Properties of hyperbolic Pascal triangles Hacene Belbachir, László Németh, and László Szalay
Citation: AIP Conference Proceedings 1867, 020031 (2017); doi: 10.1063/1.4994434 View online: http://dx.doi.org/10.1063/1.4994434 View Table of Contents: http://aip.scitation.org/toc/apc/1867/1 Published by the American Institute of Physics
Properties of Hyperbolic Pascal Triangles Hacene Belbachir1,b) , L´aszl´o N´emeth2,c) and L´aszl´o Szalay2,3,a) 1
RECITS Laboratory, Faculty of Mathematics, USTHB, P.O.Box32, El Alia, 16111 Algiers, Algeria. Institute of Mathematics, University of West Hungary, H-9400 Sopron, Bajcsy-Zs. utca 4., Hungary. 3 Department of Mathematics and Informatics, University J. Selye, Hradna ul. 21., 94501 Komarno, Slovakia.
[email protected]. 2
a)
Corresponding author:
[email protected] b)
[email protected] c)
[email protected]
Abstract. Here we summarize some results on a recently introduced new generalization of Pascal’s triangle called hyperbolic Pascal triangles. The name comes from the mathematical background, which goes back to regular mosaics on the hyperbolic plane. A few open questions will also be posed.
INTRODUCTION There exist several generalizations and extensions of Pascal’s arithmetic triangle (see, for instance, [1] and [3]). In the article [2] we introduced the so called hyperbolic Pascal triangles (in short HPT s) by extending the link between the infinite graph corresponding to the regular square Euclidean mosaic and classical Pascal’s triangle to the hyperbolic plane. The purpose of this paper to make an overview about the results on HPT s and to pose some open problems. In the hyperbolic plane there are infinite types of regular mosaics, they are denoted by Schl¨afli’s symbol {p, q}, where the positive integers p and q satisfy (p − 2)(q − 2) > 4. The two parameters p and q mean that in one node of the mosaic exacly q regular p-gon meet. Each regular mosaic induces a HPT based on the mosaic {p, q}, and they 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 the value which gives the number of different shortest paths from the base vertex. Figure 1 illustrates the hyperbolic Pascal triangle HPT p,q , when {p, q} = {4, 5}. In the sequel, we fix a type of mosaic given by {4, q}, q ≥ 5. Generally, for {4, q} the base vertex has two edges, the leftmost and the rightmost vertices have three, the others have q edges. The square shaped cells surrounded by appropriate edges are corresponding to the regular squares in the mosaic. Apart from the winger elements, certain vertices (called “Type A” for convenience) have two ascendants and q − 2 descendants, the others (“Type B”) have 1 ascendant and q − 1 descendants. In the figures we denote the vertices type A by red circles and the vertices type B by cyan diamonds, further the wingers by white diamonds. The vertices which are n-edge-long far from the base vertex are in row n. The general method of drawing is the following. Going along the vertices of the jth row, according to type of the elements (winger, A, B), we draw appropriate number of edges downwards (2, q − 2, q − 1, respectively). Neighbor edges of two neighbor vertices of the jth row meet in the ( j + 1)th row, constructing a vertex with type A. The other descendants of row j in row j + 1 have type B. In the sequel, ) nk ( denotes the kth 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. Note that the hyperbolic Pascal pyramid has also been constructed by N´emeth [4, 5].
International Conference on Mathematics: Pure, Applied and Computation AIP Conf. Proc. 1867, 020031-1–020031-5; doi: 10.1063/1.4994434 Published by AIP Publishing. 978-0-7354-1547-8/$30.00
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FIGURE 1. Hyperbolic Pascal triangle linked to {4, 5} up to row 5
QUANTITATIVE PROPERTIES OF ROWS Recall that the type HPT {4,q} has been fixed (q ≥ 5). Even if we put q = 5, the explicit formula for ) nk ( is not known. Question 1. Give a “good” explicit formula for ) nk ( if q = 5 and generally, if q ≥ 5. The vain of explicit formula makes harder to study HPT {4,q} . Using the theory of linear recurrences, further the construction rule of the triangle, we were able to determine a couple of properties. Denote by an and bn the number of vertices of type A and B, respectively, further let sn = an + bn + 2, which gives the total number of the vertices of row n ≥ 1. The next part describes sn , an and bn . Theorem 1. (see [2]) The three sequences {an }, {bn } and {sn } can be described by the same ternary homogeneous recurrence relation xn = (q − 1)xn−1 − (q − 1)xn−2 + xn−3 (n ≥ 4), the initial values are a1 = 0, a2 = 1, a3 = 2,
b1 = 0, b2 = 0, b3 = q − 4,
s1 = 2, s2 = 3, s3 = q.
Moreover, the explicit formulae ! ! 2 − q q2 − 4q + 2 √ 2 − q q2 − 4q + 2 √ + D αnq + − D βnq + 1, 2 2q(q − 4) 2 2q(q − 4) ! ! √ √ q−3 1−q q−3 1−q bn = + D αnq + − D βnq − 1, 2 2q 2 2q ! ! √ 1 q−2 1 q−2 √ sn = − + D αnq + − − D βnq + 2 2 2q(q − 4) 2 2q(q − 4) an =
are valid for n ≥ 1, where D = q − 4q, 2
(q − 2) + αq = 2
√ D
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,
(q − 2) − βq = 2
√ D
.
After having the number of elements in rows, consider the kth -power sum (sk )n =
sX n −1 i=0
k
) ni (
of row n. For eliminating the simple sum, we use the notation (s1 )n = sˆn . Theorem 2. (see [2]) The sequence { sˆn } can be described by the ternary homogeneous recurrence relation sˆn = q sˆn−1 − (q + 1) sˆn−2 + 2 sˆn−3 the initial values are
(n ≥ 4),
sˆ1 = 2, sˆ2 = 4, sˆ3 = 2q. Moreover, ! ! √ √ 1 1 q−1 q−1 n sˆn = − + D α ˆ + − D βˆ nq + 2 − q 2 2(q2 − 2q − 7) 2 2(q2 − 2q − 7)
hold, where D = q2 − 2q − 7,
αˆ q =
(q − 1) + 2
√ D
,
(q − 1) − βˆ q = 2
√ D
.
Theorem 3. (see [7]) The sequence (s2 )n of sum of squares satisfies the 4th order recurrence relation (s2 )n = (q + 2)(s2 )n−1 − (q + 7)(s2 )n−2 + 8(s2 )n−3 − 2(s2 )n−4
(n ≥ 5)
with initial values (s2 )1 = 2, (s2 )2 = 6, (s2 )3 = 4q + 4, (s3 )4 = 4q2 + 6q − 20. Theorem 4. (see [7]) The sequence (s3 )n of sum of cubes satisfies the 4th order recurrence relation (s3 )n = (q + 4)xn−1 + (q − 19)(s3 )n−2 − 2(q − 9)(s3 )n−3 − 2(s3 )n−4
(n ≥ 5)
with initial values (s3 )1 = 2, (s3 )2 = 10, (s3 )3 = 8q + 24, (s3 )4 = 8q2 + 62q − 60. Observe, that the recurrence rules make it possible to find explicit formula for the sums (s2 )n and (s3 )n . We have a general method to describe the sums (sk )n for small k values by linear recurrences, but for arbitrary k we are unable to provide general recurrence relation. Note that this question is also a hard problem in Pascal’s original arithmetical triangle. Question 2. Describe the sum (sk )n for general value k. At the end of this section, let e sn denote the alternating sum of elements of HPT {4,q} in row n. Theorem 5. Let q be even. Then sX n −1
n e sn = (−1) = i i=0 i
) (
(
0, −2(5 − q)t−1 + 2,
if if
n = 2t + 1, n = 2t,
n ≥ 1, n ≥ 2,
hold, further e s0 = 1. Theorem 6. Let q ≥ 5 be odd. Then e s0 = 1, further 0, if n t t−1 (q − 5) + 2, if (−2) e = sn = (−1) i 2(−2)t (q − 5)t−1 + 2, if i=0 sX n −1
i
) (
n = 3t + 1, n ≥ 1, n = 3t − 1, n ≥ n1 , n = 3t, n ≥ n2 ,
where (n1 , n2 ) = (2, 3) and (5, 6) if n > 5 and n = 5, respectively. In the latter case e s2 = 0, e s3 = −2.
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Location of elements type A and B in the hyperbolic Pascal triangle HPT 4,5 A fractal-type repetition can be observed as follows. Theorem 7. Let n , 1. The pattern (of characters A and B, where the wingers count type B) given by the first sn elements of row n + 1 coincide the pattern of row n. Now we can phrase the main result of this subsection. Let encode the type of the elements A and B by 0 and 1, respectively, and consider the code of row n as a binary expansion of the integer ϕn . For example, if n = 3, then the pattern BABAB of row 3 is transformed into 10101, hence ϕ3 = 101012 = 21. Theorem 8. Put S n = 2 sn+1 −sn , and Φ(n) = ϕn+1 − ϕn . The terms of the sequence Φn satisfy ! Sn 2 + S n + S n−1 Φn−1 − S n−1 Φn−2 , Φn = S n−1
(n ≥ 3).
APPEARANCE OF BINARY RECURRENCES IN HPT {4,5} Figure 2 shows how the Fibonacci sequence (defined by F0 = 0, F1 = 1 and Fn = Fn−1 + Fn−2 for n ≥ 2) and the Pell sequence (defined by P0 = 0, P1 = 1 and Pn = 2Pn−1 + Pn−2 for n ≥ 2) appear in the hyperbolic Pascal triangle with the parameter q = 5.
FIGURE 2. Apparence of the Fibonacci and Pell sequences in HPT {4,5}
It is easy to show (see [8]) that each pair i < j ∈ Z+ can be found next to each other in HPT {4,5} , such that type of j is A (and type of i is B or winger). Assume now T ∈ Z+ , and •
f0 < f1 ∈ Z+ ,
•
f0 (in the second case f1 − f0 ) is the left neighbor of f1 (having type A).
Table 1 shows the information about the location of the terms of the binary recurrences fn = T fn−1 ± fn−2 (see [2, 8]). Here, for example, LRT −1 means that going down from a given element having type A, via elements type A, first turn left, and then (T − 1)-times right. For illustration, see Figure 3.
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TABLE 1. Location of the sequences
condition distance of fn and fn−1 pattern of steps
fn = T fn−1 + fn−2
fn = T fn−1 − fn−2
T ≥1
T ≥2
T
T −1
LRT −1 , and RLT −1 , alternately
LRT −2
FIGURE 3. Apparence of fn = 4 fn−1 − fn−2 , f0 = 1, f1 = 2 in HPT {4,5}
REFERENCES [1] [2] [3] [4] [5] [6] [7] [8]
Bondarenko, B. A., Generalized Pascal Triangles and Pyramids, www.fq.math.ca/pascal.html Belbachir, H., N´emeth, L. and Szalay, L., Hyperbolic Pascal triangles, Appl. Math. Comp., 273 (2016), pp. 453-464. ˘ Belbachir, H. and Szalay, L., On the arithmetic triangles, Siauliai Math. Sem., 9 (17) (2014), pp. 15-26. N´emeth, L., On the hyperbolic Pascal pyramid, Beitr. Algebra Geom., 57 (2016), pp. 913-927. N´emeth, L., Pascal pyramid in the space H2 ×R, Mathematical Communications, (accepted). N´emeth, L., Szalay, L., Alternating sums in hyperbolic Pascal triangles, Miskolc Mathematical Notes, (accepted). N´emeth, L., Szalay, L., Power sums in hyperbolic Pascal triangles, (submitted). N´emeth, L., Szalay, L., Recurrence sequences in the hyperbolic Pascal triangle corresponding to the regular mosaic {4, 5}, Ann. Math. Inf., 46 (2016), pp. 165-173.
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