Ramanujan's theta functions and linear combinations of four triangular numbers
aa r X i v : . [ m a t h . N T ] O c t Ramanujan’s theta functions and linear combinations of four triangular numbers
Zhi-Hong SunSchool of Mathematics and StatisticsHuaiyin Normal UniversityHuaian, Jiangsu 223300, P.R. ChinaEmail: [email protected]: http://maths.hytc.edu.cn/szh1.htm
Abstract
Let Z and Z + be the set of integers and the set of positive integers, respectively.For a, b, c, d, n ∈ Z + let t ( a, b, c, d ; n ) be the number of representations of n by ax ( x +1) / by ( y + 1) / cz ( z + 1) / dw ( w + 1) / x, y, z, w ∈ Z ). In this paper, by usingRamanujan’s theta functions ϕ ( q ) and ψ ( q ) we present many formulas and conjectures on t ( a, b, c, d ; n ).Keywords: theta function; triangular number; quadratic formMathematics Subject Classification 2010: 11D85, 11E25, 30B10, 33E20
1. Introduction
Let Z , Z + and N be the set of integers, the set of positive integers and the set of nonneg-ative integers, respectively, and let Z k = Z × Z × · · · × Z | {z } k times . For a , a , . . . , a k ∈ Z + ( k ≥ n ∈ N set N ( a , a , . . . , a k ; n ) = (cid:12)(cid:12) { ( x , . . . , x k ) ∈ Z k | n = a x + a x + · · · + a k x k } (cid:12)(cid:12) ,t ( a , a , . . . , a k ; n )= (cid:12)(cid:12)(cid:12)n ( x , . . . , x k ) ∈ Z k (cid:12)(cid:12) n = a x ( x − a x ( x − · · · + a k x k ( x k − o(cid:12)(cid:12)(cid:12) and C ( a , . . . , a k ) = i ( i − i − i − i ( i − i i i , where i j denotes the number of elements in { a , . . . , a k } which are equal to j . For conve-nience we also define t ( a , a , . . . , a k ; n ) = N ( a , a , . . . , a k ; n ) = 0 for n N . In 2005 Adiga, Cooper and Han [ACH] showed that(1.1) t ( a , a , . . . , a k ; n )= 22 + C ( a , . . . , a k ) N ( a , . . . , a k ; 8 n + a + · · · + a k ) for a + · · · + a k ≤ . n 2008 Baruah, Cooper and Hirschhorn [BCH] proved that(1.2) t ( a , a , . . . , a k ; n )= 22 + C ( a , . . . , a k ) ( N ( a , . . . , a k ; 8 n + 8) − N ( a , . . . , a k ; 2 n + 2))for a + · · · + a k = 8 . Ramanujan’s theta functions ϕ ( q ) and ψ ( q ) are defined by ϕ ( q ) = ∞ X n = −∞ q n = 1 + 2 ∞ X n =1 q n and ψ ( q ) = ∞ X n =0 q n ( n +1) / ( | q | < . It is evident that for positive integers a , . . . , a k and | q | < ∞ X n =0 N ( a , . . . , a k ; n ) q n = ϕ ( q a ) · · · ϕ ( q a k ) , (1.3) ∞ X n =0 t ( a , . . . , a k ; n ) q n = 2 k ψ ( q a ) · · · ψ ( q a k ) . (1.4)There are many identities involving ϕ ( q ) and ψ ( q ). From [BCH, Lemma 4.1] or [Be] weknow that for | q | < ψ ( q ) = ϕ ( q ) ψ ( q ) , (1.5) ϕ ( q ) = ϕ ( q ) + 2 qψ ( q ) = ϕ ( q ) + 2 q ψ ( q ) + 2 qψ ( q ) , (1.6) ϕ ( q ) = ϕ ( q ) + 4 qψ ( q ) = ϕ ( q ) + 4 q ψ ( q ) + 4 qψ ( q ) , (1.7) ψ ( q ) ψ ( q ) = ϕ ( q ) ψ ( q ) + qϕ ( q ) ψ ( q ) . (1.8)By [S1, Lemma 2.4],(1.9) ϕ ( q ) = ϕ ( q ) + 4 q ψ ( q ) + 4 q ψ ( q ) + 4 qϕ ( q ) ψ ( q ) + 8 q ψ ( q ) ψ ( q ) . By [S1, Lemma 2.3], for | q | < ϕ ( q ) ϕ ( q ) = ϕ ( q ) ϕ ( q ) + 4 q ψ ( q ) ψ ( q ) + 2 qϕ ( q ) ψ ( q ) + 2 q ϕ ( q ) ψ ( q )+ 6 q ψ ( q ) ψ ( q ) + 4 q ψ ( q ) ψ ( q ) + 4 q ψ ( q ) ψ ( q ) . Let a, b, c, d, n ∈ Z + . From 1859 to 1866 Liouville made about 90 conjectures on N ( a, b, c, d ; n ) in a series of papers. Most conjectures of Liouville have been proved. SeeCooper’s survey paper [C], Dickson’s historical comments [D] and Williams’ book [W].Recently, some connections between t ( a, b, c, d ; n ) and N ( a, b, c, d ; 8 n + a + b + c + d ) havebeen found. See [ACH,BCH,S1,S3,WS2]. More recently Yao [Y] confirmed some con-jectures posed by the author in [S1]. We also note that the evaluations of t ( a, b, c, d ; n )( a + b + c + d ≥
8) have been given for some special values of ( a, b, c, d ). In [C] Cooper deter-mined t ( a, b, c, d ; n ) for ( a, b, c, d ) = (1 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , , , , t ( a, b, c, d ; n ) for ( a, b, c, d ) = (1 , , , , (1 , , , , (1 , , , , (1 , , , , , , , , , , (1 , , , , , , , , , , , , t ( a, b, c, d ; n ) for ( a, b, c, d ) = (1 , , , , , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , t ( a, b, c, d ; n ) for ( a, b, c, d ) = (1 , , , , (1 , , , , (1 , , , a , t ( a, a, a, b ; 4 n + 3 a ) = 4 t ( a, a, a, b ; n ) , t ( a, a, a, b ; 4 n + 3 a ) = 2 t ( a, a, a, b ; n ) ,t ( a, a, a, b ; 2 n ) = t ( a, a, a, b ; n ) , t ( a, a, a, b ; 2 n + a ) = 2 t ( a, a, a, b ; n ) . In Section 2, using theta function identities we establish new general results for t ( a, b, c, d ; n ). Let a, b, n ∈ Z + and k ∈ N . We show that for odd integers a and b , t ( a, a, a, b ; n ) = 12 N ( a, a, a, b ; 8 n + 5 a + 2 b ) ,t ( a, a, a, b ; n ) = N (3 a, a, a, b ; 8 n + 7 a + 2 b ) ,t ( a, a, a, b ; n ) = 2 N ( a, a, a, b ; 8 n + 4 a + b ) for a ≡ − b (mod 4) ,t (2 a, a, a, b ; n ) = 13 N ( a, a, a, b ; 32 n + 28 a + 4 b ) for a ≡ − b (mod 4) ,t ( a, a, a, b ; n ) = 13 N ( a, a, a, b ; 32 n + 52 a + 4 b ) for a ≡ b (mod 4) ,t ( a, a, b, b ; n ) = 4 N ( a, a, b, b ; 2 n + a + b ) for n ≡ ( a − b ) / . For a, b ∈ Z + let ( a, b ) be the greatest common divisor of a and b . For an odd prime p and a ∈ Z let ( ap ) be the Legendre symbol. Suppose that a, b, n ∈ Z + , ( a, b ) = 1, b , − a (mod 4) and there is an odd prime divisor p of b such that ( a (8 n +9 a ) p ) = −
1. We provethat t ( a, a, a, b ; n ) = 12 N ( a, a, a, b ; 8 n + 9 a + b ) . In Section 3, using theta function identities we establish 31 transformation formulas for t ( a, b, c, d ; n ). As typical examples, for a, b, c, n ∈ Z + with 2 ∤ a we have t ( a, a, b, c ; 2 n + a ) = 2 t ( a, a, b, c ; n ) ,t ( a, a, b, c ; 4 n + 3 a ) = 2 t (3 a, a, b, c ; n ) ,t ( a, a, b, c ; 4 n + 6 a ) = 2 t ( a, a, b, c ; n ) ,t ( a, a, b, c ; 2 n + a ) = t ( a, a, b, c ; n ) ,t (3 a, a, b, c ; 2 n + 3 a ) = t ( a, a, b, c ; n ) ,t ( a, a, b, c ; 2 n ) = t (3 a, a, b, c ; n ) ,t ( a, a, a, b ; 8 n + 6 a ) = 4 t (2 a, a, a, b ; n ) ,t (2 a, a, a, b ; 8 n + 12 a ) = 4 t ( a, a, a, b ; n ) . In Section 4, we completely determine t (2 , , , n ), t (1 , , , n ) and t (1 , , , n ) forany positive integer n . In Section 5 we prove some special relations between t ( a, b, c, d ; n )and N ( a, b, c, d ; n ), and pose many challenging conjectures based on calculations on Maple.
2. New general formulas for t ( a, b, c, d ; n ) In this section we present several general formulas for t ( a, b, c, d ; n ), which were found bycalculations on Maple and proved by using Ramanujan’s theta functions. heorem 2.1. Let a, b ∈ Z + with ∤ ab . For n = 0 , , , . . . we have t ( a, a, a, b ; n ) = 12 N ( a, a, a, b ; 8 n + 5 a + 2 b ) . Proof. By (1.6) and (1.9), ∞ X n =0 N ( a, a, a, b ; n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q b )= ( ϕ ( q a ) + 4 q a ψ ( q a ) + 4 q a ψ ( q a ) + 4 q a ϕ ( q a ) ψ ( q a ) + 8 q a ψ ( q a ) ψ ( q a )) × ( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q b ) + 2 q b ψ ( q b )) . Collecting the terms of the form q n +5 a +2 b we get ∞ X n =0 N ( a, a, a, b ; 8 n + 5 a + 2 b ) q n +5 a +2 b = 8 q a ψ ( q a ) ψ ( q a ) · ϕ ( q a ) · q b ψ ( q b )+ 4 q a ϕ ( q a ) ψ ( q a ) · q a ψ ( q a ) · q b ψ ( q b )= 32 q a +2 b ψ ( q a ) ψ ( q b ) ϕ ( q a ) ψ ( q a ) = 32 q a +2 b ψ ( q a ) ψ ( q a ) ψ ( q b ) . Therefore, ∞ X n =0 N ( a, a, a, b ; 8 n + 5 a + 2 b ) q n = 32 ψ ( q a ) ψ ( q b ) ϕ ( q a ) ψ ( q a ) = 2 ∞ X n =0 t ( a, a, a, b ; n ) q n . Hence the result follows.
Theorem 2.2.
Let a, b ∈ Z + with ab ≡ − . For n ∈ Z + we have t ( a, a, a, b ; n ) = 2 N ( a, a, a, b ; 8 n + 4 a + b ) . Proof. It is clear that ∞ X n =0 N ( a, a, a, b ; n ) q n = ϕ ( q a ) ϕ ( q b ) ϕ ( q a ) ϕ ( q a )= ( ϕ ( q a ) + 2 q a ψ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q b ) + 2 q b ψ ( q b ) + 2 q b ψ ( q b )) × ( ϕ ( q a ) + 2 q a ψ ( q a )) ϕ ( q a ) . Thus, ∞ X n =0 N ( a, a, a, b ; 8 n + 4 a + b ) q n +4 a + b = 2 q b ψ ( q b )( ϕ ( q a ) · q a ψ ( q a ) + 2 q a ψ ( q a ) · ϕ ( q a )) ϕ ( q a ) nd so ∞ X n =0 N ( a, a, a, b ; 8 n + 4 a + b ) q n = 8 ψ ( q b ) ϕ ( q a ) ψ ( q a ) ϕ ( q a ) = 8 ψ ( q b ) ψ ( q a ) ϕ ( q a ) = 8 ψ ( q a ) ψ ( q a ) ψ ( q b )= 12 ∞ X n =0 t ( a, a, a, b ; n ) q n . This yields the result.
Corollary 2.1.
Suppose a, b, n ∈ Z + and ab ≡ − . Then N ( a, a, a, b ; 8 n + 4 a + b ) = N ( a, a, a, b ; 8 n + 4 a + b ) − N ( a, a, a, b ; 8 n + 4 a + b ) . Proof. From [S3, Theorem 3.1] we know that t ( a, a, a, b ; n ) = 23 ( N ( a, a, a, b ; 8 n + 4 a + b ) − N ( a, a, a, b ; 8 n + 4 a + b )) . This together with Theorem 2.2 yields the result.
Corollary 2.2.
Suppose that n ∈ { , , , . . . } . Then N (1 , , ,
8; 8 n + 7) = 15 N (1 , , ,
3; 8 n + 7) . Proof. By Theorem 2.2, t (1 , , , n ) = 2 N (1 , , ,
3; 8 n +7). By [ACH], t (1 , , , n ) = N (1 , , ,
3; 8 n + 7). Thus the result follows. Theorem 2.3.
Let a, b ∈ Z + with ∤ ab . For n ∈ Z + we have t ( a, a, a, b ; n ) = N (3 a, a, a, b ; 8 n + 7 a + 2 b ) . Proof. Clearly, ∞ X n =0 N (3 a, a, a, b ; n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q b )= ( ϕ ( q a ) + 4 q a ψ ( q a ) + 4 q a ψ ( q a ) )( ϕ ( q b ) + 2 q b ψ ( q b )) ϕ ( q a ) . Thus, ∞ X n =0 N (3 a, a, a, b ; 4 n + 3 a + 2 b ) q n +3 a +2 b = 4 q a ψ ( q a ) · q b ψ ( q b ) · ϕ ( q a ) = 8 q a +2 b ϕ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 N (3 a, a, a, b ; 4 n + 3 a + 2 b ) q n = 8 ϕ ( q a ) ψ ( q a ) ψ ( q b ) = 8 ϕ ( q a ) ϕ ( q a ) ψ ( q a ) ψ ( q b ) . pplying (1.6) and (1.8) we see that ∞ X n =0 N (3 a, a, a, b ; 4 n + 3 a + 2 b ) q n = 8( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b )= 8( ϕ ( q a ) ϕ ( q a ) + 4 q a ψ ( q a ) ψ ( q a ) + 2 q a ψ ( q a ) ψ ( q a )) ψ ( q a ) ψ ( q b ) . Hence ∞ X n =0 N (3 a, a, a, b ; 4(2 n + a ) + 3 a + 2 b ) q n + a = 16 q a ψ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 N (3 a, a, a, b ; 8 n + 7 a + 2 b ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) = ∞ X n =0 t ( a, a, a, b ; n ) q n , which gives the result. Corollary 2.3.
Suppose n ∈ Z + . Then N (3 , , ,
6; 8 n + 13) = 2 N (1 , , ,
24; 8 n + 13) = 25 N (1 , , ,
6; 8 n + 13) . Proof. By Theorem 2.3, t (1 , , , n ) = N (3 , , ,
6; 8 n + 13). By Theorem 2.2, t (1 , , , n ) = 2 N (1 , , ,
24; 8 n +13). By [S1, Theorem 2.10], t (1 , , , n ) = N (1 , , , n + 13). Thus, the result follows. Corollary 2.4.
Suppose a, b, n ∈ Z + and ∤ ab . Then N (3 a, a, a, b ; 8 n + 7 a + 2 b ) = 12 N (3 a, a, a, b ; 8 n + 7 a + 2 b ) , (i) N ( a, a, a, b ; 8 n + 7 a + 2 b )(ii) = N ( a, a, a, b ; 8 n + 7 a + 2 b ) − N (3 a, a, a, b ; 8 n + 7 a + 2 b ) ,N ( a, a, a, b ; 8 n + 7 a + 2 b )(iii) = N ( a, a, a, b ; 8 n + 7 a + 2 b ) − N (3 a, a, a, b ; 8 n + 7 a + 2 b ) . Proof. By [S3, Corollary 4.2], t ( a, a, a, b ; n ) = 2 N (4 a, a, a, b ; 8 n + 7 a + 2 b ).This together with Theorem 2.3 proves (i). By [S3,Theorem 3.1] and Theorem 2.3, N ( a, a, a, b ; 8 n + 7 a + 2 b ) − N ( a, a, a, b ; 8 n + 7 a + 2 b )= 2 t ( a, a, a, b ; n ) = 2 N (3 a, a, a, b ; 8 n + 7 a + 2 b ) . This yields (ii). By [S3, Theorem 4.5] and Theorem 2.3,23 ( N ( a, a, a, b ; 8 n + 7 a + 2 b ) − N ( a, a, a, b ; 8 n + 7 a + 2 b ))= t ( a, a, a, b ; n ) = N (3 a, a, a, b ; 8 n + 7 a + 2 b ) , hich yields part(iii). Hence the theorem is proved. Theorem 2.4.
Let a, b, n ∈ Z + with ∤ ab . For n ∈ Z + with n ≡ a − b (mod 2) wehave t ( a, a, b, b ; n ) = 4 N ( a, a, b, b ; 2 n + a + b ) . Proof. Clearly, ∞ X n =0 t ( a, a, b, b ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q b )= 16( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ))( ϕ ( q b ) ψ ( q b ) + q b ϕ ( q b ) ψ ( q b )) . Thus, ∞ X n =0 t ( a, a, b, b ; 2 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b ) + 16 q a + b ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b )and ∞ X n =0 t ( a, a, b, b ; 2 n + 1) q n +1 = 16 q a ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b ) + 16 q b ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b ) . Therefore,(2.1) ∞ X n =0 t ( a, a, b, b ; 2 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b ) + 16 q a + b ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b )and(2.2) ∞ X n =0 t ( a, a, b, b ; 2 n + 1) q n = 16 q a − ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b ) + 16 q b − ϕ ( q a ) ψ ( q a ) ϕ ( q b ) ψ ( q b ) . On the other hand, ∞ X n =0 N ( a, a, b, b ; n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q b ) ϕ ( q b )= ( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q a ) + 2 q a ψ ( q a )) × ( ϕ ( q b ) + 2 q b ψ ( q b ))( ϕ ( q b ) + 2 q b ψ ( q b )) . Collecting the terms of the form q n +2 yields for a ≡ − b (mod 4), ∞ X n =0 N ( a, a, b, b ; 4 n + 2) q n +2 q a ψ ( q a ) · ϕ ( q a ) · ϕ ( q b ) · q b ψ ( q b )+ 2 q b ψ ( q b ) · ϕ ( q b ) · ϕ ( q a ) · q a ψ ( q a )= 4 q a + b +2 ( q b − ϕ ( q b ) ψ ( q b ) ψ ( q a ) ϕ ( q a ) + q a − ϕ ( q a ) ψ ( q a ) ψ ( q b ) ϕ ( q b )) . Hence(2.3) ∞ X n =0 N ( a, a, b, b ; 4 n + 2) q n = 4 q a + b ( q b − ϕ ( q b ) ψ ( q b ) ψ ( q a ) ϕ ( q a ) + q a − ϕ ( q a ) ψ ( q a ) ψ ( q b ) ϕ ( q b )) . Similarly, for a ≡ b (mod 4) we have ∞ X n =0 N ( a, a, b, b ; 4 n + 2) q n +2 = 2 q a ψ ( q a ) · ϕ ( q a ) · q b ψ ( q b ) · ϕ ( q b ) + ϕ ( q a ) · q a ψ ( q a ) · ϕ ( q b ) · q b ψ ( q b )= 4 q a + b ( ϕ ( q a ) ϕ ( q b ) ψ ( q a ) ψ ( q b ) + q a +2 b ϕ ( q a ) ϕ ( q b ) ψ ( q a ) ψ ( q b ))and so(2.4) ∞ X n =0 N ( a, a, b, b ; 4 n + 2) q n = 4 q a + b − ( ϕ ( q a ) ϕ ( q b ) ψ ( q a ) ψ ( q b ) + q a + b ϕ ( q a ) ϕ ( q b ) ψ ( q a ) ψ ( q b )) . For a ≡ − b (mod 4) combining (2.2) with (2.3) gives ∞ X n =0 N ( a, a, b, b ; 4 n + 2) q n = 416 q a + b ∞ X n =0 t ( a, a, b, b ; 2 n + 1) q n and so t ( a, a, b, b ; 2 n + 1) = 4 N ( a, a, b, b ; 4 n + 2 + a + b ) . For a ≡ b (mod 4) combining (2.1) with (2.4) gives ∞ X n =0 N ( a, a, b, b ; 4 n + 2) q n = 416 q a + b − ∞ X n =0 t ( a, a, b, b ; 2 n ) q n and so t ( a, a, b, b ; 2 n ) = 4 N ( a, a, b, b ; 4 n + a + b ) . This completes the proof.
Theorem 2.5.
Let a, b ∈ Z + with ab ≡ − . For n ∈ Z + we have t (2 a, a, a, b ; n ) = 13 N ( a, a, a, b ; 32 n + 28 a + 4 b ) . Proof. By (1.10), ∞ X n =0 N ( a, a, a, b ; n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q a ) ϕ ( q b ) ( ϕ ( q a ) ϕ ( q a ) + 4 q a ψ ( q a ) ψ ( q a ) + 2 q a ϕ ( q a ) ψ ( q a ) + 2 q a ϕ ( q a ) ψ ( q a )+ 6 q a ψ ( q a ) ψ ( q a ) + 4 q a ψ ( q a ) ψ ( q a ) + 4 q a ψ ( q a ) ψ ( q a )) × ϕ ( q a )( ϕ ( q b ) + 2 q b ψ ( q b )) . Collecting the terms of the form q n yields ∞ X n =0 N ( a, a, a, b ; 8 n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q b ) + 4 q a ϕ ( q a ) ψ ( q a ) ψ ( q a ) ϕ ( q b )+ 12 q a +4 b ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 N ( a, a, a, b ; 8 n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q b ) + 4 q a ϕ ( q a ) ψ ( q a ) ψ ( q a ) ϕ ( q b )+ 12 q ( a + b ) / ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b )= ϕ ( q a ) ϕ ( q a ) ϕ ( q b ) + 4 q a ϕ ( q a ) ψ ( q a ) ψ ( q a ) ϕ ( q b )+ 12 q ( a + b ) / ϕ ( q a ) ψ ( q b )( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) . Therefore ∞ X n =0 N ( a, a, a, b ; 8(2 n + 1)) q n +1 = 12 q ( a + b ) / a ϕ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 N ( a, a, a, b ; 8(2 n + 1)) q n = 12 q ( a + b ) / a − / ϕ ( q a ) ψ ( q a ) ψ ( q b )= 12 q ( a + b ) / a − / ( ϕ ( q a ) + 4 q a ψ ( q a ) ) ψ ( q a ) ψ ( q b ) . This yields ∞ X n =0 N ( a, a, a, b ; 16 n +12 a +4 b ) q n = 12 ϕ ( q a ) ψ ( q a ) ψ ( q b )+48 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) . Hence ∞ X n =0 N ( a, a, a, b ; 32 n + 12 a + b ) q n = 12 ϕ ( q a ) ψ ( q a ) ψ ( q b ) , ∞ X n =0 N ( a, a, a, b ; 16(2 n + 1) + 12 a + 4 b ) q n +1 = 48 q a ψ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 N ( a, a, a, b ; 32 n + 12 a + 4 b ) q n = 12 ϕ ( q a ) ψ ( q a ) ψ ( q b ) , X n =0 N ( a, a, a, b ; 16(2 n + 1) + 12 a + 4 b ) q n = 48 q ( a − / ψ ( q a ) ψ ( q a ) ψ ( q b ) . Thus, ∞ X n =0 N ( a, a, a, b ; 32 n + 28 a + 4 b ) q n = 48 ψ ( q a ) ψ ( q a ) ψ ( q b ) . This yields N ( a, a, a, b ; 32 n + 28 a + 4 b ) = 3 t (2 a, a, a, b ; n ) . Putting all the above together proves the theorem.
Theorem 2.6.
Let a, b ∈ Z + with ab ≡ . For n ∈ Z + we have t ( a, a, a, b ; n ) = 13 N ( a, a, a, b ; 32 n + 52 a + 4 b ) . Proof. By (1.6), ∞ X n =0 N ( a, a, a, b ; n ) q n = ϕ ( q a ) ϕ ( q a ) ϕ ( q a ) ϕ ( q b )= ( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q a ) + 2 q a ψ ( q a )) ϕ ( q a ) ϕ ( q b )= ( ϕ ( q a ) ϕ ( q a ) + 4 q a ψ ( q a ) ψ ( q a ) + 2 q a ( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ))) × ϕ ( q a ) ϕ ( q b ) . Thus, ∞ X n =0 N ( a, a, a, b ; 4 n ) q n = ( ϕ ( q a ) ϕ ( q a ) + 4 q a ψ ( q a ) ψ ( q a )) ϕ ( q a ) ϕ ( q b ) . Appealing to (1.6) and (1.8) we see that ∞ X n =0 N ( a, a, a, b ; 4 n ) q n = ( ϕ ( q a ) ϕ ( q a ) + 4 q a ψ ( q a ) ψ ( q a )) ϕ ( q a ) ϕ ( q b )= ( ϕ ( q a ) ϕ ( q a ) + 4 q a ψ ( q a ) ψ ( q a ) + 6 q a ( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ))) × ϕ ( q a )( ϕ ( q b ) + 2 q b ψ ( q b )) . Therefore, ∞ X n =0 N ( a, a, a, b ; 4(4 n + a + b )) q n + a + b = 6 q a ϕ ( q a ) ψ ( q a ) · ϕ ( q a ) · q b ψ ( q b )and so ∞ X n =0 N ( a, a, a, b ; 4(4 n + a + b )) q n = 12 ϕ ( q a ) ψ ( q a ) ψ ( q b ) = 12( ϕ ( q a ) + 4 q a ψ ( q a ) ) ψ ( q a ) ψ ( q b ) . t then follows that ∞ X n =0 N ( a, a, a, b ; 4(4(2 n + 3 a ) + a + b )) q n +3 a = 48 q a ψ ( q a ) ψ ( q a ) ψ ( q b )and therefore ∞ X n =0 N ( a, a, a, b ; 4(4(2 n +3 a )+ a + b )) q n = 48 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 3 ∞ X n =0 t ( a, a, a, b ; n ) q n . This yields the result.For n ∈ { , , , . . . } let r ( n ) = N (1 , , n ) = |{ ( x, y, z ) ∈ Z (cid:12)(cid:12) n = x + y + z }| . Theorem 2.7.
For n ∈ Z + we have t (1 , , n ) − N (1 , ,
4; 8 n + 9)= t (1 , , n ) − r (8 n + 9)= ( ( − m +12 m if n + 9 = m for some m ∈ Z + , otherwise. Proof. Since ∞ X n =0 t (1 , , n ) q n = 8 ψ ( q ) ψ ( q ) = 8 ϕ ( q ) ψ ( q ) ψ ( q )= 8( ϕ ( q ) + 2 qψ ( q )) ψ ( q ) ψ ( q ) , we see that ∞ X n =0 t (1 , ,
8; 2 n + 1) q n +1 = 16 qψ ( q ) ψ ( q ) and so ∞ X n =0 t (1 , ,
8; 2 n + 1) q n = 16 ψ ( q ) ψ ( q ) = 2 ∞ X n =0 t (1 , , n ) q n . Thus, t (1 , , n ) = 12 t (1 , ,
8; 2 n + 1) . By [S2, Theorem 3.1], t (1 , ,
8; 2 n + 1) − r (8 n + 9) = ( − m +12 m if 8 n + 9 = m for some m ∈ Z + ,0 otherwise.Thus,(2.5) t (1 , , n ) − r (8 n + 9) = ( ( − m +12 m if 8 n + 9 = m for some m ∈ Z + ,0 otherwise. n the other hand, ∞ X n =0 N (1 , , n ) q n = ϕ ( q ) ϕ ( q ) = ( ϕ ( q ) + 2 q ψ ( q ) + 2 qψ ( q ))( ϕ ( q ) + 4 q ψ ( q ) ) . Thus, ∞ X n =0 N (1 , ,
4; 8 n + 1) q n +1 = 2 qϕ ( q ) ψ ( q )and so ∞ X n =0 N (1 , ,
4; 8 n + 1) q n = 2 ϕ ( q ) ψ ( q ) . Hence(2.6) ∞ X n =0 N (1 , ,
4; 8 n + 9) q n = 2 ϕ ( q ) ψ ( q ) − q . By [S2, (2.13)], ∞ X n =0 r (4 n + 1) q n = 6 ϕ ( q ) ψ ( q ) = 6 ϕ ( q ) ψ ( q ) = 6( ϕ ( q ) + 4 qψ ( q ) ) ψ ( q ) . Hence ∞ X n =0 r (8 n + 1) q n = 6 ϕ ( q ) ψ ( q )and so ∞ X n =0 r (8 n + 1) q n = 6 ϕ ( q ) ψ ( q ) . It then follows that(2.7) ∞ X n =0 r (8 n + 9) q n = 6 ϕ ( q ) ψ ( q ) − q . Combining (2.6) with (2.7) gives r (8 n + 9) = 3 N (1 , ,
4; 8 n + 9). Now putting all theabove together proves the theorem. Corollary 2.5.
Let n ∈ Z + with n ≡ . Then t (1 , , n ) = 12 N (1 , ,
4; 8 n + 9) . Theorem 2.8.
Suppose a, b, n ∈ Z + , ( a, b ) = 1 and there is an odd prime divisor p of b such that ( a (8 n +9 a ) p ) = − . Then t ( a, a, a, b ; n ) = 12 ( N ( a, a, a, b ; 8 n + 9 a + b ) − N ( a, a, a, b ; 8 n + 9 a + b )) . Proof. By Theorem 2.7, t ( a, a, a, b ; n ) X w ∈ Z t ( a, a, a ; n − bw ( w − /
2) = X w ∈ Z t (cid:16) , , n − bw ( w − / a (cid:17) = 12 X w ∈ Z N (cid:16) , ,
4; 8 n − bw ( w − / a + 9 (cid:17) = 12 X w ∈ Z N ( a, a, a ; 8 n − bw ( w −
1) + 9 a )= 12 X w ∈ Z N ( a, a, a ; 8 n + 9 a + b − b (2 w − )= 12 X w ∈ Z ( N ( a, a, a ; 8 n + 9 a + b − bw ) − N ( a, a, a ; 8 n + 9 a + b − b (2 w ) )= 12 ( N ( a, a, a, b ; 8 n + 9 a + b ) − N ( a, a, a, b ; 8 n + 9 a + b )) . This proves the theorem.
Theorem 2.9.
Suppose a, b, n ∈ Z + , ( a, b ) = 1 , b , − a (mod 4) and there is anodd prime divisor p of b such that ( a (8 n +9 a ) p ) = − . Then t ( a, a, a, b ; n ) = 12 N ( a, a, a, b ; 8 n + 9 a + b ) . Proof. Suppose 8 n + 9 a + b = ax + 4 ay + 4 az + 4 bw for some x, y, z, w ∈ Z . Then ax ≡ a + b ≡ a + b (mod 4) and so b ≡ a ( x − ≡ , − a (mod 4). But, b , − a (mod 4). Hence, N ( a, a, a, b ; 8 n + 9 a + b ) = 0. Now the result follows from Theorem2.8. Corollary 2.6.
Suppose a, b, n ∈ Z + , ∤ a , | b , b , a (mod 4) and | n − a .Then t ( a, a, a, b ; n ) = 12 N ( a, a, a, b ; 8 n + 9 a + b ) . Corollary 2.7.
Suppose m, n ∈ Z + , m ≡ , and n ≡ , . Then t (1 , , , m ; n ) = 12 N (1 , , , m ; 8 n + 5 m + 9) . Theorem 2.10.
Suppose a, b, n ∈ Z + , ( a, b ) = 1 and there is an odd prime divisor p of b such that ( a (4 n +5 a ) p ) = − . Then t ( a, a, a, b ; n ) = 12 ( N ( a, a, a, b ; 8 n + 10 a + b ) − N ( a, a, a, b ; 8 n + 10 a + b )) . Proof. By [S2, Theorem 3.1], t ( a, a, a, b ; n )= X w ∈ Z t ( a, a, a ; n − bw ( w − /
2) = X w ∈ Z t (cid:16) , , n − bw ( w − / a (cid:17) = 12 X w ∈ Z N (cid:16) , ,
8; 8 n − bw ( w − / a + 10 (cid:17) X w ∈ Z N ( a, a, a ; 8 n − bw ( w −
1) + 10 a )= 12 X w ∈ Z N ( a, a, a ; 8 n + 10 a + b − b (2 w − )= 12 X w ∈ Z ( N ( a, a, a ; 8 n + 10 a + b − bw ) − N ( a, a, a ; 8 n + 10 a + b − b (2 w ) )= 12 ( N ( a, a, a, b ; 8 n + 10 a + b ) − N ( a, a, a, b ; 8 n + 10 a + b )) . This proves the theorem.
Theorem 2.11.
Suppose a, b, n ∈ Z + , ( a, b ) = 1 and there is an odd prime divisor p of b such that ( a (4 n +5 a ) p ) = − . Assume that a is even or ab ≡ , , for odd a .Then t ( a, a, a, b ; n ) = 12 N ( a, a, a, b ; 8 n + 10 a + b ) . Proof. Suppose 8 n + 10 a + b = ax + ay + 8 az + 4 bw for some x, y, z, w ∈ Z . Then a ( x + y ) ≡ a + b (mod 4). If 2 | a , then 2 ∤ b and so a ( x + y ) a + b (mod 4). Weget a contradiction. If 4 | ab −
1, then a ≡ b (mod 4) and so a ( x + y ) ≡ a (mod 4).Hence x + y ≡ t ≡ , t ∈ Z . If2 ∤ a and ab ≡ | b and so a ( x + y ) ≡ a + b (mod 8). This yields x + y ≡ a + ab ≡ ab + 2 ≡ t ≡ , , t ∈ Z we see that x + y N ( a, a, a, b ; 8 n +10 a + b ) = 0.Now the result follows from Theorem 2.10. Remark 2.1
Theorem 2.11 is a generalization of [S2, Theorem 3.2].
3. New transformation formulas for t ( a, b, c, d ; n ) In this section we present 31 transformation formulas for t ( a, b, c, d ; n ). Theorem 3.1.
Let a, b, c ∈ Z + and ∤ a . For n ∈ Z + we have t ( a, a, b, c ; 2 n + a ) = 2 t ( a, a, b, c ; n ) , (3.1) t ( a, a, a, b ; 2 n + a ) = t ( a, a, a, b ; n ) , (3.2) t ( a, a, a, b ; 2 n ) = t (2 a, a, a, b ; n ) . (3.3)Proof. Since ∞ X n =0 t ( a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q b ) ψ ( q c )= 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c )we see that ∞ X n =0 t ( a, a, b, c ; 2 n + a ) q n + a = 32 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c ) nd so ∞ X n =0 t ( a, a, b, c ; 2 n + a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) = 2 ∞ X n =0 t ( a, a, b, c ; n ) q n , which yields t ( a, a, b, c ; 2 n + a ) = 2 t ( a, a, b, c ; n ).It is clearly that ∞ X n =0 t ( a, a, a, b ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q a ) ψ ( q b ) . Thus, ∞ X n =0 t ( a, a, a, b ; 2 n + a ) q n + a = 16 q a ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16 q a ψ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 t ( a, a, a, b ; 2 n + a ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) = ∞ X n =0 t ( a, a, a, b ; n ) q n . This yields t ( a, a, a, b ; 2 n + a ) = t ( a, a, a, b ; n ).Note that ∞ X n =0 t ( a, a, a, b ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q a ) ψ ( q b ) . We have ∞ X n =0 t ( a, a, a, b ; 2 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16 ψ ( q a ) ψ ( q a ) ψ ( q b )and so ∞ X n =0 t ( a, a, a, b ; 2 n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) = ∞ X n =0 t (2 a, a, a, b ; n ) q n , which gives t ( a, a, a, b ; 2 n ) = t (2 a, a, a, b ; n ). This completes the proof. Theorem 3.2.
Suppose a, b, c ∈ Z + and ∤ a . For n ∈ Z + we have t ( a, a, b, c ; 4 n + 3 a ) = 2 t (3 a, a, b, c ; n ) , (3.4) t ( a, a, b, c ; 4 n + 6 a ) = 2 t ( a, a, b, c ; n ) , (3.5) t ( a, a, a, b ; 4 n ) = t ( a, a, a, b ; n ) , (3.6) t ( a, a, a, b ; 4 n + a ) = t (2 a, a, a, b ; n ) , (3.7) t (2 a, a, a, b ; 4 n + 3 a ) = 2 t (2 a, a, a, b ; n ) . (3.8) roof. It is clear that for | q | < ∞ X n =0 t ( a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) = 16( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c )= 16(( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) + q a ( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a )) ψ ( q b ) ψ ( q c ) . Thus, ∞ X n =0 t ( a, a, b, c ; 4 n + 6 a ) q n +6 a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 4 n + 3 a ) q n +3 a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, a ; 4 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q a ) , ∞ X n =0 t ( a, a, b, a ; 4 n + a ) q n + a = 16 q a ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q a )and so ∞ X n =0 t ( a, a, b, c ; 4 n + 6 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) = 2 ∞ X n =0 t ( a, a, b, c ; n ) q n , ∞ X n =0 t ( a, a, b, c ; 4 n + 3 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) = 2 ∞ X n =0 t (3 a, a, b, c ; n ) q n , ∞ X n =0 t ( a, a, b, a ; 4 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) , ∞ X n =0 t ( a, a, b, a ; 4 n + a ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q a ) = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) . This yields (3.4)-(3.7). To prove (3.8), appealing to Theorem 3.1 we see that t (2 a, a, a, b ; 4 n + 3 a ) = 2 t ( a, a, a, b ; 2 n ) = 2 t (2 a, a, a, b ; n ) . Theorem 3.3.
Suppose a, b ∈ Z + with ∤ a . For n ∈ Z + we have t (2 a, a, a, b ; 8 n + 9 a ) = 4 t (3 a, a, a, b ; n ) , (3.9) t ( a, a, a, b ; 8 n + 13 a ) = 4 t ( a, a, a, b ; n ) , (3.10) t ( a, a, a, b ; 8 n + 4 a ) = 2 t ( a, a, a, b ; n ) , (3.11) t ( a, a, a, b ; 8 n + 6 a ) = 4 t (2 a, a, a, b ; n ) , (3.12) t (2 a, a, a, b ; 8 n + 12 a ) = 4 t ( a, a, a, b ; n ) , (3.13) t (2 a, a, a, b ; 8 n + 6 a ) = 2 t ( a, a, a, b ; n ) . (3.14) roof. By Theorems 3.1 and 3.2, t (2 a, a, a, b ; 8 n + 9 a ) = 2 t ( a, a, a, b ; 4 n + 3 a ) = 4 t (3 a, a, a, b ; n ) ,t ( a, a, a, b ; 8 n + 13 a ) = 2 t ( a, a, a, b ; 4 n + 6 a ) = 4 t ( a, a, a, b ; n ) . This proves (3.9) and (3.10). Since ψ ( q ) = ϕ ( q ) ψ ( q ) and ϕ ( q ) = ϕ ( q ) + 2 qψ ( q ) we seethat ∞ X n =0 t ( a, a, a, b ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16 ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b )= 16( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b )= 16( ϕ ( q a ) + 2 q a ψ ( q a ) + 2 q a ψ ( q a )) × ( ϕ ( q a ) ψ ( q a ) + 2 q a ψ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + 2 q a ψ ( q a ) ψ ( q a )) ψ ( q b ) . Thus, ∞ X n =0 t ( a, a, a, b ; 8 n + 6 a ) q n +6 a = 16 ψ ( q b )( ϕ ( q a ) · q a ψ ( q a ) ψ ( q a ) + 2 q a ψ ( q a ) · q a ϕ ( q a ) ψ ( q a ))= 64 q a ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 64 q a ψ ( q a ) ψ ( q a ) ψ ( q b )and ∞ X n =0 t ( a, a, a, b ; 8 n + 4 a ) q n +4 a = 16 · q a ( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q a ) ψ ( q b )= 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) . It then follows that ∞ X n =0 t ( a, a, a, b ; 8 n + 6 a ) q n = 64 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 4 ∞ X n =0 t (2 a, a, a, b ; n ) q n , ∞ X n =0 t ( a, a, a, b ; 8 n + 4 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 2 ∞ X n =0 t ( a, a, a, b ; n ) q n , which yields (3.11) and (3.12). By (1.5), (1.6) and (1.8), ∞ X n =0 t (2 a, a, a, b ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16 ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b )= 16( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b ) . Thus ∞ X n =0 t (2 a, a, a, b ; 4 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) , X n =0 t (2 a, a, a, b ; 4 n + 2 a ) q n +2 a = 16 q a ϕ ( q a ) ψ ( q b ) ϕ ( q a ) ψ ( q a )and so ∞ X n =0 t (2 a, a, a, b ; 4 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b )= 16( ϕ ( q a ) + 4 q a ψ ( q a ) ) ψ ( q a ) ψ ( q b ) , ∞ X n =0 t (2 a, a, a, b ; 4 n + 2 a ) q n = 16 ϕ ( q a ) ψ ( q b ) ϕ ( q a ) ψ ( q a )= 16( ϕ ( q a ) + 2 q a ψ ( q a ))( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) . Hence, ∞ X n =0 t (2 a, a, a, b ; 4(2 n + 3 a )) q n +3 a = 64 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) , ∞ X n =0 t (2 a, a, a, b ; 4(2 n + a ) + 2 a ) q n + a = 32 q a ( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q a ) ψ ( q b ) = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) . This yields ∞ X n =0 t (2 a, a, a, b ; 8 n + 12 a ) q n = 64 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 4 ∞ X n =0 t ( a, a, a, b ; n ) q n ∞ X n =0 t (2 a, a, a, b ; 8 n + 6 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) = 2 ∞ X n =0 t ( a, a, a, b ; n ) q n , which yields (3.13) and (3.14). The proof is now complete. Theorem 3.4.
Suppose a, b, c ∈ Z + with ∤ a . For n ∈ Z + we have t ( a, a, b, c ; 2 n + a ) = t ( a, a, b, c ; n ) , (3.15) t ( a, a, b, c ; 8 n + 10 a ) = 2 t (4 a, a, b, c ; n ) , (3.16) t ( a, a, b, c ; 8 n + 28 a ) = 2 t ( a, a, b, c ; n ) , (3.17) t ( a, a, a, b ; 4 n + 6 a ) = t ( a, a, a, b ; n ) , (3.18) t ( a, a, a, b ; 4 n ) = t (2 a, a, a, b ; n ) . (3.19)Proof. By [Be, p.315],(3.20) ψ ( q ) ψ ( q ) = ψ ( q ) ϕ ( q ) + qψ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . Thus,(3.21) ψ ( q ) ψ ( q ) = ψ ( q ) ϕ ( q ) + q ψ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . ombining (3.20) with (3.21) gives(3.22) ψ ( q ) ψ ( q ) = ψ ( q ) ϕ ( q ) + q ϕ ( q ) ψ ( q ) + qψ ( q ) ϕ ( q )+ q ψ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . Using (3.20) we see that ∞ X n =0 t ( a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )= 16( ψ ( q a ) ϕ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + q a ψ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c )and so ∞ X n =0 t ( a, a, b, c ; 2 n + a ) q n + a = 16 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . This yields ∞ X n =0 t ( a, a, b, c ; 2 n + a ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . Hence (3.15) is true. Applying (3.22) we see that ∞ X n =0 t ( a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )= 16( ψ ( q a ) ϕ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + q a ψ ( q a ) ϕ ( q a )+ q a ψ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c ) . Thus, ∞ X n =0 t ( a, a, b, c ; 4 n ) q n = 16 ψ ( q a ) ϕ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 4 n + 6 a ) q n +6 a = 16 q a ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )and so ∞ X n =0 t ( a, a, b, c ; 4 n ) q n = 16 ψ ( q a ) ϕ ( q a ) ψ ( q b ) ψ ( q c ) , (3.23) ∞ X n =0 t ( a, a, b, c ; 4 n + 6 a ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . (3.24)By (3.24) and (1.6), ∞ X n =0 t ( a, a, b, c ; 4 n + 6 a ) q n = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c ) . Therefore ∞ X n =0 t ( a, a, b, c ; 4(2 n + a ) + 6 a ) q n + a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) nd so ∞ X n =0 t ( a, a, b, c ; 4(2 n + a ) + 6 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . This yields (3.16). Similarly, from (3.23) and (1.6), ∞ X n =0 t ( a, a, b, c ; 4 n ) q n = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c ) . Therefore, ∞ X n =0 t ( a, a, b, c ; 4(2 n + 7 a )) q n +7 a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )and so ∞ X n =0 t ( a, a, b, c ; 4(2 n + 7 a )) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , which gives (3.17). Recall that ϕ ( q ) ψ ( q ) = ψ ( q ) . From (3.23) and (3.24) we see that ∞ X n =0 t ( a, a, a, b ; 4 n ) q n = 16 ψ ( q a ) ϕ ( q a ) ψ ( q a ) ψ ( q b ) = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) , ∞ X n =0 t ( a, a, a, b ; 4 n + 6 a ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q a ) ψ ( q b ) = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) , which yields (3.18) and (3.19). The proof is now complete.From [Be, p.377] we know that for | q | < ψ ( q ) ψ ( q ) = ϕ ( q ) ψ ( q ) + q ψ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . From [XZ, (2.10)] we know that(3.26) ψ ( q ) ψ ( q ) = ψ ( q ) ψ ( q ) + qϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . Thus, ψ ( q ) ψ ( q ) = ψ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . This together with (3.25) yields(3.27) ψ ( q ) ψ ( q ) = ϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) + q ψ ( q ) ψ ( q )+ q ϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . By (3.25), ψ ( q ) ψ ( q ) = ϕ ( q ) ψ ( q ) + q ψ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . Combining this with (3.26) gives(3.28) ψ ( q ) ψ ( q ) = ϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) + q ψ ( q ) ψ ( q )+ qϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ) . heorem 3.5. Suppose a, b, c ∈ Z + with ∤ a . For n ∈ Z + we have t (3 a, a, b, c ; 2 n + 3 a ) = t ( a, a, b, c ; n ) , (3.29) t ( a, a, b, c ; 2 n ) = t (3 a, a, b, c ; n ) , (3.30) t (3 a, a, b, c ; 4 n + 3 a ) = t (3 a, a, b, c ; n ) , (3.31) t ( a, a, b, c ; 4 n + 6 a ) = t ( a, a, b, c ; n ) , (3.32) t (3 a, a, b, c ; 8 n + 18 a ) = 2 t (4 a, a, b, c ; n ) , (3.33) t (3 a, a, b, c ; 8 n + 60 a ) = 2 t ( a, a, b, c ; n ) , (3.34) t (3 a, a, a, b ; 4 n + 14 a ) = t ( a, a, a, b ; n ) , (3.35) t (3 a, a, a, b ; 4 n ) = t (2 a, a, a, b ; n ) , (3.36) t ( a, a, b, c ; 8 n + 15 a ) = 2 t (5 a, a, b, c ; n ) , (3.37) t ( a, a, b, c ; 8 n + 21 a ) = 2 t (3 a, a, b, c ; n ) , (3.38) t ( a, a, a, b ; 4 n + 3 a ) = t (3 a, a, a, b ; n ) , (3.39) t ( a, a, a, b ; 4 n + a ) = t (5 a, a, a, b ; n ) . (3.40)Proof. Using (3.25) and (3.26) we see that ∞ X n =0 t (3 a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )= 16( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + q a ψ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )= 16( ψ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c ) . Hence ∞ X n =0 t (3 a, a, b, c ; 2 n + 3 a ) q n +3 a = 16 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 2 n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . It then follows that ∞ X n =0 t (3 a, a, b, c ; 2 n + 3 a ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 2 n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , which implies (3.29) and (3.30). By (3.29) and (3.30), t (3 a, a, b, c ; 4 n + 3 a ) = t ( a, a, b, c ; 2 n ) = t (3 a, a, b, c ; n ) ,t ( a, a, b, c ; 4 n + 6 a ) = t (3 a, a, b, c ; 2 n + 3 a ) = t ( a, a, b, c ; n ) . hus, (3.31) and (3.32) hold. Appealing to (3.27), ∞ X n =0 t (3 a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )= 16( ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + q a ψ ( q a ) ψ ( q a )+ q a ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c ) . Thus, ∞ X n =0 t (3 a, a, b, c ; 4 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t (3 a, a, b, c ; 4 n + 14 a ) q n +14 a = 16 q a ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , which yields ∞ X n =0 t (3 a, a, b, c ; 4 n ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , (3.41) ∞ X n =0 t (3 a, a, b, c ; 4 n + 14 a ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . (3.42)From (3.42), (3.41) and (1.6) we see that ∞ X n =0 t (3 a, a, b, c ; 4 n + 14 a ) q n = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t (3 a, a, b, c ; 4 n ) q n = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c ) . Thus, ∞ X n =0 t (3 a, a, b, c ; 4(2 n + a ) + 14 a ) q n + a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t (3 a, a, b, c ; 4(2 n + 15 a )) q n +15 a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )and so ∞ X n =0 t (3 a, a, b, c ; 4(2 n + a ) + 14 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t (3 a, a, b, c ; 4(2 n + 15 a )) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , which yields (3.33) and (3.34). Note that ϕ ( q ) ψ ( q ) = ψ ( q ) . Taking c = 2 a in (3.42) and c = 30 a in (3.41) yields (3.35) and (3.36). Using (3.28) we see that ∞ X n =0 t ( a, a, b, c ; n ) q n = 16 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a ) + q a ψ ( q a ) ψ ( q a )+ q a ϕ ( q a ) ψ ( q a ) + q a ϕ ( q a ) ψ ( q a )) ψ ( q b ) ψ ( q c ) . From this it follows that ∞ X n =0 t ( a, a, b, c ; 4 n + a ) q n + a = 16 q a ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 4 n + 3 a ) q n +3 a = 16 q a ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c )and so ∞ X n =0 t ( a, a, b, c ; 4 n + a ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , (3.43) ∞ X n =0 t ( a, a, b, c ; 4 n + 3 a ) q n = 16 ϕ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . (3.44)Hence, applying (1.6) we get ∞ X n =0 t ( a, a, b, c ; 4 n + a ) q n = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 4 n + 3 a ) q n = 16( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) ψ ( q b ) ψ ( q c )and so ∞ X n =0 t ( a, a, b, c ; 4(2 n + 5 a ) + a ) q n +5 a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 4(2 n + 3 a ) + 3 a ) q n +3 a = 32 q a ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) . It then follows that ∞ X n =0 t ( a, a, b, c ; 4(2 n + 5 a ) + a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , ∞ X n =0 t ( a, a, b, c ; 4(2 n + 3 a ) + 3 a ) q n = 32 ψ ( q a ) ψ ( q a ) ψ ( q b ) ψ ( q c ) , which yields (3.37) and (3.38). Finally, taking c = 6 a in (3.44) and c = 10 a in (3.43) andthen applying (1.5) we deduce (3.39) and (3.40). The proof is now complete.
4. Evaluation of t (2 , , , n ) , t (1 , , , n ) and t (1 , , , n ) In this section we determine t (2 , , , n ), t (1 , , , n ) and t (1 , , , n ) for any positiveinteger n . heorem 4.1. Let n be a positive integer. (i) If n + 5 = 3 β n (3 ∤ n ) , then t (1 , , ,
24; 2 n + 1) = t (2 , , ,
8; 2 n + 3) = 4( σ ( n ) − ( − n a (2 n + 5)) , where { a ( n ) } is given by q ∞ Y k =0 (1 − q k )(1 − q k )(1 − q k )(1 − q k ) = ∞ X n =1 a ( n ) q n ( | q | < . (ii) If n + 1 = 2 α β n with ∤ n and ∤ n , then t (1 , , ,
24; 2 n −
2) + t (2 , , ,
8; 2 n ) = 2 α +4 σ ( n ) . Proof. By (3.1), t (1 , , ,
24; 2 n + 1) = 2 t (1 , , , n ) = t (2 , , ,
8; 2 n + 3). By [WS2,Theorem 3.4], if 2 n +5 = 3 β n with 3 ∤ n , then t (1 , , , n ) = 2( σ ( n ) − ( − n a (2 n +5)) . Thus (i) is true.Now we prove (ii). Since ψ ( q ) = ϕ ( q ) ψ ( q ) = ( ϕ ( q ) + 2 qψ ( q )) ψ ( q ) we see that ∞ X n =0 t (1 , , , n ) q n = 16 ψ ( q ) ψ ( q ) ψ ( q ) = (16 ϕ ( q ) + 32 qψ ( q )) ψ ( q ) ψ ( q ) ψ ( q ) . Thus, ∞ X n =0 t (1 , , ,
24; 2 n ) q n = 16 ϕ ( q ) ψ ( q ) ψ ( q ) ψ ( q )and so(4.1) ∞ X n =0 t (1 , , ,
24; 2 n ) q n = 16 ϕ ( q ) ψ ( q ) ψ ( q ) ψ ( q ) . Similarly, ∞ X n =0 t (2 , , , n ) q n = 16 ψ ( q ) ψ ( q ) ψ ( q ) = (16 ϕ ( q ) + 32 q ψ ( q )) ψ ( q ) ψ ( q ) ψ ( q ) . Thus, ∞ X n =0 t (2 , , ,
8; 2 n ) q n = 16 ψ ( q ) ψ ( q ) ψ ( q ) ϕ ( q )and so(4.2) ∞ X n =0 t (2 , , ,
8; 2 n ) q n = 16 ψ ( q ) ψ ( q ) ψ ( q ) ϕ ( q ) . From (4.1) and (4.2) we see that ∞ X n =0 ( t (2 , , ,
8; 2 n ) + t (1 , , ,
24; 2 n − q n ψ ( q ) ψ ( q )( ϕ ( q ) ψ ( q ) + qϕ ( q ) ψ ( q )) = 16 ψ ( q ) ψ ( q ) . Hence(4.3) t (2 , , ,
8; 2 n ) + t (1 , , ,
24; 2 n −
2) = t (1 , , , n ) . By [WS1, Lemma 4.1], if n + 1 = 2 α β n with 2 ∤ n and 3 ∤ n , then(4.4) t (1 , , , n ) = 2 α +4 σ ( n ) . Thus (ii) holds and the proof is complete.
Theorem 4.2.
Let n be a positive integer. (i) If n + 1 = 2 α β n with ∤ n and ∤ n , then t (2 , , ,
8; 4 n + 2) = t (1 , , ,
24; 4 n ) = 2 α +4 σ ( n ) . (ii) If n + 1 = 3 β n (3 ∤ n ) , then t (2 , , ,
8; 4 n ) = 8( σ ( n ) + a (2 n + 1)) ,t (1 , , ,
24; 4 n −
2) = 8( σ ( n ) − a (2 n + 1)) . Proof. If n + 1 = 2 α β n with 2 ∤ n and 3 ∤ n , from Theorem 4.1 we have t (1 , , ,
24; 4 n ) + t (2 , , ,
8; 4 n + 2) = 2 α +5 σ ( n ) . By [S1, Theorem 2.14], t (1 , , ,
24; 4 n ) = 2 α +4 σ ( n ) . Thus, t (2 , , ,
8; 4 n +2) = 2 α +4 σ ( n ) . This proves (i).Now we consider (ii). Suppose 2 n + 1 = 3 β n (3 ∤ n ). We first assume that n is oddand n = 2 m + 1. By (3.13), t (2 , , ,
8; 8 m + 12) = 4 t (1 , , , m ) . This together with [S3,Theorem 4.15] yields t (2 , , ,
8; 8 m + 4) = 4 t (1 , , , m −
1) = 8( σ ( n ) + a (4 m + 3)) . Now, appealing to Theorem 4.1(ii) we get t (1 , , ,
24; 8 m + 2) = 16 σ ( n ) − t (2 , , ,
8; 8 m + 4) = 8( σ ( n ) − a (4 m + 3)) . From now on suppose that n is even and n = 2 m . From (3.12) and [S3, Theorem 4.15]we see that t (1 , , ,
24; 8 m −
2) = 4 t (2 , , , m −
1) = 8( σ ( n ) − a (4 m + 1)) . Now applying Theorem 4.1(ii) gives t (2 , , ,
8; 8 m ) = 16 σ ( n ) − t (1 , , ,
24; 8 m −
2) = 8( σ ( n ) + a (4 m + 1)) . Putting all the above together proves the theorem.
Theorem 4.3.
Let n ∈ Z + . i) If n + 1 = 2 α β n with ∤ n and ∤ n , then t (1 , , ,
8; 2 n ) = 2 α +2 (cid:16) β +1 (cid:16) n (cid:17) + ( − α + β + n − (cid:17) X d | n d (cid:16) d (cid:17) . (ii) If n + 3 = 3 β n with ∤ n , then t (1 , , ,
8; 2 n + 1) = 2 (cid:16) β +1 (cid:16) n (cid:17) + ( − n (cid:17) X d | n d (cid:16) d (cid:17) − X a,b ∈ Z + , ∤ a n +12= a +3 b ( − ( a − / a. Proof. From [S1, Theorem 2.7], t (1 , , ,
8; 2 n ) = t (1 , , , n ) and t (1 , , ,
8; 2 n + 1) = 2 t (1 , , , n ) . By [C, (5.6) and Theorem 5.4], if n + 1 = 2 α β n with 2 ∤ n and 3 ∤ n , then t (1 , , , n ) = 2 α +2 (cid:16) β +1 + ( − α + β (cid:16) − n (cid:17)(cid:17) X d | n n d (cid:16) d (cid:17) = 2 α +2 (cid:16) β +1 + ( − α + β (cid:16) − n (cid:17)(cid:17) X d | n d (cid:16) n /d (cid:17) = 2 α +2 (cid:16) β +1 (cid:16) n (cid:17) + ( − α + β +( n − / (cid:17) X d | n d (cid:16) d (cid:17) . Thus part(i) is true.Let us consider part(ii). By [C, (5.6) and Theorem 5.6], if 2 n + 3 = 3 β n with 3 ∤ n ,then t (1 , , , n ) = (cid:16) β +1 − ( − β (cid:16) − n (cid:17)(cid:17) X d | n n d (cid:16) d (cid:17) − g (2 n + 3) , where { g ( n ) } is given by qψ ( q ) ∞ Y k =1 (1 − q k ) = ∞ X n =1 g ( n ) q n . Using Jacobi’s identity, qψ ( q ) ∞ Y k =1 (1 − q k ) = q (cid:16) ∞ X m =0 q m ( m +1) (cid:17)(cid:16) ∞ X k =0 ( − k (2 k + 1) q k ( k +1) (cid:17) = (cid:16) ∞ X m =0 q m +1) / (cid:17)(cid:16) ∞ X k =0 ( − k (2 k + 1) q (2 k +1) / (cid:17) = ∞ X n =0 (cid:16) X k,m ∈{ , , ,... } n =(2 k +1) +3(2 m +1) ( − k (2 k + 1) (cid:17) q n = ∞ X n =0 (cid:16) X a,b ∈ Z + , ∤ a n = a +3 b ( − ( a − / a (cid:17) q n . ence g ( n ) = X a,b ∈ Z + , ∤ a n = a +3 b ( − ( a − / a. It then follows that t (1 , , ,
8; 2 n + 1) = 2 t (1 , , , n ) = 2 (cid:16) β +1 − ( − β (cid:16) − n (cid:17)(cid:17) X d | n n d (cid:16) d (cid:17) − g (2 n + 3)= 2 (cid:16) β +1 (cid:16) n (cid:17) − ( − β (cid:16) − n (cid:17)(cid:17) X d | n d (cid:16) d (cid:17) − X a,b ∈ Z + , ∤ a n +12= a +3 b ( − ( a − / a. Observe that ( − β (cid:0) − n (cid:1) ≡ β n = 2 n + 3 ≡ ( − n − (mod 4). We then obtain part(ii).Now the proof is complete.
5. Special relations between t ( a, b, c, d ; n ) and N ( a, b, c, d ; n ) In this section, we present some special relations between t ( a, b, c, d ; n ) and N ( a, b, c, d ; n )and pose a lot of related conjectures. Theorem 5.1.
Let n ∈ Z + Then t (1 , , , n ) = 2 N (1 , , ,
8; 2 n + 5) for n ≡ ,t (1 , , , n ) = 2 N (1 , , ,
15; 2 n + 7) for n ≡ ,t (3 , , , n ) = 2 N (3 , , ,
24; 2 n + 11) for n ≡ ,t (3 , , , n ) = 2 N (3 , , ,
40; 2 n + 17) for n ≡ . Proof. From (3.15), (3.31) and (3.32) we see that t (1 , , ,
8; 4 n + 3) = t (1 , , ,
4; 2 n + 1) = t (1 , , , n ) ,t (1 , , ,
15; 4 n + 6) = t (1 , , , n ) ,t (3 , , ,
24; 4 n + 3) = t (3 , , , n ) ,t (3 , , ,
40; 4 n + 3) = t (3 , , , n ) . Now applying Theorem 2.2 yields the result.
Theorem 5.2.
For n ∈ Z + we have t (2 , , , n ) = 2 N (2 , , ,
4; 2 n + 3) for n ≡ , ,t (2 , , , n ) = 2 N (2 , , ,
12; 2 n + 5) for n ≡ , ,t (2 , , , n ) = 4 N (2 , , ,
24; 2 n + 8) for n ≡ ,t (2 , , , n ) = 2 N (2 , , ,
36; 2 n + 11) for n ≡ , ,t (1 , , , n ) = 2 N (1 , , ,
12; 2 n + 5) for n ≡ , ,t (1 , , , n ) = (cid:26) N (1 , , , n + 3) if n ≡ , N (1 , , , n + 3) if n ≡ . roof. By Theorem 2.3 and (3.8), N (2 , , ,
4; 8 n + 9) = t (1 , , , n ) = 12 t (2 , , ,
4; 4 n + 3) . By (1.6) and (1.9), ∞ X n =0 N (2 , , , n ) q n = ϕ ( q ) ϕ ( q ) ϕ ( q ) = ( ϕ ( q ) + 2 q ψ ( q ))( ϕ ( q ) + 2 q ψ ( q )) × ( ϕ ( q ) + 4 q ψ ( q ) + 4 q ψ ( q ) + 4 q ϕ ( q ) ψ ( q ) + 8 q ψ ( q ) ψ ( q )) . Thus ∞ X n =0 N (2 , , ,
4; 8 n + 7) q n +7 = ϕ ( q ) ϕ ( q ) · q ψ ( q ) ψ ( q ) + 2 q ϕ ( q ) ψ ( q ) · q ϕ ( q ) ψ ( q )= 8 q ϕ ( q ) ψ ( q )( ϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q ))= 8 q ϕ ( q ) ψ ( q ) ψ ( q ) ψ ( q )and so(5.1) ∞ X n =0 N (2 , , ,
4; 8 n + 7) q n = 8 ϕ ( q ) ψ ( q ) ψ ( q ) . On the other hand, using (1.5), (1.6) and (1.8) we see that for b ∈ Z + ,(5.2) ∞ X n =0 t (2 , , , b ; n ) q n = 16 ψ ( q ) ψ ( q ) ψ ( q b ) = 16 ϕ ( q ) ψ ( q ) ψ ( q ) ψ ( q b )= 16( ϕ ( q ) + 2 q ψ ( q ))( ϕ ( q ) ψ ( q ) + q ϕ ( q ) ψ ( q )) ψ ( q b ) . Thus ∞ X n =0 t (2 , , , b ; 4 n + 2) q n +2 = 16 q ϕ ( q ) ψ ( q b ) ϕ ( q ) ψ ( q )and so ∞ X n =0 t (2 , , , b ; 4 n + 2) q n = 16 ϕ ( q ) ψ ( q b ) ϕ ( q ) ψ ( q ) = 16 ϕ ( q ) ψ ( q ) ψ ( q b ) . This together with (5.1) yields t (2 , , ,
4; 4 n + 2) = 2 N (2 , , ,
4; 8 n + 7). Therefore t (2 , , , n ) = 2 N (2 , , ,
4; 2 n + 3) for n ≡ , Lemma 5.1.
Suppose a, b ∈ Z + and ab ≡ − . Then ∞ X n =0 N ( a, a, a, b ; 8 n + 5 a ) q n = 24 ϕ ( q b ) ψ ( q a ) ψ ( q a ) X n =0 N ( a, a, a, b ; 8 n + a + 2 b ) q n = 12 ϕ ( q a ) ψ ( q a ) ψ ( q b ) . Proof. It is evident that ∞ X n =0 N ( a, a, a, b ; n ) q n = ϕ ( q a ) ϕ ( q b ) = ( ϕ ( q a ) + 2 q a ψ ( q a ) + 2 q a ψ ( q a )) ( ϕ ( q b ) + 2 q b ψ ( q b ))= (cid:0) ( ϕ ( q a ) + 2 q a ψ ( q a )) + 6 q a ( ϕ ( q a ) + 2 q a ψ ( q a )) ψ ( q a ) + 12 q a ( ϕ ( q a )+ 2 q a ψ ( q a )) ψ ( q a ) + 8 q a ψ ( q a ) (cid:1) ( ϕ ( q b ) + 2 q b ψ ( q b )) . Thus, ∞ X n =0 N ( a, a, a, b ; 8 n + 5 a ) q n +5 a = 6 q a ϕ ( q a ) · q a ψ ( q a ) ψ ( q a ) ϕ ( q b ) , ∞ X n =0 N ( a, a, a, b ; 8 n + a + 2 b ) q n + a +2 b = 3( ϕ ( q a ) + 4 q a ψ ( q a ) ) · q a ψ ( q a ) · q b ψ ( q b ) = 12 q a +2 b ϕ ( q a ) ψ ( q a ) ψ ( q b )and so the result follows. Theorem 5.3.
For n ∈ Z + with n ≡ , , t (1 , , , n ) = 4 N (1 , , , n + 2) = 83 N (1 , , , n + 2) t (3 , , , n ) = 83 N (2 , , , n + 2) . Proof. By (3.1) and [S1, Theorem 2.11], t (1 , , ,
12; 8 n + 3) = 2 t (1 , , ,
6; 4 n + 1) = 4 t (1 , , ,
4; 2 n ) = 4 N (1 , , ,
6; 8 n + 5) ,t (1 , , ,
12; 8 n + 5) = 2 t (1 , , ,
6; 4 n + 2) = 4 N (1 , , ,
6; 8 n + 7) . From [S1, p.283] and Lemma 5.1 we know that ∞ X n =0 t (1 , , ,
6; 4 n + 1) q n = 32 ϕ ( q ) ψ ( q ) ψ ( q ) = 43 ∞ X n =0 N (1 , , ,
6; 8 n + 5) q n , ∞ X n =0 t (1 , , ,
6; 4 n + 2) q n = 16 ϕ ( q ) ψ ( q ) ψ ( q ) = 43 ∞ X n =0 N (1 , , ,
6; 8 n + 7) q n . Hence the formula for t (1 , , , n ) is true. From (5.2) we see that ∞ X n =0 t (2 , , ,
12; 4 n ) q n = 16 ϕ ( q ) ψ ( q ) ψ ( q ) , ∞ X n =0 t (2 , , ,
12; 4 n + 5) q n +5 = 32 q ψ ( q ) ϕ ( q ) ψ ( q ) ψ ( q ) . hus, appealing to Lemma 5.1 we get ∞ X n =0 t (2 , , ,
12; 4 n ) q n = 16 ϕ ( q ) ψ ( q ) ψ ( q ) = 43 ∞ X n =0 N (2 , , ,
3; 8 n + 5) q n , ∞ X n =0 t (2 , , ,
12; 4 n + 5) q n = 32 ϕ ( q ) ψ ( q ) ψ ( q ) = 43 ∞ X n =0 N (2 , , ,
3; 8 n + 15) q n . This together with (3.1) gives t (3 , , ,
6; 8 n + 3) = 2 t (2 , , ,
12; 4 n ) = 83 ∞ X n =0 N (2 , , ,
3; 8 n + 5) ,t (3 , , ,
6; 8 n + 5) = 2 t (2 , , ,
12; 4 n + 1) = 4 t (1 , , ,
12; 2 n − ∞ X n =0 N (2 , , ,
3; 8 n + 15) , which completes the proof. Corollary 5.1.
For n ∈ Z + with n ≡ , , N (1 , , ,
6; 4 n ) = 5 N (1 , , , n ) and N (2 , , ,
3; 4 n ) = 5 N (2 , , , n ) . Proof. From Theorem 5.3 and [S1, Theorem 2.1],83 N (1 , , , n ) = t (1 , , , n −
2) = 23 ( N (1 , , ,
6; 4 n ) − N (1 , , , n )) , N (2 , , , n ) = t (3 , , , n −
2) = 23 ( N (2 , , ,
3; 4 n ) − N (2 , , , n )) . This yields the result.Using similar method one can prove the following results: t (3 , , , n ) = 2 N (3 , , ,
18; 2 n + 7) for n ≡ , , (5.3) t (1 , , , n ) = 4 N (1 , , , n + 3) for n ≡ , , (5.4) t (1 , , , n ) = 4 N (1 , , , n + 4) for n ≡ , , (5.5) t (1 , , , n ) = 4 N (1 , , , n + 4) for n ≡ , , (5.6) t (2 , , , n ) = N (2 , , ,
48; 2 n + 14) for n ≡ , (5.7) t (1 , , , n ) = 8 N (1 , , , n + 3) for n ≡ , (5.8) t (1 , , , n ) = 4 N (1 , , , n + 4) for n ≡ , (5.9) t (1 , , , n ) = 4 N (1 , , , n + 4) for n ≡ , (5.10) t (1 , , , n ) = 8 N (1 , , , n + 3) for n ≡ , (5.11) t (1 , , , n ) = 4 N (1 , , ,
30; 2 n + 10) for n ≡ , (5.12) t (1 , , , n ) = 43 N (1 , , ,
16; 2 n + 6) for n ≡ , (5.13) t (3 , , , n ) = 4 N (3 , , ,
48; 2 n + 16) for n ≡ , (5.14) t (1 , , , n ) = 4 N (1 , , ,
14; 2 n + 6) for n ≡ , (5.15) (2 , , , n ) = 4 N (2 , , ,
24; 2 n + 14) for n ≡ , (5.16) t (5 , , , n ) = 4 N (5 , , ,
8; 2 n + 6) for n ≡ , (5.17) t (1 , , , n ) = 43 N (1 , , ,
48; 2 n + 16) for n ≡ , (5.18) t (2 , , , n ) = 4 N (2 , , , n + 2) for n ≡ , (5.19) t (4 , , , n ) = 4 N (4 , , , n + 4) for n ≡ , (5.20) t (1 , , , n ) = 4 N (1 , , ,
30; 2 n + 12) for n ≡ , (5.21) t (1 , , , n ) = 4 N (1 , , ,
40; 2 n + 18) for n ≡ , (5.22) t (1 , , , n ) = 43 N (1 , , ,
36; 2 n + 14) for n ≡ , (5.23) t (2 , , , n ) = 4 N (2 , , ,
32; 2 n + 10) for n ≡ , (5.24) t (2 , , , n ) = 4 N (2 , , ,
24; 2 n + 10) for n ≡ , (5.25) t (3 , , , n ) = 4 N (3 , , ,
24; 2 n + 10) for n ≡ , (5.26) t (1 , , , n ) = 4 N (1 , , , n + 5) for n ≡ , (5.27) t (2 , , , n ) = 4 N (2 , , ,
48; 2 n + 14) for n ≡ , (5.28) t (1 , , , n ) = 4 N (1 , , , n + 4) for n ≡ , (5.29) t (2 , , , n ) = 4 N (2 , , , n + 2) for n ≡ , (5.30) t (4 , , , n ) = 4 N (4 , , , n + 4) for n ≡ , (5.31) t (1 , , , n ) = 4 N (1 , , ,
16; 2 n + 8) for n ≡ , (5.32) t (5 , , , n ) = 4 N (5 , , ,
40; 2 n + 14) for n ≡ . (5.33)By doing calculations on Maple, we pose the following conjectures. Conjecture 5.1.
Suppose n ∈ Z + . Then t (1 , , , n ) = 8 N (1 , , , n + 3) for n ≡ , t (1 , , , n ) = 8 N (1 , , , n + 4) for n ≡ , t (1 , , , n ) = 1613 N (1 , , ,
27; 2 n + 10) for n ≡ . We remark that Dongxi Ye informed the author he proved the formula for t (1 , , , n )and t (1 , , , n ) by using theta function identities and weight 2 modular forms. Conjecture 5.2.
Let n ∈ Z + . Then t (1 , , , n )= N (1 , , ,
10; 2 n + 4) = 49 N (1 , , ,
10; 8 n + 16) if | n − , N (1 , , ,
10; 2 n + 4) if | n , N (1 , , ,
10; 2 n + 4) if n ≡
10 (mod 16) , N (1 , , ,
10; 2 n + 4) if n ≡ ,
15 (mod 20) . onjecture 5.3. Let n ∈ Z + . Then t (1 , , , n ) = N (1 , , ,
18; 2 n + 6) = 49 N (1 , , ,
18; 8 n + 24) if | n, N (1 , , ,
18; 2 n + 6) if | n − , N (1 , , ,
18; 2 n + 6) if | n − , N (1 , , ,
18; 2 n + 6) if | n − , Conjecture 5.4.
Let n ∈ Z + . Then t (1 , , , n )= N (1 , , ,
30; 2 n + 10) = 49 N (1 , , ,
30; 8 n + 40) if | n , N (1 , , ,
30; 2 n + 10) = 815 N (1 , , ,
30; 8 n + 40) if | n − , N (1 , , ,
30; 2 n + 10) = 1633 N (1 , , ,
30; 8 n + 40) if | n + 1 . Conjecture 5.5.
Let n ∈ Z + . Then t (1 , , , n )= N (1 , , ,
18; 2 n + 10) = 49 N (1 , , ,
18; 8 n + 40) if | n − , N (1 , , ,
18; 2 n + 10) = 815 N (1 , , ,
18; 8 n + 40) if | n − , N (1 , , ,
18; 2 n + 10) = 1633 N (1 , , ,
18; 8 n + 40) if | n − , N (1 , , ,
18; 2 n + 10) = 47 N (1 , , ,
18; 8 n + 40) if | n − , N (1 , , ,
18; 2 n + 10) = 817 N (1 , , ,
18; 8 n + 40) if | n − . Conjecture 5.6.
Let n ∈ Z + . Then t (2 , , , n ) = N (2 , , ,
18; 2 n + 8) = 49 N (2 , , ,
18; 8 n + 32) if | n − , N (2 , , ,
18; 2 n + 8) if | n − , N (2 , , ,
18; 2 n + 8) if | n − , N (2 , , ,
18; 2 n + 8) if | n − , N (2 , , ,
18; 2 n + 8) if | n − , N (2 , , ,
18; 2 n + 8) if | n − . Conjecture 5.7.
Let n ∈ Z + . Then t (2 , , , n ) N (2 , , ,
15; 2 n + 8) = 49 N (2 , , ,
15; 8 n + 32) if | n − , N (2 , , ,
15; 2 n + 8) if | n − , N (2 , , ,
15; 2 n + 8) if | n − , N (2 , , ,
15; 2 n + 8) if n ≡ ,
81 (mod 100)
Conjecture 5.8.
Let n ∈ Z + . Then t (5 , , , n ) = N (5 , , ,
30; 2 n + 14) if | n − , N (5 , , ,
30; 2 n + 14) if | n − , N (5 , , ,
30; 2 n + 14) if | n − . Conjecture 5.9.
Let n ∈ Z + . Then t (1 , , , n ) = 2 N (1 , , ,
12; 2 n + 7) for n ≡ , . Conjecture 5.10.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 4 N (1 , , , n + 7) ,t (1 , , , n ) = 8 N (1 , , , n + 4) ,t (1 , , , n ) = 4 N (1 , , ,
24; 2 n + 12) ,t (1 , , , n ) = 4 N (1 , , ,
33; 2 n + 12) . Conjecture 5.11.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 4 N (1 , , ,
32; 2 n + 10) ,t (1 , , , n ) = 4 N (1 , , ,
24; 2 n + 10) ,t (2 , , , n ) = 4 N (2 , , ,
24; 2 n + 10) . Conjecture 5.12.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 43 N (1 , , ,
8; 2 n + 4) ,t (1 , , , n ) = 4 N (1 , , ,
24; 2 n + 8) . Conjecture 5.13.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 4 N (1 , , ,
9; 2 n + 6) ,t (1 , , , n ) = 4 N (1 , , ,
45; 2 n + 22) ,t (2 , , , n ) = 4 N (2 , , ,
27; 2 n + 14) ,t (2 , , , n ) = 43 N (2 , , ,
15; 2 n + 8) ,t (3 , , , n ) = 4 N (3 , , ,
39; 2 n + 14) . Conjecture 5.14.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 4 N (1 , , , n + 7) , (1 , , , n ) = 4 N (1 , , ,
40; 2 n + 20) ,t (2 , , , n ) = 83 N (2 , , ,
9; 2 n + 4) . Conjecture 5.15.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 4 N (1 , , , n + 6) ,t (1 , , , n ) = 43 N (1 , , ,
40; 2 n + 14) ,t (2 , , , n ) = 4 N (2 , , ,
24; 2 n + 10) . Conjecture 5.16.
Let n ∈ Z + with n ≡ . Then t (1 , , , n ) = 4 N (1 , , ,
24; 2 n + 16) ,t (3 , , , n ) = 4 N (3 , , ,
15; 2 n + 8) ,t (5 , , , n ) = 4 N (5 , , ,
45; 2 n + 16) . Conjecture 5.17.
Let n ∈ Z + with n ≡ . Then t (2 , , , n ) = 8 N (2 , , , n + 5) ,t (1 , , , n ) = 43 N (1 , , ,
24; 2 n + 10) ,t (2 , , , n ) = 4 N (2 , , ,
25; 2 n + 14) ,t (2 , , , n ) = 4 N (2 , , ,
24; 2 n + 14) ,t (3 , , , n ) = 4 N (3 , , ,
8; 2 n + 6) ,t (3 , , , n ) = 4 N (3 , , ,
40; 2 n + 22) . Conjecture 5.18.
Let n ∈ Z + . Then t (1 , , , n ) = 2 N (1 , , ,
25; 2 n + 13) for n ≡ , ,t (1 , , , n ) = 2 N (1 , , ,
21; 2 n + 7) for n ≡ , ,t (1 , , , n ) = 2 N (1 , , ,
15; 2 n + 7) for n ≡ , ,t (2 , , , n ) = 2 N (2 , , ,
5; 2 n + 3) for n ≡ , ,t (3 , , , n ) = 2 N (3 , , ,
35; 2 n + 13) for n ≡ , ,t (3 , , , n ) = 2 N (3 , , ,
15; 2 n + 7) for n ≡ , .t (1 , , , n ) = 25 N (1 , , ,
5; 8 n + 10) for n ≡ , ,t (1 , , , n ) = 25 N (1 , , ,
15; 8 n + 22) for n ≡ , . Conjecture 5.19.
Let n ∈ Z + with n ≡ , , . Then t (1 , , , n ) = 2 N (1 , , ,
7; 2 n + 3) . Conjecture 5.20.
Let n ∈ Z + with n ≡ , , . Then t (1 , , , n ) = 25 N (1 , , ,
7; 8 n + 10) . onjecture 5.21. Let n ∈ Z + . Then t (1 , , , n ) = 2 N (1 , , ,
33; 2 n + 9) for n ≡ , , , ,
10 (mod 11) ,t (1 , , , n ) = 2 N (1 , , ,
33; 2 n + 11) for n ≡ , , , , ,t (1 , , , n ) = 2 N (1 , , ,
33; 2 n + 13) for n ≡ , , , , . Remark 5.1
From [S3, Theorem 4.2] we know that if a, b, c, d, n ∈ Z + and a ≡ b ≡ c ≡ d ≡ ± t ( a, b, c, d ; n ) = N (cid:0) a, b, c, d ; 8 n + a + b + c + d (cid:1) − N ( a, b, c, d ; 2 n + ( a + b + c + d ) / . References [ACH] C. Adiga, S. Cooper and J. H. Han,
A general relation between sums of squaresand sums of triangular numbers , Int. J. Number Theory (2005), 175-182.[BCH] N. D. Baruah, S. Cooper and M. Hirschhorn, Sums of squares and sums of trian-gular numbers induced by partitions of
8, Int. J. Number Theory (2008), 525-538.[Be] B.C. Berndt, Ramanujan’s Notebooks , Part III, Springer, New York, 1991.[C] S. Cooper,
On the number of representations of integers by certain quadratic forms,II , J. Combin. Number Theory (2009), 153-182.[D] L.E. Dickson, History of the Theory of Numbers , Vol. II, Carnegie Institute ofWashington, Washington D.C., 1923. Reprinted by AMS Chelsea, 1999.[S1] Z.H. Sun,
Some relations between t ( a, b, c, d ; n ) and N ( a, b, c, d ; n ), Acta Arith. (2016), 269-289.[S2] Z.H. Sun, Ramanujan’s theta functions and sums of triangular numbers , Int. J.Number Theory (2019), 969-989.[S3] Z.H. Sun, The number of representations of n as a linear combination of triangularnumbers , Int. J. Number Theory (2019), 1191-1218.[XZ] E.X.W. Xia and Y. Zhang, Proofs of some conjectures of Sun on the relationsbetween sums of squares and sums of triangular numbers , Int. J. Number Theory (2019), 189-212.[Y] X.M. Yao, The relations between N ( a, b, c, d ; n ) and t ( a, b, c, d ; n ) and ( p, k ) -parametrization of theta functions , J. Math. Anal. Appl. (2017), 125-143.[WS1] M. Wang and Z.H. Sun, On the number of representations of n as a linear combi-nation of four triangular numbers, Int. J. Number Theory (2016), 1641-1662.[WS2] M. Wang and Z.H. Sun, On the number of representations of n as a linear combi-nation of four triangular numbers II , Int. J. Number Theory (2017), 593-617.[W] K.S. Williams, Number Theory in the Spirit of Liouville , Cambridge Univ. Press,New York, 2011., Cambridge Univ. Press,New York, 2011.