, we
say that its position is $i$. So, all possible positions range from $1$ to
$n-1$. If a dimer first appears at position $k$, the leftmost $k-1$ hexagons
can be tiled in $N_{k-1}$ ways. The rest of the strip is of length $n-k-1$ and
it can be tiled in $T_{n-k+1}$ ways. By summing over all possible positions of
the leftmost dimer we have
$T_{n+2}-N_{n}=\sum\limits_{k=1}^{n-1}N_{k-1}T_{n-k+1}$. Some rearranging of
indexes and fact that $T_{0}=T_{1}=0$ bring us to
$T_{n+2}-N_{n}=\sum\limits_{k=0}^{n}N_{k}T_{n-k}$ and our proof is over. ∎

######  Theorem 17.

For any integer $n\geq 0$ we have the identity

$T_{n+2}=\sum\limits_{k=1}^{n}P_{k+2}T_{n-k+2}+P_{n+3}.$

###### Proof.

Analogously as in two previous theorems, we prove this theorem by counting all
ways to tile a strip by using at least one monomer. The number of ways to tile
a strip of length $n$ with at least one monomer is
$g_{n}-v_{n,0}=T_{n+2}-P_{n+3}$. Now we can count the same thing by observing
the position of the leftmost monomer. All possible positions for first monomer
range from $1$ to $n$. If it first appears at position $k$, the first part of
the strip, i.e., the leftmost $k-1$ hexagons, can be tiled in $P_{k+2}$ ways.
The rest of the strip is of length $n-k$ and can be tiled in $T_{n-k+2}$ ways.
By summing over all possible positions of the leftmost monomer we have
$T_{n+2}-P_{n+3}=\sum\limits_{k=1}^{n}P_{k+2}T_{n-k+2}$, which concludes our
proof. ∎

##  7 Concluding remarks

In this paper we have considered various ways of tiling a narrow honeycomb
strip of a given length with different types of tiles. We have refined some
previously known results for the total number of tilings of a given type by
deriving formulas for the number of such tilings with prescribed number of
tiles of a given type. We have also considered tilings with colored tiles and
obtained the corresponding formulas. Along the way, we have provided
combinatorial interpretations for some known identities and established a
number of new ones. Also, we have provided closed-form expressions for several
triangles of numbers appearing in the OEIS.

In order to keep this contribution at a reasonable length, we have omitted
many interesting problems related to the considered ones. In particular, we
have not considered any jamming-related scenarios, i.e., the tilings which are
suboptimal with respect to the number of large(r) tiles. The existence of
connections of our tilings with such problems is indicated by the appearance
of Padovan numbers in both contexts [3]. Further, we have not examined
statistical properties such as the expected number of tiles in a random tiling
of a strip of a given length in a way done in ref. [2]. We have not looked at
the asymptotic behavior of the counting sequences. Each of the mentioned
omissions could be an interesting topic for further research.

Another interesting direction would be to look in more detail at triangles
$c_{n,k}$, $t_{n,k}$, $u_{n,k}$, and $v_{n,k}$. We have shown that their rows
(with one trivial exception) do not have internal zeros. By inspection of the
first few rows of $c_{n,k}$, $t_{n,k}$, and $u_{n,k}$ one can observe that the
rows seem to be also unimodal and even log-concave. It would be interesting to
investigate whether those properties hold for the whole triangles. Both
properties are violated in rows of $v_{n,k}$, but the violations seem to be
restricted to the right end. What happens if the rightmost three elements are
omitted? Also, the position of maximum presents an interesting challenge.
Finally, more interesting identities could be derived by looking at ascending
and descending diagonals of different slopes in those triangles. We hope that
at least some of the listed problems would be addressed in our future work.

## Acknowledgment

T. Došlić gratefully acknowledges partial support of Slovenian ARRS via grants
J1-3002 and P1-0383.

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