EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 2, 2021, 380-395 ISSN 1307-5543 – ejpam.com Published by New York Business Global Characterizations and Identities for Isosceles Triangular Numbers Jiramate Punpim1, Somphong Jitman1,∗ 1 Department of Mathematics, Faculty of Science, Silpakorn University, Nakhon Pathom 73000, Thailand Abstract. Triangular numbers have been of interest and continuously studied due to their beau- tiful representations, nice properties, and various links with other figurate numbers. For positive integers n and l, the nth l-isosceles triangular number is a generalization of triangular numbers defined to be the arithmetic sum of the form T (n, l) = 1 + (1 + l) + (1 + 2l) + · · ·+ (1 + (n− 1)l). In this paper, we focus on characterizations and identities for isosceles triangular numbers as well as their links with other figurate numbers. Recursive formulas for constructions of isosceles triangular numbers are given together with necessary and sufficient conditions for a positive integer to be a sum of isosceles triangular numbers. Various identities for isosceles triangular numbers are established. Results on triangular numbers can be viewed as a special case. 2020 Mathematics Subject Classifications: 11B99, 11E25 Key Words and Phrases: Integer Sequences, Triangular numbers, Isosceles triangular numbers, Polygonal Numbers, Identities 1. Introduction A triangular number is a number that can be represented as an equilateral triangular arrangement of points equally spaced. Precisely, the nth triangular number is the number of points composing an equilateral triangle with n points on a side which equals the sum of the n natural numbers of the form T (n) = 1 + 2 + 3 + · · ·+ n = n(n + 1) 2 . (1) The nth triangular number can be represented as points in an equilateral triangle as in Figure 1. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i2.3952 Email addresses: punpim j@su.ac.th (J. Punpim), sjitman@gmail.com (S. Jitman) http://www.ejpam.com 380 c© 2021 EJPAM All rights reserved. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 381 n points n points Figure 1: Triangular Number T (n) Triangular numbers have been introduced and studied since the 6th century BC. Due to their nice properties and wide links with other mathematical objects, such numbers have been extensively studied (see [1–3] and references therein). Subsequently, characterizations of triangular numbers and interesting relationships among triangular numbers and other figurate numbers have been established (see [2, 4, 10]). Various subsequences of triangular numbers with nice properties have been presented in [2, 8, 11]. Identities of triangular numbers have been of interest and studied in [1–3, 7, 9, 12]. In [5], an isosceles triangular number has been introduced as a generalization of tri- angular numbers and it is defined to be a number that can be represented as an isosceles triangular arrangement of points. Precisely, the nth l-isosceles triangular number, denoted by T (n, l), is defined to be the arithmetic sum of the form T (n, l) = 1 + (1 + l) + (1 + 2l) + · · ·+ (1 + (n− 1)l) = n + n(n− 1)l 2 = n + lT (n− 1). Alternatively, an l-isosceles triangular number can be viewed as a special case of generalized trapezoidal numbers in [6]. Clearly, the isosceles triangular number T (n, l) becomes the nth triangular number T (n) whenever l = 1. In the case where l = 2, it can be easily seen that T (n, 2) = n2 is a square number. For convenience, the notions of the l-isosceles triangular number T (0, l) and the triangular number T (0) are used in some contexts and they are set to be zero. In [5], parity and some properties of isosceles triangular numbers have been studied. The nth l-isosceles triangular number T (n, l) can be represented as an isosceles triangular arrangement of points in Figure 2. Illustrative examples of isosceles triangular numbers T (4, 2) = 16 and T (4, 3) = 22 are given in Example 1. Example 1. The positive integers 16 and 22 are 2-isosceles and 3-isosceles triangular numbers, respectively. They can be represented as isosceles triangular shapes as follows. A polygonal number is a number represented as points or pebbles arranged in the shape of a regular polygon (see [2]). For positive integers n and m, the nth m-gonal number is defined to be J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 382 1 + l points 1 + 2l points 1 + (n− 3)l points 1 + (n− 2)l points 1 + (n− 1)l points n points Figure 2: Isosceles triangular number T (n, l) (a) T (4, 2) = 16 (b) T (4, 3) = 22 Figure 3: Isosceles triangular numbers T (4, 2) = 16 and T (4, 3) = 22 P (n,m) = 1 + (1 + (m− 2)) + (1 + 2(m− 2)) + · · ·+ (1 + (m− 2)(n− 1)) = n + n(n− 1)(m− 2) 2 . It is not difficult to see that T (n, l) = P (n, l + 2), i.e., an isosceles triangular number is actually a shifted version of a polygonal number. However, in this manuscript, the notion of isosceles triangular numbers is used to present the generality of triangular numbers and nice appearance of some identities. As mentioned above, various characterizations and identities for triangular numbers have been established. For isosceles triangular numbers, only a few works have been done on their characterizations and identities. It is therefore of interest to investigate such problems. The first goal of this paper is to generalize the facts (see [3]) that “a positive integer N is a triangular number if and only if 9N + 1 is a triangular number” and “N is a triangular number if and only if 8N + 1 is square” to isosceles triangular numbers. Secondly, we aim to generalize some identities for triangular numbers in [4] and [7] to isosceles triangular numbers. To the best of our knowledge, the characterizations and identities for isosceles triangular numbers presented in this paper have not been established either in terms of isosceles triangular numbers or polygonal numbers. The paper is organized as follows. Characterizations of isosceles triangular numbers are presented in Section 2 as well as their links with other figurate numbers. In Section 3, some identities for isosceles triangular numbers are established as generalizations of triangular numbers. Summary and remarks are provided in Section 4. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 383 2. Characterizations of Isosceles Triangular Numbers In this section, some characterizations of isosceles triangular numbers are given as well as links with other figurate numbers. Some classical results on the characterizations of triangular numbers can be viewed as a special case. 2.1. Characterizations and Recursive Constructions of Isosceles Triangu- lar Numbers In this subsection, we focus on a generalization of the well-known fact “a positive integer N is a triangular number if and only if 9N + 1 is a triangular number” in [3]. In the following theorem, a characterization of l-isosceles triangular numbers is pre- sented. A recursive construction of l-isosceles triangular numbers can be deduced directly from the theorem. Theorem 1. Let N and l be positive integers. Let r and s be positive integers of the form r = (2l+1)2 and s = l3 − 3l2 + 4 2 . Then N is an l-isosceles triangular number if and only if rN + s is an l-isosceles triangular number. Proof. Assume that N is an l-isosceles triangular number. Then N = n + n(n− 1)l 2 for some positive integer n. It follows that rN + s = (2l + 1)2 ( n + n(n− 1)l 2 ) + l3 − 3l2 + 4 2 = n2l(2l + 1)2 + (2− l)n(2l + 1)2 + l3 − 3l2 + 4 2 = 4l3n2 + 4l2n2 + ln2 + 8l2n + 8ln + 2n− 4l3n− 4l2n− ln + l3 − 3l2 + 4 2 = (4l3 + 4l2 + l)n2 − (4l3 − 4l2 − 7l − 2)n + (l3 − 3l2 + 4) 2 = l ( (4l2 + 4l + 1)n2 − (4l2 − 6l − 4)n + (l2 − 4l + 4) ) − (l − 2) ((2l + 1)n− (l − 2)) 2 = l ((2l + 1)n− (l − 2))2 − (l − 2) ((2l + 1)n− (l − 2)) 2 = 2 ((2l + 1)n− (l − 2)) + l ((2l + 1)n− (l − 2))2 − l ((2l + 1)n− (l − 2)) 2 = ((2l + 1)n− (l − 2)) + ((2l + 1)n− (l − 2)) (((2l + 1)n− (l − 2))− 1) l 2 = T ((2l + 1)n− (l − 2), l). Hence, rN + s is an l-isosceles triangular number. Conversely, assume that rN + s is an l-isosceles triangular number. Then rN + s = n + n(n− 1)l 2 J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 384 for some positive number n. Equivalently, we have N = 2n + ln2 − ln− 2s 2r = 2n + ln2 − ln− (l3 − 3l2 + 4) 2(2l + 1)2 = l ( n2 + 2n(l − 2) + (l − 2)2 ) − (l − 2)(2l + 1)(n + l − 2) 2(2l + 1)2 = l(n + l − 2)2 − (l − 2)(2l + 1)(n + l − 2) 2(2l + 1)2 = l(n + l − 2)2 (2l + 1)2 − (l − 2)(n + l − 2) 2l + 1 2 = 2(n + l − 2) 2l + 1 + l(n + l − 2)2 (2l + 1)2 − l(n + l − 2) 2l + 1 2 = n + l − 2 2l + 1 + ( n + l − 2 2l + 1 )( n + l − 2 2l + 1 − 1 ) l 2 . Since N = 2n + ln2 − ln− (l3 − 3l2 + 4) 2(2l + 1)2 = (ln− l2 + l + 2)(n + l − 2) 2(2l + 1)2 is a positive inte- ger, we have (2l+1)2|(ln− l2+ l+2)(n+ l−2). If (2l+1)|(n+ l−2), then (n + l − 2) (2l + 1) ∈ N. Suppose that (2l + 1) - (n+ l− 2). Since ln− l2 + l + 2 = l(n+ l− 2)− (2l + 1)(l− 2) and gcd(l, 2l+1) = 1, we have (2l+1) - l(n+ l−2) which implies that (2l+1) - (ln− l2 + l+2). Hence, (2l+1)2 - (ln− l2+ l+2)(n+ l−2) which is a contradiction. Then (2l+1)|(n+ l−2) which implies that (n + l − 2) (2l + 1) is a positive integer. Therefore, N = n + l − 2 2l + 1 + ( n + l − 2 2l + 1 )( n + l − 2 2l + 1 − 1 ) l 2 = T ( n + l − 2 2l + 1 , l ) is an l-isosceles triangular number. By setting l = 1, we have the classical fact “N is a triangular number if and only if 9N + 1 is a triangular number”. For l ∈ {2, 3, 4}, we have the following results. • A positive integer N is 2-isosceles triangular if and only if 25N is a 2-isosceles triangular number. Equivalently, N is a square number. • A positive integer N is 3-isosceles triangular if and only if 49N + 2 is a 3-isosceles triangular number. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 385 • A positive integer N is 4-isosceles triangular if and only if 81N + 10 is a 4-isosceles triangular number. In general, each positive integer l, a recursive construction of l-isosceles triangular numbers can be deduced directly from Theorem 1. 2.2. Isosceles Triangular Numbers and Square Numbers In this subsection, we focus on a generalization of the classical fact “N is a triangular number if and only if 8N + 1 is square” (see [3]). Necessary and sufficient conditions for an l-isosceles triangular number to be square are given in the following theorem. Theorem 2. Let N and l be positive integers. Then N is an l-isosceles triangular number if and only if 8Nl + (l − 2)2 is square and √ 8Nl + (l − 2)2 ≡ (l + 2) (mod 2l). Proof. Assume that N is an l-isosceles triangular number. Then N = n + n(n− 1)l 2 for some positive integer n. It follows that 8Nl + (l − 2)2 = 8nl + 4n(n− 1)l2 + (l − 2)2 = (2nl)2 − 2(2nl)(l − 2) + (l − 2)2 = (2nl − l + 2)2 is square and √ 8Nl + (l − 2)2 ≡ 2nl − l + 2 ≡ l + 2 (mod 2l). Conversely, assume that 8Nl+(l−2)2 is square and √ 8Nl + (l − 2)2 ≡ (l+2) (mod 2l). Then there exists a positive integer m such that √ 8Nl + (l − 2)2 = 2lm+(l+2). It follows that 8Nl + (l − 2)2 = (2lm + (l + 2))2 = 4l2m2 + 4lm(l + 2) + (l + 2)2, and hence, N = 4l2m2 + 4lm(l + 2) + (l + 2)2 − (l − 2)2 8l = 8ml + 8l + 4(m + 1)ml2 8l = (m + 1) + (m + 1)ml 2 = T (m + 1, l). Therefore, N is an l-isosceles triangular number as desired. For l = 1, 8Nl + (l − 2)2 = 8N + 1 is always odd. Hence, if 8Nl + (l − 2)2 is square, then √ 8Nl + (l − 2)2 ≡ √ 8N + 1 ≡ 1 ≡ (l+2) (mod 2l). Using Theorem 2, the following well-know result can be derived immediately. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 386 Corollary 1 ([3]). Let N be a positive integer. Then N is a triangular number if and only if 8N + 1 is square. In general, we have the following results on the sum of k l-isosceles triangular numbers. Theorem 3. Let l, k, and N be positive integers. Then N is a sum of k l-isosceles triangular numbers if and only if there exist positive integers u1, u2, . . . , uk such that (i) 8lN + k(l − 2)2 = u21 + u22 + · · ·+ u2k, and (ii) ui ≡ l + 2 (mod 2l) for all i = 1, 2, . . . , k. Proof. Assume that N is a sum of k l-isosceles triangular numbers. Then N = k∑ i=1 T (mi, l) for some positive integers mi. It follows that N = k∑ i=1 ( mi + mi(mi − 1)l 2 ) = k∑ i=1 2mi + m2 i l −mil 2 and 8lN + k(l − 2)2 = k(l − 2)2 + 4l k∑ i=1 ( 2mi + m2 i l −mil ) = k∑ i=1 ( 4m2 i l 2 − 4mil(l − 2) + (l − 2)2 ) = k∑ i=1 (2mil − (l − 2))2. For each i ∈ {1, 2, . . . , k}, let ui = 2mil − (l − 2). It follows that 8lN + k(l − 2)2 = u21 + u22 + · · ·+ u2k and ui ≡ 2lmi − (l − 2) ≡ l + 2 (mod 2l) for all i ∈ {1, 2, . . . , k}. Conversely, assume that there exist positive integers u1, u2, . . . , uk such that 8lN + k(l − 2)2 = u21 + u22 + · · ·+ u2k and ui ≡ l + 2 (mod 2l) for all i = 1, 2, . . . , k. For each i ∈ {1, 2, . . . , k}, let mi = ui + (l − 2) 2l . Since ui ≡ l + 2 (mod 2l), it follows that mi is a positive integer. It is not difficult to verify that N = k∑ i=1 T (mi, l). For k = 1, Theorem 3 becomes Theorem 2. By setting l = 1 in Theorem 3, we have the following well-known result. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 387 Corollary 2 ([3]). Let k and N be positive integers. Then N is a sum of k triangular numbers if and only if 8N + k is a sum of k odd squares. From Theorem 3, a positive integer N is a sum of two l-isosceles triangular numbers if and only if there exist positive integers u and v such that 8lN + 2(l− 2)2 = u2 + v2 and u ≡ v ≡ l + 2 (mod 2l). An alternative characterization for this case is given in the next theorem. Theorem 4. Let l and N be positive integers. Then N is a sum of two l-isosceles trian- gular numbers if and only if there exist positive integers u and v such that (i) 4lN + (l − 2)2 = u2 + v2, and (ii) u + v ≡ l + 2 (mod 2l) and u− v ≡ l + 2 (mod 2l). Proof. Assume that N is a sum of two l-isosceles triangular numbers. Then N = T (m, l) + T (n, l) for some positive integers m and n, i.e., N = m + m(m− 1)l 2 + n + n(n− 1)l 2 = 2m + m(m− 1)l + 2n + n(n− 1)l 2 . It follows that 4lN + (l − 2)2 = 2l (2m + m(m− 1)l + 2n + n(n− 1)l) + (l − 2)2 = 4ml + 2m2l2 − 2ml2 + 4nl + 2n2l2 − 2nl2 + (l − 2)2 = (lm)2 + (ln)2 + (l − 2)2 + 2l2mn− 2(l − 2)lm − 2(l − 2)ln + (lm)2 − 2mnl2 + (ln)2 = (l(m + n)− (l − 2))2 + (l(m− n))2. Let u = l(m + n)− (l− 2) and v = l(m− n). We therefore have 4lN + (l− 2)2 = u2 + v2, u + v ≡ 2lm− l + 2 ≡ l + 2 (mod 2l), and u− v ≡ 2ln− l + 2 ≡ l + 2 (mod 2l). Conversely, assume that there exist positive integers u and v such that 4lN + (l − 2)2 = u2 + v2, u+v ≡ l+2 (mod 2l), and u−v ≡ l+2 (mod 2l). Without loss of generality, assume that u ≥ v. Let m = u + v + (l − 2) 2l and n = u− v + (l − 2) 2l . Since u + v ≡ l + 2 (mod 2l) and u− v ≡ l+ 2 (mod 2l), it follows that m and n are positive integers. It is not difficult to verify that N = T (m, l) + T (n, l). J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 388 3. Identities In this section, we focus on generalizations of classical identities for triangular numbers (see, for example, [1–3, 7, 9, 12]). Each identity for l-isosceles triangular numbers is given and followed by the corresponding identity for triangular numbers. The identity concerning the (m + n)th l-isosceles triangular numbers is given in the next theorem. Theorem 5. T (m + n, l) = T (m, l) + T (n, l) + lmn for all positive integers l, m, and n. Proof. Let l,m, and n be positive integers. Then T (m + n, l) = m + n + m + n(m + n− 1)l 2 = m + n + (m2 −m + n2 − n + 2mn)l 2 = 2m + 2n + m2l −ml + n2l − nl + 2lmn 2 = 2m + m(m− 1)l + 2n + n(n− 1)l + 2lmn 2 = m + m(m− 1)l 2 + n + n(n− 1)l 2 + lmn = T (m, l) + T (n, l) + lmn. As desired, we have T (m + n, l) = T (m, l) + T (n, l) + lmn. By setting l = 1, we have the classical identity T (m + n) = T (m) + T (n) + mn in [12, Equation (4a)]. In the following theorem, the identity concerning the mnth l-isosceles triangular num- bers is presented. Theorem 6. T (mn, l) = T (m, l)T (n, l)+(2−l)lT (m− 1)T (n− 1) for all positive integers m,n, and l. Proof. Let m, n, and l be positive integers. Using a direct calculation, we have T (mn, l) = mn + mn(mn− 1)l 2 = 4mn + 2m2n2l − 2mnl 4 = (2m + m2l −ml)(2n + n2l − nl) 4 + (2l − l2)(m2 −m)(n2 − n) 4 = ( m + m(m− 1)l 2 )( n + n(n− 1)l 2 ) + (2− l)l ( (m− 1)m 2 )( (n− 1)n 2 ) = T (m, l)T (n, l) + (2− l)lT (m− 1)T (n− 1). J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 389 This completes the proof. The classical identity T (mn) = T (m)T (n) + T (m− 1)T (n− 1) for triangular numbers in [12, Equation (17a)] follows easily when l = 1. One of the classical facts about triangular numbers is that a sum of two consecutive triangular numbers is square (see [12, Equation (1)]). Precisely, T (n)+T (n+1) = (n+1)2 is square for all positive integers n. In the following theorem, we prove that l = 1 is the necessary and sufficient condition for the sum T (n, l) + T (n + 1, l) to be square for all positive integers n. Theorem 7. Let l be a positive integer. Then T (n, l)+T (n+1, l) is square for all positive integers n if and only if l = 1. Proof. Assume that l ≥ 2. First, we note that T (n, l) + T (n + 1, l) = n2l + 2n + 1 for all positive integers n. It is not difficult to see that T (1, l) + T (1 + 1, l) = l + 3 is square if and only if 22(l + 3) is square. For l ≥ 2, if T (1, l) + T (1 + 1, l) = l + 3 is square, then l ≥ 6 and 22(l + 3) ≥ 36. Hence, T (2, l) + T (2 + 1, l) = 4l + 5 = 22(l + 3) − 7 cannot be square. Therefore, T (n, l) + T (n + 1, l) is non-square for n = 1 or n = 2. The converse follows directly from [12, Equation (1)]. Next, we focus on some identities for isosceles triangular numbers induced by a recur- rence relation. Theorem 8. nT (n + 1, l)− (l − 1)n = (n + 2)T (n, l) for all positive integers n and l. Proof. Let n and l be positive integers. Then nT (n + 1, l)− (l − 1)n = n ( (n + 1) + (n + 1)nl 2 ) − nl + n = 2n2 + 2n + n3l + n2l − 2nl + 2n 2 = 2n2 + n3l − n2l + 4n + 2n2l − 2nl 2 = (n + 2) ( 2n + n2l − nl 2 ) = (n + 2) ( n + n(n− 1)l 2 ) = (n + 2)T (n, l). Hence, nT (n + 1, l)− (l − 1)n = (n + 2)T (n, l) as desired. By setting l = 1, the identity nT (n + 1) = (n + 2)T (n) for triangular numbers in [7, Equation (1.7)] follows. Theorem 9. T (2n + 1, l) − T (2n, l) = T (n + 1, l) − T (n− 1, l) + (l − 1) for all positive integers n and l. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 390 Proof. Let n and l be positive integers. Then T (2n + 1, l)− T (2n, l) = 2n + 1 + (2n + 1)(2n)l 2 − 2n− (2n)(2n− 1)l 2 = 4n + 2 + 4n2l + 2nl − 4n− 4n2l + 2nl 2 = 2n + 2 + n2l + nl − 2n + 2− n2l + 3nl − 2l + 2l − 2 2 = n + 1 + (n + 1)(n)l 2 − (n− 1)− (n− 1)(n− 2)l 2 + l − 1 = T (n + 1, l)− T (n− 1, l) + (l − 1). This completes the proof. From the theorem above, the relation T (2n + 1)− T (2n) = T (n + 1)− T (n− 1) in [9, Equation (7.14)] can be obtained directly when l = 1. Theorem 10. T (2n, l) = 3T (n, l) + lT (n− 1) + (l− 1)n for all positive integers n and l. Proof. Let n and l be positive integers. Then 3T (n, l) + lT (n− 1) + (l − 1)n = 3 ( n + n(n− 1)l 2 ) + l ( (n− 1)n 2 ) + (l − 1)n = 6n + 3n2l − 3nl + n2l − nl + 2nl − 2n 2 = 4n + (4n2 − 2n)l 2 = 2n + 2n(2n− 1)l 2 = T (2n, l). As desired, we have T (2n, l) = 3T (n, l) + lT (n− 1) + (l − 1)n. Lemma 1. lT (n− 1) + (l− 1)n = T (n− 1) + (l− 1)T (n) for all positive integers n and l. Proof. Let n and l be positive integers. Then we have T (n− 1) + (l − 1)T (n) = (n− 1)n 2 + (l − 1) ( n(n + 1) 2 ) = n2 − n + n2l + nl − n2 − n 2 = n2l − nl + 2nl − 2n 2 = l ( (n− 1)n 2 ) + (l − 1)n J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 391 = lT (n− 1) + (l − 1)n as desired. From Theorem 10 and Lemma 1, we have the following identity. Corollary 3. T (2n, l) = 3T (n, l) + T (n− 1) + (l − 1)T (n) for all positive integers n and l. By setting l = 1, we have T (2n) = 3T (n) + T (n − 1) for all positive integers n as in [7, Equation (1.12)]. Theorem 11. T (2n + 1, l) = 3T (n, l) + T (n + 1, l) + 2(l − 1)n for all positive integers n and l. Proof. Let n and l be positive integers. Then T (2n + 1, l) = 2n + 1 + (2n + 1)(2n)l 2 = 4n + 2 + 4n2l + 2nl 2 = 6n + 3n2l − 3nl + 2n + 2 + n2l + nl + 4nl − 4n 2 = 3 ( n + n(n− 1)l 2 ) + ( n + 1 + (n + 1)(n)l 2 ) + 2ln− 2n = 3T (n, l) + T (n + 1, l) + 2(l − 1)n. Hence, the proof is completed. Theorem 12. lT (2n + 1, l) = 3lT (n, l)+ lT (n + 1, l)+4nT (l − 1) for all positive integers n and l. Proof. Let n and l be positive integers. Then we have lT (2n + 1, l) = l ( 2n + 1 + (2n + 1)(2n)l 2 ) = 4nl + 2l + 4n2l2 + 2nl2 2 = 6nl + 3n2l2 − 3nl2 + 2nl + 2l + n2l2 + nl2 + 4nl2 − 4nl 2 = 3 ( n + n(n− 1)l 2 ) + l ( n + 1 + (n + 1)(n)l 2 ) + 4n ( (l − 1)l 2 ) = 3lT (n, l) + lT (n + 1, l) + 4nT (l − 1) as required. From the two theorems above, the well-known relation T (2n + 1) = 3T (n) + T (n + 1) derived from [9, Equations (7.14) and (7.22)] can be obtained by setting l = 1. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 392 Theorem 13. n2T (k − 1, l)+kT (n, l) = T (nk, l)−n2(k−1)(l−1) for all positive integers n, k, and l. Proof. Let n, k, and l be positive integers. Then n2T (k − 1, l) + kT (n, l) = n2 ( k − 1 + (k − 1)(k − 2)l 2 ) + k ( n + n(n− 1)l 2 ) = 2n2k − 2n2 + n2k2l − 3n2kl + 2n2l + 2nk + n2kl − nkl 2 = 2nk + n2k2l − nkl − 2n2kl + 2n2k + 2n2l − 2n2 2 = 2nk + n2k2l − nkl 2 − n2kl + n2k + n2l − n2 = nk + nk(nk − 1)l 2 − n2(k − 1)(l − 1) = T (nk, l)− n2(k − 1)(l − 1). Therefore, we have n2T (k − 1, l) + kT (n, l) = T (nk, l)− n2(k − 1)(l − 1) as desired. By setting l = 1, the identity n2T (k − 1) + kT (n) = T (nk) for triangular numbers follows. Theorem 14. n2T (k − 1, l) + kT (n− 1, l) = T (nk − 1, l)− (n2 − 1)(k − 1)(l − 1) for all positive integers n, k, and l. Proof. Let n, k, and l be positive integers. Then n2T (k − 1, l) + kT (n− 1, l) = n2 ( k − 1 + (k − 1)(k − 2)l 2 ) + k ( n− 1 + (n− 1)(n− 2)l 2 ) = 2n2k − 2n2 + n2k2l − 3n2kl + 2n2l + 2nk − 2k + n2kl − 3nkl + 2kl 2 = 2nk − 2 + n2k2l − 3nkl + 2l − 2n2kl + 2n2l + 2kl − 2l + 2nk − 2n2 − 2k + 2 2 = 2nk − 2 + n2k2l − 3nkl + 2l 2 − (n2kl − n2l − kl + l − n2k + n2 + k − 1) = (nk − 1) + (nk − 1)(nk − 2)l 2 − (n2 − 1)(k − 1)(l − 1) = T (nk − 1, l)− (n2 − 1)(k − 1)(l − 1) as desired. For l = 1, we have the identity n2T (k − 1) + kT (n− 1) = T (nk − 1) for all positive integers k and n. In the following theorems, some identities concerning squares of l-isosceles triangular numbers are presented. J. Punpim, S. Jitman / Eur. J. Pure Appl. Math, 14 (2) (2021), 380-395 393 Theorem 15. T (n + 1, l)2 − T (n, l)2 = (n + 1)3 + (l − 1)n ( (l + 1)n2 + 3n + 1 ) for all positive integers l and n. Proof. Let l and n be positive integers. Then T (n + 1, l)2 − T (n, l)2 = ( (n + 1) + (n + 1)nl 2 )2 − ( n + n(n− 1)l 2 )2 = 4n2 + 8n + 4n3l + 8n2l + 4n + 4 + 4nl + n4l2 + 2n3l2 + n2l2 4 − 4n2 + 4n3l − 4n2l + n4l2 − 2n3l2 + n2l2 4 = 8n + 12n2l + 4 + 4nl + 4n3l2 4 = 2n + 3n2l + 1 + nl + n3l2 = (n3 + 3n2 + 3n + 1) + (nl − n)(n2l + n2 + 3n + 1) = (n + 1)3 + (l − 1)n((l + 1)n2 + 3n + 1). Hence, T (n + 1, l)2 − T (n, l)2 = (n + 1)3 + (l − 1)n ( (l + 1)n2 + 3n + 1 ) as desired. The identity T (n + 1)2−T (n)2 = (n+1)3 for triangular numbers in [7, Equation (1.5)] is a special case of the above theorem where l = 1. Theorem 16. T (n, l)2 + T (n− 1, l)2 = lT (n2, l)− (2n2 − 2n + 1)(2T (l, n− 1)− (l + 1)) for all positive integers n and l. Proof. Let n and l be positive integers. Then T (n, l)2 + T (n− 1, l)2 = ( n + n(n− 1)l 2 )2 + ( n− 1 + (n− 1)(n− 2)l 2 )2 = ( 2n + n2l − nl 2 )2 + ( 2n− 2 + n2l − 3nl + 2l 2 )2 = 8n2 + 8n3l − 20n2l + 2n4l2 − 8n3l2 + 14n2l2 + 20nl − 8n + 4− 8l − 12nl2 + 4l2 4 = 4n2 + 4n3l − 10n2l + n4l2 − 4n3l2 + 7n2l2 + 10nl − 4n + 2− 4l − 6nl2 + 2l2 2 = ( 2n2l + n4l2 − n2l2 2 ) − (2n2 − 2n + 1) ( 4l + 2nl2 − 2l2 − 2nl + 2l − 2l − 2 2 ) = l ( n2 + n2(n2 − 1)l 2 ) − (2n2 − 2n + 1) ( 2(l + l(l − 1)(n− 1) 2 )− l − 1 ) = lT (n2, l)− (2n2 − 2n + 1)(2T (l, n− 1)− (l + 1)). The proof is therefore completed. The identity T (n)2 + T (n− 1)2 = T (n2) in [4, Equation (19a)] is a special case of the theorem where l = 1. REFERENCES 394 4. Conclusion and Remarks Generalizations of triangular numbers have been studied in terms of isosceles triangular numbers. Some characterizations of such numbers are given as well as links with square numbers. 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