Balances in the Set of Arithmetic Progressions †

: This article focuses on searching and classifying balancing numbers in a set of arithmetic progressions. The sufﬁcient and necessary conditions for the existence of balancing numbers are presented. Moreover, explicit formulae of balancing numbers and various relations are included.


Introduction
An integer n is called a balancing number if there exists another integer r, called a balancer, corresponding to n, such that the following Diophantine equation holds 1 + 2 + · · · + (n − 1) = (n + 1) + (n + 2) + · · · + (n + r). The problem of determining all balancing numbers in the set of natural numbers leads to a second order linear recursive sequence or a Pell equation. In [1], Behera and Panda prove that the square of any balancing number is a triangular number. In fact, it is easy to derive that n is a balancing number if and only if n 2 is a triangular number, and if and only if 8n 2 + 1 is a perfect square. They also proved that the balancing sequence {B m } m≥1 fulfilled the recursive relation for all m ≥ 2 with initials B 1 = 1, B 2 = 6. According to the recursive relation, it forces that B 0 = 0. From (1), we have r 2 + (2n + 1)r − n(n − 1) = 0, or r = −2n − 1 + √ 8n 2 + 1 2 .
Since r is an integer, then 8n 2 + 1 must be an odd square, say 8n 2 + 1 = t 2 with t odd. We have In addition, one can check that C m+1 = 6C m − C m−1 and the relation R m = C m −2B m −1 2 for any m ≥ 1 .
Similarly, one can define the cobalancing numbers with cobalancers as the solutions in n, r to the Diophantine equation 1 + 2 + · · · + n = (n + 1) + (n + 2) + · · · + (n + r). ( From (2), it implies that n is a cobalancing number if and only if n(n + 1) is a triangular number, and if and only if 8n 2 + 8n + 1 is a perfect square. Throughout this paper, let b m , r m and c m := 8b 2 m + 8b m + 1 be the m-th cobalancing number, the m-th cobalancer, and the m-th Lucas-balancing number, respectively. For example, b 1 = r 1 = 0 and c 1 = 1.
Various possible generalizations of balancing numbers, balancers, and Lucas-balancing numbers have been studied by several authors from many aspects. In [2], Panda defined sequence balancing numbers with sequence balancers as follows. Let {u m } m≥1 be a sequence of real numbers. We call a pair (u m , k) a sequence balancing number with sequence balancer if Kovács, Liptai and Olajos [3] introduced the concept of sequence balancing numbers to the sequence of arithmetic progressions. They defined the (a, b)-type balancing numbers with the (a, b)-type balancer as solutions of the Diophantine equation where a > 0 and b ≥ 0 are coprime integers. In [3], several effective finiteness and explicit results about (a, b)-type balancing numbers had been given. In particular, Kovács et al. proved the following theorem.
Thus, a question naturally arises: can we determine all (a, b)-type balancing numbers in the set of arithmetic progressions? In this paper, we will provide an answer to this question. In other words, we search "few" balances in the set of arithmetic progressions. First, we must be clear about the terminology and notation. We denote the m-th (a, b)-type balancing number, if it exists for infinitely many m, by B (a,b) m . Note that when (a, b) = (1, 0), we get nothing but the original balancing number B m . Instead of requesting integers a > 0 and b ≥ 0, our definition of (a, b)-type balancing should only exclude from the cases a = 0 or gcd(a, b) = 1. For the former case a = 0, it is trivial. For the later case, we just notice for Without loss of generality, we may assume that a > 0, for otherwise, use the above identity by putting d = −1. For our convenience, let B exist for infinitely many m if and only if a | 2b. Moreover, we sort and classify all (a, b)-type balancing numbers and provide explicit formulae for them. In Section 3, we discuss the (a, b)-type cobalancing numbers, the (a, b)-type cobalancers, and the (a, b)-type Lucas-cobalancing numbers. We summarize the paper in the conclusion section.

Main Results
The problem of determining all (a, b)-type balancing numbers in the set of arithmetic progressions leads to the equation x 2 − 8y 2 = c. In the following, we derive the condition a | 2b as a sufficient and necessary condition for the existence of the (a, b)-type balancing number B (a,b) m for infinitely many m. On one hand, from (3) we have or We solve the equation in r directly to get It "forces" that the number 8(an + b) 2 + (a 2 − 4ab − 4b 2 ) must be a perfect square. We then define the m-th (a, b)-type Lucas-balancing number by Note that both a and C (a,b) m have the same parity. Now we have, for all m ≥ 1, On the other hand, multiplying Equation (4) by 4a (note that a = 0), we get Let x = 2a(n + r) + a + 2b and y = an + b. We may rewrite (5) as where To solve Equation (6), we solve the following Pell equation in two variables u, v: The fundamental solution is (u, v) = (3, 1) and, thus, all the solutions can be determined by is an initial solution of the Equation (6). Hence, all solutions of the Equation (6) takes the form It is easy to prove that both sequences {x m } m≥1 , {y m } m≥1 satisfy the same recurrence relation as below for m ≥ 2 and with initials x 1 = 3, x 2 = 17 and y 1 = 1, y 2 = 6, respectively. Therefore, we conclude that x m = C m and y m = B m , the m-th Lucas-balancing number, and the m-th balancing number. Now, we get It immediately implies that Whether the positive or negative sign it takes, the key to guarantee is B In light of C m = 3B m − B m−1 (this follows easily by strong induction on m), we may rewrite the explicit formulae for the (a, b)-type balancing number as For example, when (a, b) = (2, 1), we have B Another example, when a = b = 1, we relabel the (1, 1)-type balancing numbers according to Equation (7). That is, for k ≥ 1, Actually the number B (1,1) m is an integer n, such that 8n 2 − 7 is a perfect square. Thus, n 2 − 1 is a triangular number. In [4], Subramaniam studied such an integer n, which he called the almost square triangular number (ASTN) of type β, such that n 2 − 1 is triangular. See also Theorem 2.2.2 in [2].
In addition, we see that the x-solutions of Equation (5) are given by In light of B and the recursive relation R (a,b) Recall that 8n 2 + 1 is an odd square if n is a balancing number. Thus, the number C m must be an odd integer for each positive integer m. Hence, we conclude that That is, under the restriction of gcd(a, b) = 1, the positive integer a is taken either 1 or 2. We summarize the above discussion in the following theorem. Hence, we just relabel and conclude that the explicit formula for the (1, b)-type balancing numbers is given by, for all k ≥ 1, In light of (8) and (9), the corresponding (1, b)-type balancers R for all k ≥ 1.
The above identity holds by the definition of balancing numbers with the balancer.
The corresponding (1, −c)-type balancers R (1,−c) m are given as below. By definition, In addition, we have that C and for m ≥ 3, In viewing of (8) and (9), the corresponding (2, b)-type balancers R for all k ≥ 1.  1, 7, 41, 239, 1393, 8119, . . ., which are related to the square order of simple groups and these numbers are called NSW numbers [5]. The NSW numbers have nice properties similar to Mersenne numbers M p = 2 p − 1 with p prime. It is easy to see that NSW numbers verify the recurrence relation Barcucci et al. [6] gave a combinatorial interpretation of the sequence of NSW numbers. Actually, they proved that the cardinality of the set of words of L having length m is equal to B For c > 1 is an odd integer, we note that B The corresponding (2, −c)-type balancers R (2,−c) m are given as below. By definition, and C

Corollary 1.
For any integer k ≥ 0, we have Of course, some explicit formulae of cobalancing and Lucas-cobalancing numbers, and some modular relations among them, can be obtained in a similar way. We leave it to the interested reader.