Recurrence sequences in the hyperbolic Pascal triangle corresponding to the regular mosaic $\{4,5\}

0.35 MB·

by  László Németh

Recurrence sequences in the hyperbolic
Pascal triangle corresponding to the regular
mosaic {4, 5}
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0 La´szl´o N´emeth∗, L´aszlo´ Szalay †‡
2

n
a
J

1
2 Abstract

] Recently, a new generalization of Pascal’s triangle, the so-called hyperbolic Pascal
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triangles were introduced. The mathematical background goes back to the regular
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mosaics in the hyperbolic plane. In this article, we investigate the paths in the hyper-
.
h
bolic Pascal triangle corresponding to the regular mosaic {4,5}, in which the binary
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a recursive sequences f = αf ±f are represented (α ∈ N+).
m n n−1 n−2
Key Words: Pascal triangle, hyperbolic Pascal triangle, binary recurrences.
[
MSC code: 11B37, 05A10.

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v
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7 1 Introduction
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0 In the hyperbolic plane there are an infinite number of types of regular mosaics (see, for
1. example [4]), they are assigned by Schla¨fli’s symbol {p,q}, where the positive integers p
0 and q satisfy (p−2)(q −2) > 4. Each regular mosaic induces a so-called hyperbolic Pascal
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triangle (see [1]), following and generalizing the connection between the classical Pascal’s
1
: triangle and the Euclidean regular square mosaic {4,4}. For more details see [1], but here
v
i we also collect some necessary information.
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There are several approaches to generalize Pascal’s arithmetic triangle (see, for instance
r
a [3]). The hyperbolic Pascal triangle based on the mosaic {p,q} can be figured as a digraph,
where the vertices and the edges are the vertices and the edges of a well defined part of the
lattice {p,q}, respectively, further each vertex possesses a value, say label, giving the number
of different shortest paths from the fixed base vertex. Figure 1 illustrates the hyperbolic
Pascal triangle linked to {p,q} = {4,5}. Generally, for {4,q}, the quadrilateral shape cells
surrounded by appropriate edges are corresponding to the squares in the mosaic. The base
vertex has two edges (both are outgoing), the leftmost and the rightmost vertices have three
(oneingoingandtwooutgoing), theothershaveq edges(eithertwoingoingandq−2outgoing
∗Institute of Mathematics, University of West Hungary, Sopron. [email protected]
†Institute of Mathematics,University of West Hungary, Sopron. [email protected]
‡Department of Mathematics and Informatics, J. Selye University, Hradna ul. 21., 94501 Komarno,
Slovakia.
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2 L. N´emeth, L. Szalay
(type A) or one ingoing and q−1 outgoing (type B)). In other words, apart from the winger
elements, vertices of type A have two ascendants and q −2 descendants, vertices of type B
do one ascendant and q − 1 descendants. In the figures, we denote the A-type vertices by
red circle and B-type vertices by cyan diamond, further the wingers by white diamond. The
vertices having distance n from the base vertex are located in row n. The general method
of drawing is the following. Going along the vertices of the nth row, according to type of the
elements (winger, A, B), we draw appropriate number of edges downward (2, q −2, q −1,
respectively). Neighbor edges of two neighbor vertices of the nth row meet in the (n+1)th
row, constructing a vertex of type A. The other descendants of row n in row n+1 have type
B, except the two wingers. In the sequel, )n( denotes the kth element in row n, which is either
k
the sum of the labels of its two ascendants or coincide the label of its unique ascendant. For
instance, if {p,q} = {4,5}, then
)4( )3( )3( )4( )3(
= 5 = 2+3 = + and = 2 =
6 2 3 5 2
hold (see Figure 1). We note, that the hyperbolic Pascal triangle has the property of vertical
symmetry.
Figure 1: Hyperbolic Pascal triangle linked to {4,5} up to row 6
2 Recurrence sequences linked to {4,5}
Let {p,q} = {4,5} be fixed, further we let HPT denote the hyperbolic Pascal triangle
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corresponding to the mosaic {4,5}. It was showed in [1] that all the binary recurrence
sequences (f ) which are defined by
i i≥0
f = ηf +f , (n ≥ 2), (1)
i i−1 i−2
where η and f < f are positive integers, appear in HPT .
0 1 45
In the following we describe paths corresponding to further positive integer binary re-
currence sequences. We remark that although we restrict ourselves to HPT , the methods
45
Recurrence sequences in the hyperbolic Pascal triangle ... 3
and the results have been worked out can be fitted to other hyperbolic Pascal triangles with
p = 4, q ≥ 6.
Taking a vertex of type A in row n, it has exactly two descendants of type A in the row
n+1. In order to reach and distinguish them, we denote the left-down step and right-down
step (along the appropriate edge of the graph) by L and R, respectively. For the sake of
brevity, the sequence of (cid:96)+r consecutive steps
LL···LRR···R
(cid:124) (cid:123)(cid:122) (cid:125)(cid:124) (cid:123)(cid:122) (cid:125)
(cid:96) r
will be denoted by L(cid:96)Rr. Till the end of this work, such a path is always considered on
vertices of type A. Generally, we are interested in the labels of these vertices, therefore
sometimes we call them elements (as the elements or terms of a sequence), but if it is
necessary we determine the location of the element, too.
This paper will use the next theorem (Theorem 5 in [1]), which states that any two
positive integers can be found next to each other somewhere in HPT .
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Theorem 1. Given u,v ∈ N+, then there exist n,k ∈ N+ such that u = )n( and v = ) n (.
k k+1
Using Theorem 1, Corollary 1 provides an immediate consequence of the properties of
HPT .
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Corollary 1. If u = )n( < v = ) n ( holds for some positive integers u and v, then ) n ( =
k k+1 k+2
v −u, moreover the type of ) n ( is A, while the types of )n( and ) n ( are not A (i.e., either
k+1 k k+2
B or winger).
Remark 1. Clearly, by the symmetry we also have the construction u = )n( > v = ) n ( and
k k+1
) n ( = u−v. Further, the type of )n( is A.
k−1 k
2.1 Recurrence sequences and paths
Let (f ) be a recurrence sequence defined by
i i≥0
f = αf −f , (i ≥ 2), (2)
i i−1 i−2
where α ∈ N+, α ≥ 2, and f < f are positive integers with gcd(f ,f ) = 1. If α = 2 then
0 1 0 1
(f ) is an arithmetic progression given by f = f +(f −f ).
i i≥0 i i−1 1 0
From Theorem 1 and Corollary 1 we know that in case of any positive integers f < f ,
0 1
there exist an element in HPT with value f , and with neighbors in the same row valued
45 1
by f and f −f . In Theorem 2 we give a path in HPT (analogously to Theorem 6 in [1])
0 1 0 45
contains all the elements of (2).
Theorem 2. There exists a path in HPT crossing vertices of type A, such that the vertices
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are labelled with the terms of (f ) as follows. Assume that )n( = f , and ) n ( = f −f .
i i≥1 k 1 k−1 1 0
Then the first element of the path is f and the pattern of the steps from f to f (i ≥ 2)
1 i−1 i
is LRα−2.
4 L. N´emeth, L. Szalay
Proof. According to Theorem 1, any f and f −f can be neighbours in HPT , where type
1 1 0 45
of f is A (and the type of f −f is not A).
1 1 0
If α = 2, then the statment is easy to show, since no R steps. Indeed, the difference of
an element type A, and its immediate left descendant having type A is the constant f −f .
1 0
Assume now α ≥ 3. By the construction rule of HPT , we can follow the way from
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any f to f (i ≥ 2) in Figure 2, which justifies the theorem (the type of the rectangle
i−1 i
shaped elements is A). In the last row of the figure we use, among others, the equality
f −f = (α−1)f −f .
i i−1 i−1 i−2
f −f f f
i−1 i−2 i−1 i−2
2f −f f
i−1 i−2 i−1
3f −f f
i−1 i−2 i−1
. .
. .
. .
(α−1)f −f f
i−1 i−2 i−1
f −f f = αf −f f
i i−1 i i−1 i−2 i−1
Figure 2: Path LRα−2 between f and f
i−1 i
Remark 2. Theorem 2 can be extended for the whole sequence (f ) if and only if (α −
i i≥0
1)f < f < αf . Under these conditions one can follow the path back from the bottom of
0 1 0
Figure 2 to the top, from f to f .
1 0
The path showed on the right hand side of Figure 4 (cf. Figure 1) is an example for the
binary recurrence f = 4f −f with f = 1, f = 2.
i i−1 i−2 0 1
Theorem 2 finds a path to the sequence (2). Considering the opposite direction, now
we decribe the sequence corresponding to a given pattern of steps. The expression “corner
element” means a labelled vertex where the direction of the sequence of steps changes. For
example, the first corner element of the path L3R2 is the vertex reached after three left steps,
the second corner element comes after further two right steps, etc.
Theorem 3. Suppose that the A-type vertex )n( = U = u is a starting point of the path
k 1 1
L(cid:96)Rr. We let U , and u (i = 1,...) denote the label of the corner elements, and the label of
i i
every second corner elements of the path, respectively. Then we have
u = ((cid:96)r+2)u −u , (i ≥ 3). (3)
i i−1 i−2
Recurrence sequences in the hyperbolic Pascal triangle ... 5
Moreover, if (cid:96) = r, then
U = (cid:96)U +U , (i ≥ 3).
i i−1 i−2
Obviously, u = U holds. The proof of Theorem 3 applies the following lemma (see
i 2i−1
[1], Remark 1 linked to Lemma 4).
Lemma 2.1. Let x , y , further a and b (j = 1,2) be complex numbers such that a b (cid:54)= 0.
0 0 j j 2 1
Assume that for i ≥ i the terms of the sequences (x ) and (y ) satisfy
0 i i
x = a x +b y ,
i+1 1 i 1 i
y = a x +b y .
i+1 2 i 2 i
Then for both sequences
z = (a +b )z +(−a b +a b )z
i+2 1 2 i+1 1 2 2 1 i
holds (i ≥ i ).
0
Proof of Theorem 3. Suppose that v is the left ascendant of u . By Figure 3, which demon-
1 1
strates the path precisely from u to u (i ≥ 1) along vertices type A in HPT , we gain
i i+2 45
the system of the recursive equations
u = (r+1)u +((cid:96)+r((cid:96)−1))v , (4)
i+1 i i
v = ru +((cid:96)+(r−1)((cid:96)−1))v .
i+1 i i
Using Lemma 2.1 we receive that both u and v satisfy the equation
i i
z = ((cid:96)r+2)z −z .
i+2 i+1 i
If (cid:96) = r, then we simply obtain
U = U +(cid:96)V , (5)
i+1 i i
V = U +((cid:96)−1)V .
i+1 i i
Now Lemma 2.1 results that U and V satisfy the equation
i i
Z = (cid:96)Z +Z .
i+2 i+1 i
Remark 3. Let )n( = f be the initial element, and consider the path L(cid:96)Rr. Since every
k 1
second corner element of the path satisfies the recurrence equation (2) with α = (cid:96)r + 2,
the number of different paths belonging to different patterns but corresponding to the linear
recurrence (f )∞ is the number of the divisors of (cid:96)r = α−2.
i i=1
Figure 4 gives an example for the case when α−2 = 2 = 2·1 = 1·2 and u = f = 2,
1 1
u = f = 7. Clearly, the patters are L2R and LR2.
2 2
Now we describe the intermediate sequences located in the path given by v ,)n( = u and
1 k 1
by L(cid:96)Rr. The labels of the elements having distance ((cid:96)+r)t (t ∈ N) from the base element
)n( are given by suitable sequences {w }.
k i
6 L. N´emeth, L. Szalay
.
.
.
v
i
u
i
u +v
i i
u +2v
i i
.
.
.
V = u +((cid:96)−1)v
j i i
U = u +(cid:96)v
j i i
U +V
j j
U +2V
j j
.
.
.
v = V = U +(r−1)V
i+1 j+1 j j
u = U = U +rV
i+1 j+1 j j
u +v
i+1 i+1
u +2v
i+1 i+1
.
.
.
V = u +((cid:96)−1)v
j+2 i+1 i+1
U = u +(cid:96)v
j+2 i+1 i+1
U +V
j+2 j+2
U +2V
j+2 j+2
.
.
.
v = V = U +(r−1)V
i+2 j+3 j+2 j+2
u = U = U +rV
i+2 j+3 j+2 j+2
Figure 3: Path L(cid:96)Rr from u to u
i i+2
Recurrence sequences in the hyperbolic Pascal triangle ... 7
Figure 4: f = 1, f = 2, f = 4f −f
0 1 i i−1 i−2
Theorem 4. Put w = u +mv , where 0 ≤ m < (cid:96), or let w = U +mV = (m+1)u +
i i i i 2i 2i i
((cid:96)+m((cid:96)−1))v , where 0 ≤ m < r. Then the terms of the sequence (w ) satisfy
i i
w = ((cid:96)r+2)w −w , (i ≥ 3).
i i−1 i−2
Proof. Consider again Figure 3 to show the statement for the first type of sequences. One
can observe the labels of the path described by w = u +mv , where 0 ≤ m < (cid:96) and i ≥ 1.
i i i
From (3) we see
w = u +mv = ((cid:96)r+2)(u +kv )−(u +mv )
i+2 i+2 i+2 i+1 i+1 i i
= ((cid:96)r+2)w −w .
i+1 i
The second part of the proof is analoguous. In Figure 3 the equation j = 2i holds, but
generally it does not.
Corollary 2. In case of (cid:96) = r, W = U +mV (0 ≤ m < (cid:96)) satisfy the equation
j j j
W = (cid:96)W +W , (j ≥ 3). (6)
j j−1 j−2
In Figure 5, according to Corollary 2 we give two examples for the representation of
elements of recurrence sequence f = 3f + f . The pattern of both paths is R3L3,
i i−1 i−2
moreover, u = 3, v = 2, m = 2 and u = 4, v = 3, m = 1, respectively.
1 1 1 1
8 L. N´emeth, L. Szalay
Figure 5: f = 1, f = 2, f = 3f +f
0 1 i i−1 i−2
Theorem 5. Consider the sequence (2). If (cid:96)r = α−2, and
f −(r+1)f
j+1 j
m =
(cid:96)+r((cid:96)−1)
is an integer for some j ≥ 1, further m < f holds, then the elements f (i ≥ j) can be
j i
represented in HPT by every second corner elements of a paths given by the the pattern
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L(cid:96)Rr, and by u = f and v = m.
1 j 1
Proof. Let u = f and u = f . Then equation (4) yields v = (f − (r + 1)f )/((cid:96) +
1 j 2 j+1 1 j+1 j
r((cid:96)−1)). Since the integers v and u are neigbours in a suitable row of HPT , therefore
1 1 45
there is a path with the pattern L(cid:96)Rr from v and u such that every second corner elements
1 1
are f (i ≥ j).
i+1
Figure 4 gives examples on the paths of f = 4f −f with initial elements u = f = 7
i i−1 i−2 1 2
andu = f = 26,moreoverv = 4andv = 5,respectively,whereα−2 = 2 = 2·1 = 1·2 = lr,
2 3 1 1
and the patters are L2R and LR2.
Theorem 6. Consider now the sequence (1). If (cid:96)2 = η − 2, and m = (f − f )/(cid:96) is an
j+1 j
integer, further m < f , then the elements f (i ≥ j ≥ 1) can be represented in HPT by
j i 45
every corner elements of the paths given by the pattern L(cid:96)Rr, and by u = f and v = m.
1 j 1
Recurrence sequences in the hyperbolic Pascal triangle ... 9
Proof. The proof is similar to the proof of Theorem 5. Using (5), from u = f and u = f
1 j 2 j+1
we gain v = (f −f )/(cid:96).
1 j+1 j
References
[1] H. Belbachir, L. N´emeth, and L. Szalay, Hyperbolic Pascal triangles,
Appl. Math. Comp., 273 (2016), 453-464.
[2] L. N´emeth and L. Szalay, Alternating sums in hyperbolic Pascal triangles, Miskolc
Mathematical Notes, (accepted).
˘
[3] H. Belbachir and L. Szalay, On the arithmetic triangles, Siauliai Math. Sem., 9 (17)
(2014), 15-26.
[4] H. S. M. Coxeter, Regular honeycombs in hyperbolic space, in Proc. Int. Congress
Math., Amsterdam, Vol. III., 1954, pp. 155-169.