\documentclass{mc} %%%%% journal info -- DO NOT CHANGE %%%%%%%%% \setcounter{page}{211} \renewcommand\thisnumber{x} \renewcommand\thisyear {2017} \renewcommand\thismonth{xxx} \renewcommand\thisvolume{22} \renewcommand\datereceived{January 22, 2016} \renewcommand\dateaccepted{December 20, 2016} %%%%%%%% end journal info %%%%%%%%%%%%% \usepackage[displaymath,mathlines,pagewise]{lineno} %Rad zaprimljen dana: 22.1.2016. %Rad prihvacen dana: 20.12.2016. %%%%% author macros %%%%%%%%% \renewcommand{\arraystretch}{1.35} \def\hpp{${\cal HPP}$ } \def\hpt{${\cal HPT}$ } \def\ppHpq{${\cal PP}_{\!pq}$ } \def\ppHq{${\cal PP}_{\!4,q}$ } \def\ppH45{${\cal PP}_{\!4,5}$ } \def\H2R{$\mathbf{H}^2\!\times\!\mathbf{R}$\!\! } \def\binomh#1#2{ \scalebox{.3}[1.2]{\textbf{)}}{\genfrac{}{}{0pt}{}{#1}{#2}}\scalebox{.3}[1.2]{\textbf{(}} } \def\binomhh#1#2{ \scalebox{.4}[1.7]{\textbf{)}}{\genfrac{}{}{0pt}{}{#1}{#2}}\scalebox{.4}[1.7]{\textbf{(}} } %%%%% end %%%%%%%%% %%%%%%%%% already loaded packages -- DO NOT RELOAD %%%%% % \usepackage{amssymb} % \usepackage{amsmath,amsthm} % \usepackage{latexsym} % \usepackage{amsfonts} % \usepackage{amsbsyt} % \usepackage{graphicx} %%%%%% end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{graphics} \graphicspath{{figure_eps/}} \begin{document} %\begin{linenumbers} %%%%% title and author(s): % \markboth{Author(s)}{Short Title} % \title{Title} \markboth{L.\,N\'emeth} {Pascal pyramid in the space $\mathbf{H}^2\!\times\!\mathbf{R}$} \title{Pascal pyramid in the space $\mathbf{H}^2\!\times\!\mathbf{R}$} % single author: % \author[AUTHOR]{AUTHOR} % \address{address of AUTHOR} % \email{{\tt email address of AUTHOR} (AUTHOR)} %\author[D. Juki\'c]{Dragan Juki\'c\corrauth} %\address{Department of Mathematics, University of Osijek, Trg Ljudevita Gaja 6, HR-31 000 Osijek, Croatia} %\email{{\tt jukicd@mathos.hr} (D. Juki\'c)} \author[L.\,N\'emeth]{ L\'aszl\'o N\'emeth\corrauth} \address{University of Sopron, Institute of Mathematics, Bajcsy Zs.\,u.\,4, H9\,400, Sopron, Hungary} \email{{\tt nemeth.laszlo@uni-sopron.hu} (L.\,N\'emeth)} % multiple authors: % Please mark \corrauth after the name of the corresponding author. % different addresses: %\author[AUTHOR1 and AUTHOR2]{AUTHOR1\affil{1}\comma\corrauth and AUTHOR2\affil{2}} %\address{\affilnum{1}\ address of AUTHOR1\\ \affilnum{2}\ address of AUTHOR2} % %same address: %\author[AUTHOR1, AUTHOR2 and AUTHOR3]{AUTHOR1, AUTHOR2\corrauth and AUTHOR3} %\address{address of AUTHOR1, AUTHOR2 and AUTHOR3} % %\emails{{\tt email of AUTHOR1} (AUTHOR1), {\tt email of AUTHOR2} (AUTHOR2), {\tt email of AUTHOR3} (AUTHOR3)} % %%%%% Begin Abstract %%%%%%%%%%% \begin{abstract} In this article, we introduce a new type of Pascal pyramids. A regular squared mosaic in the hyperbolic plane yields an $(h^2r)$-cube mosaic in space $\mathbf{H}^2\!\times\!\mathbf{R}$ and the definition of the pyramid is based on this regular mosaic. The levels of the pyramid inherit some properties from the Euclidean and hyperbolic Pascal triangles. We give the growing method from level to level and show some illustrating figures. \end{abstract} %%%%% end %%%%%%%%%%% %%%%% Keywords %%%%%%%%%%% \keywords{Pascal's triangle, hyperbolic Pascal triangle, Pascal pyramid, regular mosaics, cubic honeycomb, Thurston geometries, prism tiling in space $\mathbf{H}^2\!\times\!\mathbf{R}$, recurrence sequences} %%%% AMS subject classifications %%%% \ams{52C22, 05B45, 11B99} %%%% maketitle %%%%% \maketitle %%%% Start %%%%%% \section{Introduction}\label{sec:introduction} There are several approaches to generalize the Pascal's arithmetic triangle (see, for instance \cite{AhBel,Barry,BSz,SV}). A new type of its variations is based on the hyperbolic regular mosaics denoted by Schl\"{a}fli's symbol $\{p,q\}$, where $(p-2)(q-2)>4$ (\cite{C}). Each regular mosaic induces a so-called hyperbolic Pascal triangle (see \cite{BNSz}), following and generalizing the connection between the classical Pascal's triangle and the Euclidean regular square mosaic $\{4,4\}$. For more details, see \cite{BNSz,N_pyr,NSz_alter,NSz_recu}, but here we also collect some necessary information. We use the attribute Pascal's (with the apostrophe) only to refer to the original, Euclidean arithmetic triangle and pyramid. The hyperbolic Pascal triangle based on the mosaic $\{p,q\}$ can be envisaged as a digraph, where the vertices and the edges are the vertices and the edges of a well defined part of lattice $\{p,q\}$, respectively, and the vertices possess a value that give the number of different shortest paths from the base vertex to the given vertex. In this article, we build on the hyperbolic squared mosaics, thus the other properties hold just for mosaic $\{4,q\}$. Figure~\ref{fig:Pascal_layer6} illustrates the hyperbolic Pascal triangle when $\{p,q\}=\{4,5\}$. Here the base vertex has two edges, the leftmost and the rightmost vertices have three, the others have $q$ edges. The quadrilateral shape cells surrounded by the appropriate edges correspond to the squares in the mosaic. Apart from the winger elements, certain vertices (called ``Type $A$'') have $2$ ascendants and $q-2$ descendants, while the others (``Type $B$'') have $1$ ascendant and $q-1$ descendants. In the figures, we denote vertices of type $A$ by circles and vertices of type $B$ by diamonds, while the wingers are denoted by white diamonds (according to the denotations in \cite{BNSz}). The vertices which are $n$-edge-long far from the base vertex are in row $n$. The general method of preparing the graph is the following: we go along the vertices of the $j^{\text{th}}$ row, according to the type of the elements (winger, $A$, $B$), and we draw the appropriate number of edges downwards ($2$, $q-2$, $q-1$, respectively). Neighbour edges of two neighbour vertices of the $j^{\text{th}}$ row meet in the $(j+1)^{\text{th}}$ row, constructing a new vertex of type $A$. The other descendants of row $j$ have type $B$ in row $j+1$. In the sequel, $\binomh{n}{k}$ denotes the $k^\text{th}$ element in row $n$, which is either the sum of the values of its two ascendants or the value of its unique ascendant. We note that the hyperbolic Pascal triangle has the property of vertical symmetry. \begin{figure}[h!] \centering \includegraphics[width=0.99\linewidth]{Pascal_layer6_e} \vspace*{0.2cm} \caption{\rm\em Hyperbolic Pascal triangle linked to $\{4,5\}$ up to row 6} \label{fig:Pascal_layer6} \end{figure} The 3-dimensional analogue of the original Pascal's triangle is the well-known Pascal's pyramid (or more precisely Pascal's tetrahedron). Its levels are triangles and the numbers along the three edges of the $n^{\text{th}}$ level are the numbers of the $n^{\text{th}}$ line of Pascal's triangle. Each number inside in any level is the sum of the three adjacent numbers on the level above (see \cite{ANVI,B, har,Hilton}). In the hyperbolic space based on the hyperbolic regular cube mosaic (cubic honeycomb) with Schl\"afli's symbol $\{4,3,5\}$ was defined a hyperbolic Pascal pyramid \mbox{(\hpp\!\!)} as a generalisation of the hyperbolic Pascal triangle (\hpt\!\!) linked to mosaic $\{4,5\}$ and the classical Pascal's pyramid (\cite{N_pyr}). The space \H2R is one of the eight simply connected 3-dimensional maximal homogeneous Riemannian geometries (Thurston geometries \cite{MolPSz}, \cite{Thur}). This Seifert fibre space is derived by the direct product of the hyperbolic plane $\mathbf{H}^2$ and the real line $\mathbf{R}$. For more details, see \cite{Szir}. In what follows we define the Pascal pyramids in this space based on the so-called $(h^2r)$-cube mosaics similarly to the definition of \hpp (in \cite{N_pyr}). The definition could also be extended to the other regular tiling, but we deal with the prism tilings with a square, because they are the most natural generalizations of the original Pascal's triangle and pyramid. This work was suggested by Professor Emil Moln\'{a}r. \section{Construction of the Pascal pyramid \ppHq} In the space \H2R, we define an infinite number of so-called $(h^2r)$-cube mosaics. We take a hyperbolic plane $\Pi$ as a reference plane and a regular squared mosaic with $\{4,q\}$ ($q\geq5$) on it. Denote by $d$ the common length of the sides of the squares in this mosaic. We consider the hyperbolic planes parallel to $\Pi$, where the distance between two consecutive ones is $d$. Let the same mosaic $\{4,q\}$ be defined on all planes and let the corresponding vertices of the mosaics be on the same Euclidean lines which are perpendicular to the hyperbolic planes. An $(h^2r)$-cube is the convex hull of two corresponding congruent squares on two consecutive hyperbolic mosaics. All $(h^2r)$-cubes yield an $(h^2r)$-cube mosaic in the space \H2R based on the regular hyperbolic planar mosaic $\{4,q\}$. (In \cite{Szir}, the $(h^2r)$-cube mosaics were called prism tilings with squares.) Figure \ref{fig:H2R_disk} shows three consecutive hyperbolic planes with mosaic $\{4,5\}$ and some lines perpendicular to these planes. Let $V$ be a mosaic vertex on a hyperbolic plane. The vertex figure of $V$ is a double $q$-gon based pyramid, where the vertices are the nearest mosaic vertices to $V$, all vertices are one-edge-long far from $V$. Their base vertices are on the same hyperbolic plane as $V$. The vertex figure of $V$ is illustrated in Figure \ref{fig:H2R_disk} (or Figure \ref{fig:H2R_vertex_figures}). We mention that the edges of vertex figures are not the edges of the mosaic, they are the diameters of its faces. \begin{figure}[h!] \centering \includegraphics[width=0.5\linewidth]{H2R_disk} \vspace*{0.2cm} \caption{\rm \em Construction of the $(h^2r)$-cube mosaic based on $\{4,5\}$} \label{fig:H2R_disk} \end{figure} Take the part $\mathcal{T}$ of the mosaic $\{4,q\}$ on $\Pi$ on which the hyperbolic Pascal triangle \hpt can be defined (see \cite{BNSz}) and let ${\cal P}$ be the part of the $(h^2r)$-cube mosaic which contains $\mathcal{T}$ and all its corresponding vertices on other hyperbolic planes that are "above" plane $\Pi$ (the hyperbolic planes which are in the same half space bordered by $\Pi$). (Obviously, ${\cal P}$ contains also the corresponding edges between the vertices.) The shape of this convex part of the mosaic resembles an infinite tetrahedron. This part ${\cal P}$ is darkened in Figure \ref{fig:H2R_disk}. Let $V_0$ be the base vertex of \hpt on plane $\Pi$. Let ${\cal G}_{\cal P}$ be the digraph directed according to the growing edge-distance from $V_0$, in which the vertices and edges are the vertices and edges of ${\cal P}$. We label an arbitrary vertex $V$ of ${\cal G}_{\cal P}$ by the number of different shortest paths along the edges of ${\cal P}$ from $V_0$ to $V$. (We mention that all edges of the mosaic are equivalent.) Let the labelled digraph ${\cal G}_{\cal P}$ be the Pascal pyramid (more precisely, the Pascal tetrahedron) in space $\mathbf{H}^2\!\times\!\mathbf{R}$ denoted by \mbox{\ppHq\!\!.} Some labelled vertices can be seen in Figure \ref{fig:H2R_disk_Pascal} in case \ppH45\!\!. \begin{figure}[h!] \centering \includegraphics[scale=0.6]{H2R_disk_Pascal} \vspace*{0.2cm} \caption{\rm \em Pascal piramid \ppH45 in space \H2R} \label{fig:H2R_disk_Pascal} \end{figure} Let level~0 be the vertex $V_0$. Level~$n$ consists of the vertices of \ppHq whose edge-distances from $V_0$ are an $n$-edge (the distance of the shortest path along the edges of ${\cal P}$ is $n$). It is clear that one (infinite) face of \ppHq is a \hpt in plane $\Pi$ and the other two faces are Euclidean Pascal's triangles. Figure~\ref{fig:pyramid_ppH45} shows the Pascal pyramid \ppH45\! in \H2R up to level~4. Moreover, Figures~\ref{fig:H2Rlayer2to3}--\ref{fig:H2Rlayer5to6} show the growing from a level to the next one in case of some lower levels. The colours and shapes of different types of vertices are different. (See the definitions later.) The numbers without colouring and shapes refer to vertices in the lower level in each figure. The graphs growing from a level to the new one contain graph-cycles with six nodes, which refer to convex hulls of parallel projections of the cubes from the mosaic, where the direction of the projection is not parallel to any edge of the cubes. \begin{figure}[h!] \centering \includegraphics[scale=0.85]{H2R_pyramid} \vspace*{0.2cm} \caption{\rm \em Pascal pyramid \ppH45} \label{fig:pyramid_ppH45} \end{figure} \begin{figure}[h!] \centering \includegraphics[scale=1]{H2R_level_1to2} \includegraphics[scale=1]{H2R_level_2to3} \vspace*{0.2cm} \caption{\rm \em Connection between levels one, two and three in \ppH45} \label{fig:H2Rlayer2to3} \end{figure} \begin{figure}[ht!] \centering \includegraphics[scale=1]{H2R_level_3to4} \vspace*{0.2cm} \caption{\rm \em Connection between levels three and four in \ppH45} \label{fig:H2Rlayer3to4} \end{figure} \begin{figure}[ht!] \centering \includegraphics[scale=0.8]{H2R_level_4to5} \vspace*{0.2cm} \caption{\rm \em Connection between levels four and five in \ppH45} \label{fig:H2Rlayer4to5} \end{figure} \begin{figure}[ht!] \centering \includegraphics[scale=0.78]{H2R_level_5to6} \vspace*{0.2cm} \caption{\rm \em Connection between levels five and six in \ppH45} \label{fig:H2Rlayer5to6} \end{figure} In what follows, we describe the method of the growing of \ppHq and we give the sum of the paths connecting vertex $V_0$ and level $n$. \section{Growing of \ppHq} In the classical Pascal's pyramid the number of elements on level $n$ is $(n+1)(n+2)/2$ and its growing from level $n$ to $n+1$ is $n+2$, but in the hyperbolic Pascal pyramid it is more complex (see \cite{N_pyr}). As one face of \ppHq is the hyperbolic Pascal triangle, then there are three types of vertices $A$, $B$ and $1$ corresponding to the Introduction. Their denotations are also the same. From all $A$ and $B$ only one edge starts to the inside of the pyramid; these are the Euclidean edges of the mosaic (see Figure \ref{fig:H2R_disk_Pascal}). The types of the inside vertices of these edges differ from the types $A$ and $B$; let us denote them by type $D$ and type $E$, respectively. The other two sides of the Pascal pyramid are Euclidean Pascal's triangles, which have two types of vertices, let us denote them by $C$ and $1$, respectively. A vertex $C$ connects three new vertices in the next level, two vertices $C$ on the side of \ppHq and one vertex of type $D$ inside the pyramid. Sometimes, if it is important, we distinguish the types $1$. If a vertex of type $1$ belongs to \hpt, we denote it by $1_h$. Their growing methods are illustrated in Figure~\ref{fig:H2R_graph_side} (cf. with the growing method in \cite{BNSz} and \cite{N_pyr}). In the figures, we denote vertices of type $C$ by light grey squares (not diamonds) squares. \begin{figure}[h!] \centering \includegraphics[width=0.95\textwidth]{H2R_graph_side} \vspace*{0.2cm} \caption{\rm \em Growing method in case of the faces} \label{fig:H2R_graph_side} \end{figure} For the classification and the exact definitions of inner vertices we examine the vertex figures of the inner vertices. As the structure descends from a hyperbolic plane to the consecutive one, there are two types of vertex figures. During the growing (step from level $i-1$ to level $i$) an arbitrary inner vertex $V$ on level $i$ can be reached from level $i-1$ with three or two edges. This fact allows us to classify the inner vertices. Let the type of a vertex on level $i$ be $D$ or $E$, respectively, if it has three or two joining edges to level $i-1$ (as before). Figure~\ref{fig:H2R_vertex_figures} shows the vertex figures of the inner vertices of \ppHq\!\!. Vertices $W_{i-1}$ (small grey circles) are on level $i-1$, $W_i$ and the centres are on level~$i$. We do not know the types of $W$ (or ot is not important to know them). The other vertices of the double pyramid are on level $i+1$ and their classification gives their types. An edge of the double pyramid and its centre $V$ determine a square (a side-face of an $(h^2r)$-cube) from the mosaic. (Recall that an edge of the vertex figure is a diagonal of a side-face of an $(h^2r)$-cube.) Since from a vertex of a square we can go to the opposite vertex in two ways, then a vertex $D$ of the double pyramid, where $D$ and a $W_{i-1}$ are connected by an edge, can be reached with two paths from level $i-1$. (For example, on the left-hand side in Figure~\ref{fig:H2R_vertex_figures}, between a vertex $W_{i-1}$ and $D$, there are the paths $W_{i-1}\!-\!D_i\!-\!D$ and $W_{i-1}\!-\!W_i\!-\!D$.) \noindent So, the type of the third vertex of the faces on the double pyramid whose other two vertices are $W_{i-1}$ is $D$. The others connect to only one $W_{i-1}$, they can be reached in two ways from level~$i$, their types are $E$. See Figure~\ref{fig:H2R_vertex_figures}. In the figures, we denote vertices of type $D$ by hexagons and vertices of type $E$ by pentagons. %The thick directed edges are mosaic-edges between levels $i-1$ and $i$, while the thin ones are between levels $i$ and $i+1$. \begin{figure}[h!] \centering \includegraphics[scale=0.8]{H2R_vertex_figures} \vspace*{0.2cm} \caption{\rm \em Growing method in case of the inner vertices with vertex figures} \label{fig:H2R_vertex_figures} \end{figure} In Figure~\ref{fig:H2R_graph_inner}, the growing method is presented in case of the inner vertices. These vertices are the centres and some vertices of the double pyramids are presented in Figure~\ref{fig:H2R_vertex_figures}. \begin{figure}[h!] \centering \includegraphics[scale=0.8]{H2R_graph_inner} \vspace*{0.2cm} \caption{\rm \em Growing method in case of the inner vertices} \label{fig:H2R_graph_inner} \end{figure} %Three new $D$ and $q-4$ new $E$ connect for all vertices of type $D$, but from the explanation above all new vertices $D$ and $E$ connect altogether three or two other vertices on level $i$, respectively. For the correct calculation we correspond just one third or half of them to the examined vertices $C$, respectively. By the help of the similar consideration in case of vertices $E$, we can calculate the number of vertices on level $i+1$, recursively, without multiplicity. Finally, we denote the sums of vertices of types $A$, $B$, $C$, $D$ and $E$ on level $n$ by $a_n$, $b_n$, $c_n$, $d_n$ and $e_n$, respectively. Summarising the details we prove Theorem~\ref{th:H2R_growing_type}. \begin{theorem}\label{th:H2R_growing_type} The growing of the numbers of different types of vertices are described by the system of linear inhomogeneous recurrence sequences $(n\geq1)$ \begin{equation}\label{eq:H2R_seq01} \begin{array}{ccl} a_{n+1}&=& a_n+b_n+1,\\ b_{n+1}&=& (q-4)a_n+(q-3)b_n,\\ c_{n+1}&=& c_n+2,\\ d_{n+1}&=& a_n+d_n,\\ e_{n+1}&=& b_n+e_n, \end{array} \end{equation} with zero initial values. \end{theorem} We mention that $c_n$ $(n\geq1)$ is an arithmetical sequence and $c_{n}= 2(n-1)$. \begin{lemma}\label{lem:H2R_de} For the sequences $d_n$ and $e_n$ $(n\geq1)$, there hold \begin{equation} \begin{array}{ccl} d_{n}&=& -a_n+\frac{1}{q-4}b_n+(n-1),\\ e_{n}&=& a_n-(n-1). \end{array} \end{equation} \end{lemma} \begin{proof} Obviously, for $n=1$, the equations hold. In case $n>1$, we suppose that $d_{n-1}= -a_{n-1}+\frac{1}{q-4}b_{n-1}+(n-2)$ and $e_{n-1}= a_{n-1}-(n-2)$. Firstly, from the first, second and fourth row of \eqref{eq:H2R_seq01} we have \begin{equation*} \begin{array}{ccl} a_n+d_n&=&a_{n-1}+b_{n-1}+1+a_{n-1}+d_{n-1}=a_{n-1}+b_{n-1}+1+\dfrac{1}{q-4}b_{n-1}+n-2\\ &=&\dfrac{q-4}{q-4}a_{n-1}+\dfrac{q-3}{q-4}b_{n-1}+n-1=\dfrac{1}{q-4}b_n+n-1. \end{array} \end{equation*} Secondly, from the first and fifth row we obtain $a_n-e_n=a_{n-1}-e_{n-1}+1= (n-2)+1=n-1.$ \end{proof} Moreover, let $s_n$ $(n\geq1)$ be the number of all vertices on level $n$, so that $s_0=1$ and \begin{equation}\label{eq:H2R_sn} \begin{array}{ccl} s_{n}&=& a_n+b_n+c_n+d_n+ e_n+3 \\ &=& a_n+\dfrac{q-3}{q-4}b_n+2n+1. \end{array} \end{equation} Table \ref{table:H2R_typeof_vertices} shows the numbers of vertices on levels up to 10 in case \ppH45\!\!. \begin{table}[!hb] \centering \setlength{\tabcolsep}{0.4em} \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline $n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \hline $a_n$ & 0 & 0 & 1 & 2 & 4 & 9 & 22 & 56 & 145 & 378 & 988 \\ \hline $b_n$ & 0 & 0 & 0 & 1 & 4 & 12 & 33 & 88 & 232 & 609 & 1596 \\ \hline $c_n$ & 0 & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 \\ \hline $d_n$ & 0 & 0 & 0 & 1 & 3 & 7 & 16 & 38 & 94 & 239 & 617 \\ \hline $e_n$ & 0 & 0 & 0 & 0 & 1 & 5 & 17 & 50 & 138 & 370 & 979 \\ \hline $s_n$ & 1 & 3 & 6 & 11 & 21 & 44 & 101 & 247 & 626 & 1615 & 4201 \\ \hline \end{tabular} \caption{\emph{\rm \em Number of types of vertices $(n\leq10)$ in case \ppH45}\label{table:H2R_typeof_vertices}} \end{table} \begin{theorem}\label{theorem:H2R_numvertex4q} The sequences $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\{d_n\}$, $\{e_n\}$ and $\{s_n\}$ can be described by the same fourth order linear homogeneous recurrence sequence \begin{equation}\label{eq:H2R_recurcde} x_n=qx_{n-1}+2(1-q)x_{n-2}+qx_{n-3}-x_{n-4} \qquad (n\ge5),\\ \end{equation} the initial values be can obtained from Theorem \ref{th:H2R_growing_type} (in case \ppH45 from Table \ref{table:H2R_typeof_vertices}). The sequences $\{a_n\}$, $\{b_n\}$ can also be described by \begin{equation} x_n=(q-1)x_{n-1}+(1-q)x_{n-2}+x_{n-3} \qquad (n\ge4).\label{eq:H2R_recurab} \end{equation} \noindent Moreover, the explicit formulae \begin{align*}%\label{abcn4q3} a_n =&\left(\frac{2-q}{2}+\frac{D+2}{2D}\sqrt{D}\right)\alpha_1^n+\left(\frac{2-q}{2}-\frac{D+2}{2D}\sqrt{D}\right)\alpha_2^n+1, \quad \\ b_n =&\left(\frac{q-3}{2}+\frac{1-q}{2q}\sqrt{D}\right)\alpha_1^n+\left(\frac{q-3}{2}-\frac{1-q}{2q}\sqrt{D}\right)\alpha_2^n-1, \nonumber\\ d_n =&\left(\frac{q^2-5q+5}{2(q-4)}-\frac{q^2-3q-1}{2D}\sqrt{D}\right)\alpha_1^n+\left(\frac{q^2-5q+5}{2(q-4)}+\frac{q^2-3q-1}{2D}\sqrt{D}\right)\alpha_2^n \nonumber \\ &+n-\frac{1}{q-4}+1, \nonumber \\ e_n=&\left(\frac{2-q}{2}+\frac{D+2}{2D}\sqrt{D}\right)\alpha_1^n+ \left(\frac{2-q}{2}-\frac{D+2}{2D}\sqrt{D}\right)\alpha_2^n-n+2, \nonumber \\ s_n=&\left(\frac{q}{2}-\frac{\sqrt{D}}{2D}\right)\alpha_1^n+ \left(\frac{q}{2}+\frac{\sqrt{D}}{2D}\right)\alpha_2^n+2n-\frac{1}{q-4}+ 1, \nonumber \end{align*} hold, where $D=q(q-4)$, $\alpha_1=(q-2+\sqrt{D})/2$ and $\alpha_2=(q-2-\sqrt{D})/2$. \end{theorem} \begin{proof} The sequences $\{a_n\}$ and $\{b_n\}$ are the same as $\{a_n\}$, $\{b_n\}$ in \cite{BNSz}. So they can be described by \eqref{eq:H2R_recurab} and their explicit formulae hold. According to Lemma \ref{lem:H2R_de} and \eqref{eq:H2R_sn}, more explicit formulae are derived by the combination of explicit formulae of $\{a_n\}$ and $\{b_n\}$. Let us extend \eqref{eq:H2R_recurab} to \eqref{eq:H2R_recurcde} considering the sequence $\{a_n\}$. Substitute $a_n$ into \eqref{eq:H2R_recurab} and sum $a_n$ and $a_{n-1}$. Than we obtain implicit form \eqref{eq:H2R_recurcde} for $\{a_n\}$. Similarly, \eqref{eq:H2R_recurcde} is also the implicit form of $\{b_n\}$. Now we prove that \eqref{eq:H2R_recurcde} holds for $s_n$ also. Multiply equation \eqref{eq:H2R_sn} by $q$, $2(1-q)$, $q$ and $-1$, where $n=n, n-1, n-2$ and $n-3$, respectively. If we sum them, then we obtain \begin{align*} qs_{n}&+2(1-q)s_{n-1}+qs_{n-2}-s_{n-3}\\ =& qa_{n}+2(1-q)a_{n-1}+qa_{n-2}\!-\!a_{n-3} + \frac{q-3}{q-4}\left(qb_{n} +2(1-q)b_{n-1}+qb_{n-2}\!-\!b_{n-3}\right) \\& + q(2n+1)+2(1-q)\left(2(n-1)+1\right)+q\left(2(n-2)+1\right)-\left(2(n-3)+1\right)\\ =&a_{n+1}+ \frac{q-3}{q-4}b_{n+1}+ 2n+3=s_{n+1}. \end{align*} This proves \eqref{eq:H2R_recurcde} for $\{s_n\}$. In case of $\{d_n\}$ and $\{e_n\}$, we can prove relation \eqref{eq:H2R_recurcde} similarly to $\{s_n\}$. \end{proof} \begin{remark} If $q=4$, then the sequences give results of the Euclidean Pascal's pyramid. Its all faces are Pascal's triangle, thus $b_n=0$, $e_n=0$, $a_n$ coincide $c_n$, in that way the growing equation system \eqref{eq:H2R_seq01} is just $c_{n+1}=c_n+3$, $d_{n+1}=c_n+d_n$. \end{remark} \begin{remark} The generating function of the sequence $s_n$ is given by \begin{equation*} \frac{1-(q-3)x-(q-4)x^2}{1-qx+(2q-2)x^2 -qx^3+x^4}. \end{equation*} In the case \ppH45\!\! it is \begin{equation*} \frac{1-2x-x^2}{1-5x+8x^2-5x^3+x^4}, \end{equation*} which is not in the OEIS at present. \end{remark} \section{Sum of the values on levels in \ppHq} In this section, we determine the sum of the values of elements on level $n$. Let $\hat{a}_{n}$, $\hat{b}_{n}$, $\hat{c}_{n}$, $\hat{d}_{n}$ and $\hat{e}_{n}$ denote the sums of the values of vertices of type $A$, $B$, $C$, $D$ and $E$ on level $n$, respectively. \begin{theorem} \label{th:H2R_recursum} If $n\geq1$, then \begin{equation}\label{eq:seq02} \begin{array}{ccl} \hat{a}_{n+1}&=& 2\hat{a}_n+2\hat{b}_n+2,\\[0.5ex] \hat{b}_{n+1}&=& (q-4)\hat{a}_n+(q-3)\hat{b}_n,\\[0.5ex] \hat{c}_{n+1}&=& 2\hat{c}_n+4,\\[0.5ex] \hat{d}_{n+1}&=& \hat{a}_n+c_n+3\hat{d}_n+2\hat{e}_n,\\[0.5ex] \hat{e}_{n+1}&=& \hat{b}_n+(q-4)\hat{d}_n+(q-2) \hat{e}_n \end{array} \end{equation} with zero initial values. \end{theorem} \begin{proof} From Figures \ref{fig:H2R_graph_side} and \ref{fig:H2R_graph_inner} the results can be read directly. For example, all vertices of type $A$, $B$ and $1$ on level $n$ generate two vertices of type $A$ on level $n+1$ and it follows from the first equation of \eqref{eq:seq02}. \end{proof} Table \ref{table:H2R_sumof_vertices} shows the sum of the values of the vertices on levels up to 10. Let $\hat{s}_n$ be the sum of the values of all vertices on level $n$. Then $\hat{s}_0=1$ and \begin{equation*} \hat{s}_n = \hat{a}_n+\hat{b}_n+\hat{c}_n+\hat{d}_n+\hat{e}_n+3 \qquad (n\geq1). \end{equation*} The value $\hat{s}_n$ also shows the number of paths from $V_0$ to level~$n$. \begin{table}[!hb] \centering \setlength{\tabcolsep}{0.4em} \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline $n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \hline $\hat{a}_n$ & 0 & 0 & 2 & 6 & 18 & 58 & 194 & 658 & 2242 & 7642 & 26114 \\ \hline $b_n$ & 0 & 0 & 0 & 2 & 10 & 38 & 134 & 462 & 1582 & 5406 & 18462 \\ \hline $\hat{c}_n$ & 0 & 0 & 4 & 12& 28 & 60 & 124 & 252 & 508 & 1020 & 2044 \\ \hline $d_n$ & 0 & 0 & 0 & 6 & 36 & 170& 768 & 3458 & 15596 & 70314 & 316296 \\ \hline $\hat{e}_n$ & 0 & 0 & 0 & 0 & 8 & 70 & 418 & 2156 & 10388 & 48342 & 220746 \\ \hline $\hat{s}_n$ & 1 & 3 & 9 & 29& 103& 399& 1641& 6989 & 30319 & 132735 & 583665 \\ \hline \end{tabular} \caption{\emph{\rm \em Sum of values of types of vertices $(n\leq10)$ in case \ppH45}\label{table:H2R_sumof_vertices}} \end{table} \begin{theorem} The sequences $\{\hat{a}_n\}$, $\{\hat{b}_n\}$, $\{\hat{c}_n\}$, $\{\hat{d}_n\}$, $\{\hat{e}_n\}$ and $\{\hat{s}_n\}$ can be described by the same sixth order linear homogeneous recurrence sequence \begin{equation}\begin{split}\label{eq:H2R_recurdehat} x_n=& (2q+3)x_{n-1}+(-q^2-7q-5)x_{n-2}+(4q^2+10q+9)x_{n-3}\\ &+(-5q^2-13q-10)x_{n-4}+(2q^2+12q+12)x_{n-5}+(-4q-8)x_{n-6}. \end{split} \end{equation} The initial values are from equation system \eqref{th:H2R_recursum}. The sequences $\{\hat{a}_n\}$, $\{\hat{b}_n\}$ can also be described by \begin{equation}\label{eq:H2R_recurabhat} x_n=qx_{n-1}-(q+1)x_{n-2}+2x_{n-3} \qquad (n\ge3). \end{equation} \noindent Moreover, for sequences $\{\hat{c}_n\}$, $\{\hat{s}_n\}$ \begin{align}\label{eq:H2R_recurchat} \hat{c}_n=&3\hat{c}_{n-1}-2\hat{c}_{n-2} \qquad (n\ge2),\\ \label{eq:H2R_recurshat} \hat{s}_n=&(q+3)\hat{s}_{n-1}-(3q+4)\hat{s}_{n-2}+(2q+4)\hat{s}_{n-3} \qquad (n\ge3) \end{align} and the explicit formula of $\hat{s}_n$ \begin{equation} \hat{s}_n= \left(-\frac12+(q-1)\frac{\sqrt{D}}{2D}\right)\alpha_1^n+ \left(-\frac12-(q-1)\frac{\sqrt{D}}{2D}\right)\alpha_2^n+2\cdot2^n \end{equation} holds, where $D=q^2-2q-7$, $\alpha_1= \frac12\left(1+q+\sqrt{D}\right)$ and $\alpha_2= \frac12\left(1+q-\sqrt{D}\right)$. \end{theorem} We do not give the implicit formulae for all sequences, because generally, they are complicated, but they can be calculated easily with computer. \begin{proof} Let $\hat{v}_n=3$ $(n\geq1)$ be constant sequences and $\hat{v}_0=1$. The value $\hat{v}_n$ gives the sum of the number of vertices of type ``1" on level $n$. Substitute $2=2\hat{v}_n/3$ and $4=4\hat{v}_n/3$ into the first and third equations of \eqref{th:H2R_recursum} and complete it with equation $\hat{v}_{n+1}=\hat{v}_n$. Then we have a system of linear homogeneous recurrence sequences $(n\geq1)$ \begin{equation}\label{eq:H2R_seq02v} \begin{array}{ccl} \hat{a}_{n+1}&=& 2\hat{a}_n+2\hat{b}_n+\dfrac23v_n,\\[0.5ex] \hat{b}_{n+1}&=& (q-4)\hat{a}_n+(q-3)\hat{b}_n,\\[0.5ex] \hat{c}_{n+1}&=& 2\hat{c}_n+\dfrac43v_n,\\[0.5ex] \hat{d}_{n+1}&=& \hat{a}_n+c_n+3\hat{d}_n+2\hat{e}_n,\\[0.5ex] \hat{e}_{n+1}&=& \hat{b}_n+(q-4)\hat{d}_n+(q-2) \hat{e}_n,\\[0.5ex] \hat{v}_{n+1}&=& \hat{v}_n \end{array} \end{equation} and \begin{equation*}\label{eq:snv} \hat{s}_n = \hat{a}_n+\hat{b}_n+\hat{c}_n+\hat{d}_n+\hat{e}_n+\hat{v}_n\qquad (n\geq0). \end{equation*} A shorter form of linear homogeneous recurrence sequences \eqref{eq:H2R_seq02v} is \begin{eqnarray} \vec{u}_{i+1}&=&\vec{M}\vec{u}_i,\label{eq:H2R_am} \end{eqnarray} where $\vec{u}_j=[\hat{a}_j\enskip \hat{b}_j\enskip \hat{c}_j\enskip \hat{d}_j\enskip \hat{e}_j\enskip \hat{v}_j\enskip]^T$ and \[\vec{M}=\begin{pmatrix} 2 & 2 & 0 & 0 & 0 & \frac23 \\ q-4 & q-3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & \frac43 \\ 1 & 0 & 1 & 3 & 2 & 0 \\ 0 & 1 & 0 & q-4 & q-2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}.\] The characteristic polynomial of $\vec{M}$ is \begin{equation} p_6(x)=\left( x-1 \right) \left( x-2 \right) \left( x^2-(q+1)x+q+2 \right) \left(x^2+(1-q)x+2 \right) \label{eq:H2R_karMs} \end{equation} and according to Theorem 3 from \cite{N_pyr}, equation \eqref{eq:H2R_karMs} is the characteristic equation of all sequences $\hat{r}_{i+1}=\vec{y}^T\vec{u}_i$, where $ \vec{y}=[y_1 \enskip y_2 \enskip \dots \enskip y_6]^T$. Thus \eqref{eq:H2R_karMs} is the characteristic equation of $\hat{s}_n$, $\hat{a}_n$, \ldots, $\hat{e}_n$ with $\vec{y}=(1,1,1,1,1,1)$, $\vec{y}=(1,0,0,0,0,0)$, \ldots, $\vec{y}=(0,0,0,0,1,0)$, respectively. Equation \eqref{eq:H2R_karMs} implies the six order linear recurrence sequence \eqref{eq:H2R_recurdehat} for the considered sequences, but for the lower order implicit formulae we have to examine the first elements of the sequences. One can easily check that the polynomial \begin{equation} \begin{split} p_{\hat{s}}(x)=&\left( x-2 \right) \left( x^2-(q+1)x+q+2 \right) \\ =&x^3-(q+3)x^2+(3q+4)x-(2q+4); \end{split} \label{eq:H2R_karM4s} \end{equation} moreover, in case of $\hat{a}_i$ and $\hat{b}_i$ (see \cite{BNSz}) \begin{eqnarray*} p_3(x)=(x-1)\left(x^2+(1-q)x+2 \right)=x^3-qx^2+(1+q)x-2, \label{eq:H2R_karM3} \end{eqnarray*} in case of $\hat{c}_i$ \begin{eqnarray*} p_2(x)=(x-1)(x-2)=x^2-3x+2. \label{eq:H2R_karM2} \end{eqnarray*} These polynomials imply recurrence relations \eqref{eq:H2R_recurdehat}--\eqref{eq:H2R_recurshat}, respectively. The roots of polynomial \eqref{eq:H2R_karM4s} are $\alpha_1$, $\alpha_2$, $\alpha_3=2$ and \[\hat{s}_n=\beta_1\cdot \alpha_1^n+\beta_2\cdot \alpha^n_2+\beta_3\cdot 2^n \qquad (n\geq1)\] provide the explicit formulae (see \cite{sho}). Solutions of the linear equation system from $n=1,2,3$ determine the exact values of $\beta$'s. \end{proof} \begin{remark} In the case of \ppH45\!\!, the growing ratio of values is $\lim_{n \rightarrow \infty } {\hat{s}_{n+1}}/{\hat{s}_{n}}=\alpha_1\!\approx\!4.414$, where $\alpha_1$ is the largest eigenvalue of matrix $\vec{M}$. Recall that these growing ratios are $3$ and $\approx\!10.351$ in the Euclidean and hyperbolic cases, respectively (\cite{N_pyr}). \end{remark} \begin{remark} The generating function of the sequence $\hat{s}_n$ is given by \begin{equation*} \frac{1-qx+4x^2}{1-(q+3)x+(3q+4)x^2 -(2q+4)x^3}. \end{equation*} In the case of \ppH45\!\!, it is \begin{equation*} \frac{1-5x+4x^2}{1-8x+19x^2-14x^3}, \end{equation*} which is not in the OEIS at present. \end{remark} %%%% Acknowledgment %%%%%%%% %\section*{Acknowledgement} %The author would like to thank the referees for the helpfulsuggestions. %%%% Bibliography %%%%%%%%%% \begin{thebibliography}{99} \bibitem{AhBel} {\sc M.\,Ahmia, H.\,Belbachir}, {\it Preserving log-convexity for generalized Pascal triangles}, { Electron. J. Combin.} \textbf{19}(2012), $\#$p.16. \bibitem{ANVI} {\sc G.\,Anatriello, G.\,Vincenzi}, \textit{Tribonacci-like sequences and generalized Pascal's pyramids}, Internat. J. Math. Ed. Sci. 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