Certain Somos s P Q type Dedekind η-function identities

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1 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 Certain Somos s Q type Dedekind η-function identities B R SRIVATSA KUMAR H C VIDYA Department of Mathematics, Manipal Institute of Technology, Manipal Academy of Higher Education, Manipal , India *Corresponding author. sri_vatsabr@yahoo.com; vidyaashwath@gmail.com MS received 9 July 016; revised 1 February 018; accepted 7 April 018; published online 5 July 018 Abstract. In this paper, we provide a new proof for the Dedekind η-function identities discovered by Somos. During this process, we found two new Dedekind η-function identities. Furthermore, we extract interesting partition identities from some of the η- function identities. Dedekind η-function; theta functions; modular equations; colored par- Keywords. titions. 010 Mathematics Subject Classification. rimary: 11F0, 11B65, 1183; Secondary: 14K5. 1. Introduction Ramanujan [7,15] recorded 3 identities in his unorganized pages that involved the ratio of Dedekind η-function. They are proved by Berndt Zhang [6] by utilizing Ramanujan s modular equations of various degree, mixed modular equations or through the hypothesis of modular forms. The other 14 identities of the comparative sort were found on page 55 of his lost notebook [16] are demonstrated by Berndt [10] employing the above techniques. Berndt Chan [9] Berndt et al. [8] have employed some of the above mentioned Q modular equations for the explicit evaluation of Rogers Ramanujan continued fractions Ramanujan Weber class invariants. Influenced by their work, several new Q η-function identities have been discovered employed to find the explicit evaluation of continued fractions, class invariants ratio of theta functions by numerous mathematicians. For the introduction, one can see [1 4,11,14,0 5]. Influenced by Ramanujan s work, Somos [17] used a computer to discover nearly 600 Dedekind η-function identities. Somos formulated these identities after examining thouss of Dedekind η-function identities computationally. He ran the ARI/G scripts to look at each identity to see if it is equivalent to an identity in Q form without offering the proof. Recently, Yuttanan[7] proved some of these identities established certain interesting partition identities for them. Furthermore, Vasuki Veeresha [6] demonstrated Somos s η-function identities of level 14 Srivatsa Kumar Veeresha [18] have established certain partition identities for the same. Also, Srivatsa Kumar Anuradha [19] have proved Somos s η-function identities of level 10 established certain colored partition identities for them. Indian Academy of Sciences 1

2 4 age of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 Motivated by the above work, in this paper we provide a simple proof of several Dedekind η-function identities discovered by Somos other mathematicians. Section records some preliminary results. In section 3, we prove Q type Dedekind η-function identities. Theorems are Somos s identities. Theorem 3.8 is also due to Vasuki Kahtan [5]. Theorems were due to Adiga et al. [1], Naika Dharmendra [14], Vasuki Srivatsa Kumar [] respectively. In addition, two new identities, Theorem are found. Furthermore, in section 4, by the notion of colored partitions, we are able to extract colored partitions for some of these identities.. reliminaries For q = exp(πiτ), the Dedekind η-function η(τ) is defined for Im(τ) > 0by η(τ) = q 1/4 (1 q n ). n=1 Here throughout this paper, we assume q < 1. As is customary, we define n 1 (a; q) 0 := 1, (a; q) n := (1 aq k ) (a; q) := (1 aq n ). k=0 For ab < 1, Ramanujan s general theta-function f (a, b) is given by f (a, b) := a n(n+1)/ b n(n 1)/. n= Jacobi s triple product identity [5, p. 35] is given by f (a, b) = ( a; ab) ( b; ab) (ab; ab). (.1) The three most important special cases of f (a, b) [5, p. 36] are ϕ(q) := f (q, q) = ψ(q) := f (q, q 3 ) = n= f ( q) := f ( q, q ) = q n = ( q; q ) (q ; q ) = ( q; q) (q; q), (.) q n(n+1)/ = (q ; q ) (q; q ), (.3) n= ( 1) n q n(3n 1)/ = (q; q) := q 1/4 η(τ). (.4)

3 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 3 of 18 4 Also after Ramanujan, define χ(q) := ( q; q ), where the product representation in (.) (.4) follows from (.1). For convenience, we denote f ( q n ) by f n for a positive integer n. It is easy to see that ϕ( q) = f 1, ψ(q) = f, ϕ(q) = f 5 f f 1 f1 f 4, ψ( q) = f 1 f 4, f χ(q) = f, χ( q) = f 1 f (q) = f 3. (.5) f 1 f 4 f f 1 f 4 Ramanujan recorded many modular equations in his notebooks [15,16] without offering the proof. We now define a modular equation as understood by Ramanujan. The complete elliptic integral of the first kind K (k) is defined by π/ dθ K := K (k) := 0 1 k sin θ = π = π ( 1 F 1, 1 ) ; 1; k, ( 1 ) n (n!) kn where 0 < k < 1. The series representation is found by exping the integr in a binomial series integrating termwise. Also F 1 is the basic hypergeometric function defined by F 1 (a, b; c; x) = (a) n (b) n x n, 0 x < 1 (c) n n! where a, b c are complex numbers such that c is not a nonpositive integer with (a) n := a(a + 1)(a + ) (a + n 1) for n 1. The number k is called the modulus of K := K (k) k := 1 k is called the complementary modulus. Let K, K, L L denote the complete elliptic integrals of the first kind associated with moduli k, k, l l respectively. Suppose the equality n K K = L L (.6) holds for some positive integer n. Then a modular equation of degree n is a relation between the moduli k l which is induced by (.6). Ramanujan expressed his modular equations in terms of α β, where α = k β = l. Then we say that β has degree n over α. The corresponding multiplier m is defined by m = K/L.Ifq = exp( π K /K ), one of the fundamental properties of elliptic function affirms that [5, p. 101] ϕ (q) = ( 1 π K (k) = F 1, 1 ) ; 1; k. (.7) We conclude this section by recalling some formulae for ϕ, ψ at different arguments in terms of α, q z := F 1 ( 1, 1 ; 1; α) by using (.7) recorded by Ramanujan [5, pp. 1 14].

4 4 age 4 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 ϕ(q) := z, (.8) ϕ( q) := z(1 α) 1/4, (.9) ϕ( q ) := z(1 α) 1/8, (.10) z ϕ(q ) := (1 + 1 α) 1/, (.11) z ( ψ(q) := αq 1) 1/8, (.1) z ( ψ( q) := α(1 α)q 1) 1/8, (.13) z ψ(q ) := (αq 1 ) 1/4. (.14) Lemma.1. We have ϕ( q)ϕ(q) = ϕ ( q ), (.15) ϕ(q)ψ(q ) = ψ (q), (.16) ϕ (q) + ϕ ( q) = ϕ (q ), (.17) ϕ (q) ϕ ( q) = 8qψ (q 4 ). (.18) The identities (.15) (.18) are due to Ramanujan [15] for proof, see [5]. Lemma.. We have ϕ (q) + ϕ (q 3 ) = ϕ( q )ϕ( q 3 )ϕ( q 6 ), ϕ( q) (.19) ϕ (q) 3ϕ (q 3 ) = ϕ(q3 )ϕ( q ) ϕ(q)ϕ( q 6 ), (.0) ϕ ( q) 5ϕ ( q 5 ) = 4 f ψ (q) 5qψ (q 5 ) = f 1 χ( q 5 ) χ( q) The identities (.19) (.) are due to Vasuki et al. [4]. (.1) χ( q) χ( q 5 ). (.) 3. Somos s Dedekind η-function identities Theorem 3.1. If := ϕ(q) ϕ(q 3 ) Q := ψ(q ) ψ(q 6 )

5 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 5 of 18 4 Q 3q Q = Q + q Q. roof. It can be seen from (.5) that Q = ψ (q) ψ (q 3 ) Q = ψ (q)ψ (q 6 ) ψ (q )ψ (q 3 ). We need to show that { ψ ψ (q 4 (q) ) ψ (q ) ψ4 (q 3 } { ) ψ (q 6 qψ (q 6 ) 3 ψ4 (q 3 } ) ) ψ (q 6 ) + ψ4 (q) ψ (q = 0. ) With the help of (.16), the above can be rewritten as ψ (q ) qψ (q 6 ) 3ϕ (q 3 ) + ϕ (q) ϕ (q) ϕ (q 3 = 0. (3.1) ) Now from [5, p. 3, eq.(5.1)], we have ( (1 β) 3 ) 1/8 = m + 1 ( (1 α) 3 ) 1/8, = 3 m 1 α 1 β m, ( α 3 ) 1/8 = 3 + m ( β 3 ) 1/8 = m 1. (3.) β m α where β has degree 3 over α m is the multiplier. We deduce ϕ (q) + 3ϕ (q 3 ) = 4 ϕ(q3 )ψ 3 (q) ϕ(q)ψ(q 3 ) (3.3) ϕ (q) ϕ (q 3 ) = 4q ψ3 (q 3 )ϕ(q) ϕ(q 3 )ψ(q). (3.4) by using (.8) (.1) in the third fourth terms of (3.). Now (3.1) can be established easily by employing (3.3) (3.4). Theorem 3.. If := f ( q) f (q) Q := f (q ) q 1/ f ( q 8 ), Q+ 4 Q = Q 4 Q.

6 4 age 6 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 roof. Employing (.5), we find that Q = ϕ4 ( q 4 ) q 1/ ψ 4 (q) Q = χ 4 ( q) q1/ χ 4 ( q 4 ). Substituting these in the desired equation, it suffices to prove that χ 4 ( q 4 )ϕ 4 ( q 4 ) { ψ 4 (q)χ 4 ( q 4 ) χ 4 ( q)ϕ 4 ( q 4 ) } 4qχ 4 ( q)ψ 4 (q) { ψ 4 (q)χ 4 ( q 4 ) + χ 4 ( q)ϕ 4 ( q 4 ) } = 0. Also by using (.5), one can easily see that χ 4 ( q)ϕ 4 ( q 4 ) = ψ 4 ( q)χ 4 ( q 4 ). On using this in the above, we deduce χ 4 ( q 4 )ϕ 4 ( q 4 ) 4qχ 4 ( q)ψ 4 (q) ψ4 (q) + ψ 4 ( q) ψ 4 (q) ψ 4 = 0. (3.5) ( q) Multiplying (.17) (.18) byψ (q ) using (.16), we obtain ψ 4 (q) + ψ 4 ( q) = ϕ (q )ψ (q ) (3.6) ψ 4 (q) ψ 4 ( q) = 8qψ (q )ψ (q 4 ). (3.7) Employing (3.6) (3.7) in(3.5) further using (.5), we readily arrive at (3.5). Theorem 3.3. If := ϕ( q) ϕ(q ) Q := ϕ(q) q 1/ ψ(q 4 ), Q+ 8 Q = Q + 4 Q. roof. From (.15) (.16), we note that Q = ϕ ( q ) q 1/ ψ (q ) Q = q1/ ϕ( q)ψ(q4 ) ϕ(q)ϕ(q ). Using these in the required equation employing (.15) (.16), it suffices to prove that { ϕ (q) 4qψ (q 4 ) }{ ϕ ( q) ϕ (q ) } + 4qψ 4 (q ) = 0. (3.8)

7 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 7 of 18 4 Adding (.17) (.18), we obtain ϕ (q) ϕ (q ) = 4qψ (q 4 ). (3.9) Changing q to q in the above, we get ϕ ( q) ϕ (q ) = 4qψ (q 4 ). (3.10) Employing (3.9) (3.10) in(3.8) also by using (.16), we complete the proof. Theorem 3.4. If := χ 3 (q) Q := χ 3 (q 3 ), Q 8q Q = ( ) Q + q ( ). Q roof. Employing (.5) in Q, we obtain Q = ϕ(q)ϕ(q3 ) ψ( q)ψ( q 3 ) Q = ϕ(q)ψ( q3 ) ϕ(q 3 )ψ( q). Using these in the desired equation, it suffices to prove that ϕ 3 (q 3 )ψ( q) qϕ(q)ψ 3 ( q 3 ) ϕ3 (q)ψ( q 3 ) + 8ϕ(q 3 )ψ 3 ( q) ϕ 3 (q)ψ( q 3 ) ϕ(q 3 )ψ 3 = 0. (3.11) ( q) Now from (.8) (.13), we have ψ( q) ϕ(q) ψ 3 ( q 3 ) ϕ 3 (q 3 ) = 1 ( ) α(1 α) 1/8 q = 1 ( ) β(1 β) 3/8, q 3 where α β have degrees 1 3 respectively. Substituting the above two equations in (3.11) simplifying, we complete the proof. Theorem 3.5. If := f (q) f ( q) Q := f (q3 ) f ( q 3 ),

8 4 age 8 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 ( Q 1 ) = Q ( ) Q ( ) Q. roof. From (.5), one can see that We need to show Q = ϕ( q )ϕ( q 6 ) ϕ( q)ϕ( q 3 ) Q = ϕ( q )ϕ( q 3 ) ϕ( q)ϕ( q 6 ). ϕ 3 ( q )ϕ( q 3 ) + ϕ( q 6 )ϕ 3 ( q) = 0. (3.1) From (.9) (.10), we have ϕ( q) = z 1 (1 α) 1/4, ϕ( q ) = z 1 (1 α) 1/8 ϕ( q 3 ) = z 3 (1 β) 1/4, ϕ( q 6 ) = z 3 (1 β) 1/8, (3.13) where β has degree 3 over α. Now using the above identities in (3.1), we deduce z 1 z1 z 3 { ((1 α) 3 1 β ) 1/8 ( (1 α) 3 ) 1/4 } + = 0. 1 β On employing (3.) in the above, we complete the proof. Theorem 3.6. If := f (q) f ( q 4 ) Q := f (q3 ) f ( q 1 ), Q 4q Q = ( ) Q q roof. It follows from (.5) that ( ). Q Q = ψ(q)ψ(q3 ) ψ(q )ψ(q 6 ) Now it suffices to prove that Q = ψ(q)ψ(q6 ) ψ(q )ψ(q 3 ). ψ(q )ψ 3 (q 3 ) qψ(q)ψ 3 (q 6 ) 4 ψ 3 (q ) ψ(q 6 ) ψ3 (q) ψ(q 3 ) ψ 3 (q) ψ(q 3 ) ψ3 (q ) ψ(q 6 ) = 0. (3.14)

9 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 9 of 18 4 Employing (.1) (.14) in(3.14) simplifying, we obtain ψ(q )ψ 3 (q 3 ) qψ(q)ψ 3 (q 6 ) 4 ψ 3 (q ) ψ(q 6 ) ψ3 (q) ψ(q 3 ) ψ 3 (q) ψ(q 3 ) ψ3 (q ) ψ(q 6 ) ( ) α 1/8 = β 3 ( ) α 3 1/8 β ( ) α 3 1/8, (3.15) β where β has degree 3 over α. Finally, using (3.) in(3.15), we obtain (3.14). Theorem 3.7. If := χ( q4 ) χ(q) Q := χ( q1 ) χ(q 3 ), Q+ q Q = roof. From (.5), we note that ( ) Q q ( ). Q Q = ϕ( q4 )ϕ( q 1 ) ψ(q)ψ(q 3 ) We need to prove that Q = ϕ( q4 )ψ(q 3 ) ϕ( q 1 )ψ(q). ψ(q) ϕ( q 1 ) qψ 3 (q 3 ) ϕ( q 4 ) ψ 3 (q) ψ(q 3 ) + ϕ3 ( q 4 ) ϕ( q 1 ) ψ 3 (q) ψ(q 3 ) ϕ3 ( q 4 ) ϕ( q 1 ) Using (.10) (.11) in(.15), we have ϕ( q 4 ) = z(1 α) 1/16 ( 1 + ) 1/4 1 α. = 0. (3.16) Now employing (.1) the above in the left-h side of (3.16), we obtain ψ(q) ϕ( q 1 ) qψ 3 (q 3 ) = ( α ϕ( q 4 ) β 3 ( α 3 β ( 1 α 3 β ψ 3 (q) ψ(q 3 ) + ϕ3 ( q 4 ) ϕ( q 1 ) ψ 3 (q) ψ(q 3 ) ϕ3 ( q 4 ) ϕ( q 1 ) ) 1/8 ( (1 β) 3 ) 1/16 ( (1 + 1 β) 3 1 α β ) 1/8 ( ) 1/16 ( ) 1 β (1 α) + (1+ 1 α) 3 1/ β ) 1/8 ( ) 1/16 ( 1 β (1 α) (1+ 1 α) β ) 1/4 ) 1/4. (3.17)

10 4 age 10 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 Now, from [5, p. 33, eq. (5.)], we have (m 1)(3 + m)3 α = 16m 3, β = (m 1)3 (m + 3). 16m where β has degree 3 over α m is the multiplier. Also it follows that α = 4m m + 16m 3 (m 1)(3 + m) 3 4m m (3.18) β = 4 m + 16m (m 1) 3 (3 + m) 4. (3.19) m With the help of (3.18) (3.19), (3.17) can be rewritten as ψ(q) ϕ( q 1 ) qψ 3 (q 3 ) ϕ( q 4 ) ψ 3 (q) ψ(q 3 ) + ϕ3 ( q 4 ) ϕ( q 1 ) ψ 3 (q) ψ(q 3 ) ϕ3 ( q 4 ) ϕ( q 1 ) ( m m + ) 3 16m (m 1) 3 (3 + m) = m 1 4m m + 16m 3 (m 1)(3 + m) m + 3 m 3+m 3 m ( ((4m m+ 16m 3 (m 1)(3+m) 3 ) 3 1/4 ) 1/4 16m 4 (4m+ 16m (m 1) 3 (3+m)) ) 3 ) 1/4 ( ( (4m m+ 16m 3 (m 1)(3+m) 3 ) 16m 4 (4m+ 16m (m 1) 3 (3+m)) which is valid only for m = 3. We thus complete the proof. Theorem 3.8 [5]. If := ϕ( q) ϕ( q 3 ) Q := ϕ(q) ϕ(q 3 ), Q + Q = 3 Q Q. roof. The above identity is equivalent to (1 + Q ) = 3 Q

11 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 11 of 18 4 Using (.15), (.19) (.0), we observe that Theorem 3.9. If (1 + Q ) = ϕ ( q){ϕ (q) + ϕ (q 3 )} ϕ ( q 3 )ϕ (q 3 ) = ϕ( q)ϕ( q )ϕ( q 6 ) ϕ( q 3 )ϕ(q 3 )ϕ(q 3 ) = ϕ( q)ϕ( q ) ϕ( q 6 )ϕ(q 3 ) = 3 Q. := ψ(q) ϕ( q 8 ) Q := χ( q)χ( q 8 ), (Q) 4q (Q) = Q q ( ). Q roof. From (.5), we observe that Q = f f 8 Q = f 3 f 16 f1 f 8 3. Using these in the desired equation, it suffices to prove that A(q) = ϕ( q) { f ψ (q) f 8 ϕ ( q 8 ) } qψ(q 8 ) { f 8 ϕ ( q 8 ) f ψ (q) } = 0. Employing (.), (.3) (.4) in the above, A(q) = q 4( 6q 8q 6q 3 10q 4 18q 5 + 1q 6 + ). By analytic continuation, A(q) 0asq 0. Hence the proof. Theorem If := f ( q) f ( q ) Q := f ( q5 ) f ( q 10 ), (Q) 4q (Q) = ( ) Q 3 + q ( ) 3. (3.0) Q

12 4 age 1 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 roof. Employing (.5) in(3.0) after simplification, it is sufficient to prove that χ 5 ( q) + 4χ( q 5 ) χ 5 ( q) χ( q 5 ) + χ 5 ( q 5 ) = 0. (3.1) qχ( q) On employing (.5)in(.1) (.), we deduce χ 5 ( q) + 4χ( q 5 ) = 5 f 5 f χ ( q 5 )χ( q) (3.) χ 5 ( q) χ( q 5 ) = 5q f 10 f 1 χ 4 ( q) χ( q 5 ). (3.3) Using (3.) (3.3) in(3.1), we complete the proof. Theorem 3.11 [1]. If := ψ( q) q 1/4 ψ( q 3 ) Q := ϕ(q) ϕ(q 3 ), ( ) Q ( ) Q = (Q) 9 (Q). roof. The desired equation is equivalent to 4 Q 4 = (Q) 4 9. Employing (.5) in the above after simplification, it is sufficient to prove that ψ 4 ( q) qψ 4 ( q 3 ) ϕ4 (q) 9ϕ 4 (q 3 ) ϕ 4 (q 3 ) ϕ 4 = 0. (3.4) (q) Now multiplying (.19) (3.4), we have ϕ 4 (q 3 ) ϕ 4 (q) = 8q ψ3 (q 3 )ϕ(q)ϕ( q )ϕ( q 3 )ϕ( q 6 ) ψ(q)ϕ( q)ϕ(q 3. (3.5) ) Also multiplying (.0) (3.3), we obtain ϕ 4 (q) 9ϕ 4 (q 3 ) = 8 ϕ (q 3 )ϕ 3 ( q )ψ 3 (q) ϕ( q 6 )ϕ (q)ψ(q 3 ). (3.6) Employing (3.5) (3.6) in(3.4) using the identity (.16) repeatedly, we complete the proof.

13 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 13 of 18 4 Theorem 3.1 [14]. If := ϕ( q)ϕ( q ) ϕ(q)ϕ(q ) Q := ϕ( q)ϕ(q ) ϕ(q)ϕ( q ), Q+ 1 Q = Q. roof. It is easy to see that Q = ϕ ( q) ϕ (q) Q = ϕ ( q ) ϕ (q ). Using (.15) (.17), we observe that 1 Q+ = ϕ( q) Q ϕ(q) + ϕ(q) ϕ( q) = ϕ ( q) + ϕ (q) ϕ(q)ϕ( q) = Q. Theorem 3.13 []. If := ϕ( q) ϕ(q) Q := ϕ( q3 ) ϕ(q 3 ), ( ) + Q ( ) Q ( = 4 Q+ 1 ) 6. (3.7) Q roof. For 0 < t < 1, set Then or (1 t) 1/4 = ϕ( q) ϕ(q). (3.8) = (1 α) 1/4 Q = (1 β) 1/4. α = 1 4 β = 1 Q 4. (3.9)

14 4 age 14 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 From Entry 5(ii) of [5, p. 30], we have (αβ) 1/4 +{(1 α)(1 β)} 1/4 = 1, (3.30) where β has degrees 3 over α. Employing (3.9) in the above dividing throughout by (Q), we readily arrive at (3.7). Theorem If := ϕ( q) ϕ(q) Q := ϕ( q7 ) ϕ(q 7 ), ( ) ( ) Q ( ) ( + 8 (Q) 3/ 1 + Q (Q) 3/ + 8 Q+ 1 ) Q ( Q+ ) = 0. (3.31) Q roof. If β has degree 7 over α, we have from Entry 19(i) of [5, p.314] (αβ) 1/8 +{(1 α)(1 β)} 1/8 = 1. (3.3) Now employing (3.8) (3.9) in the above, we obtain (1 4 )(1 Q 4 ) = (1 Q) 8. On simplifying the above dividing throughout by (Q), we easily arrive at (3.31). Theorem If := ψ(q)ψ(q ) ψ( q)ψ( q ) Q := ψ(q)ψ( q ) ψ( q)ψ(q ), Q+ 1 Q = Q. roof. It is easy to see that Q = ψ (q) ψ ( q) Q = ψ (q ) ψ ( q ).

15 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 15 of 18 4 Using (3.6) the identity ψ(q)ψ( q) = ψ(q )ϕ( q ),wehave Q+ 1 Q = ψ4 (q) + ψ 4 ( q) ψ (q)ψ ( q) = ϕ (q ) ϕ ( q ) = ψ4 (q ) ψ 4 ( q ) = Q. 4. Application to the theory of partitions The identities stated in section 3 have application to the theory of colored partitions. Haung [13] Chen Haung [1] introduced the concept of colored partitions to Göllnitz Gordon functions. We demonstrate this by giving combinatorial interpretations for Theorem For simplicity, we adopt the stard notation (q 1, q,...,q n ; q) := n (q j ; q), j=1 define (q r± ; q s ) := (q r, q s r ; q s ), (r < s); r, s N. For example, (q ± ; q 8 ) means (q, q 6 ; q 8 ) which is (q ; q 8 ) (q 6 ; q 8 ). A positive integer n has l colors if there are l copies of n available colors all of them are viewed as distinct objects. artitions of a positive integer into parts with colors are colored partitions. For example, if is allowed to have two colors, say b (black) y (yellow), the colored partitions of 4 are 4, 3 + 1, y + b, b + b, y + y, b , y , An important fact is that, (q a ; q b ) k is the generating function for the number of partitions of n, where all the parts are congruent to a(mod b) having k colors. Theorem 4.1. Let p 1 (n) denote the number of partitions of n into parts congruent to ±, ±4(mod 1) with twelve four colors respectively. Let p (n) denote the number of partitions of n into parts congruent to ±1, ±5(mod 1) with twelve colors each, ±4(mod 1) with four colors. Let p 3 (n) denote the number of partitions of n into parts congruent to ±1, ±5(mod 1) with eight colors each. Let p 4 (n) denote the number of partitions of n into parts congruent to ±1, ±5(mod 1) with four colors each ±, ±4(mod 1) with twelve, eight colors respectively. Then for any positive integer n 1, the following relation holds true:

16 4 age 16 of 18 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 p 1 (n) p (n 1) = p 3 (n) 9p 4 (n 1). roof. Rewriting the products of Theorem 3.11 subject to the common base q 1, we obtain 1 q (q1 ±, q4± 4 ; q1 ) (q1 1±, q4± 4, q5± 1 ; q1 ) 1 = (q8 1±, q5± 8 ; q1 ) 9q (q4 1±, q± 1, q4± 8, q5± 4 ;. q1 ) The quotients of the above identity represent the generating functions for p 1 (n), p (n), p 3 (n) p 4 (n) respectively. Hence the above identity is equivalent to p 1 (n)q n q p (n)q n = p 3 (n)q n 9q p 4 (n)q n, where we set p 1 (0) = p (0) = p 3 (0) = p 4 (0) = 1. Now equating the coefficients of q n on both sides, we lead to the desired result. The following table verifies the case for n = in the above theorem. p 1 () = 1 : r, w, b, g, y, o, m, p, gr, bl, br, i p (1) = 1 : 1 r, 1 w, 1 b, 1 g, 1 y, 1 o, 1 m, 1 p, 1 gr, 1 bl, 1 br, 1 i p 3 () = 36 : 1 r + 1 r, 1 w + 1 w, 1 b + 1 b, 1 g + 1 g, 1 y + 1 y, 1 o + 1 o, 1 m + 1 m, 1 p + 1 p, 1 r + 1 w, 1 r + 1 b, 1 r + 1 g, 1 r + 1 y, 1 r + 1 o, 1 r + 1 m, 1 r + 1 p, 1 w + 1 b, 1 w + 1 g, 1 w + 1 y, 1 w + 1 o, 1 w + 1 m, 1 w + 1 p, 1 b + 1 g, 1 b + 1 y, 1 b + 1 o, 1 b + 1 m, 1 b + 1 p, 1 g + 1 y, 1 g + 1 o, 1 g + 1 m, 1 g + 1 p, 1 y + 1 o, 1 y + 1 m, 1 y + 1 p, 1 o + 1 m, 1 o + 1 p, 1 m + 1 p. p 4 (1) = 4 : 1 r, 1 w, 1 b, 1 g Theorem 4.. Let p 1 (n) denote the number of partitions of n into parts congruent to ±1, ±3, +4(mod 8) with eight, eight eighteen colors respectively. Let p (n) denote the number of partitions of n into parts congruent to ±, +4(mod 8) with four eighteen colors respectively. Let p 3 (n) denote the number of partitions of n into parts congruent to ±1, ±, ±3(mod 8) with four, ten four colors respectively. Then for any positive integer n 0, the following equality holds true: p 1 (n) + p (n) = p 3 (n). roof. On rewriting the products of Theorem 3.15 subject to the common base q 8, we obtain

17 roc. Indian Acad. Sci. (Math. Sci.) (018) 18:4 age 17 of (q8 1±, q3± 8, q4+ 18 ; + q8 ) 1 (q4 ±, q4+ 18 ; = q8 ) (q4 1±, q± 10, q3± 4 ;. q8 ) The quotients of the above identity represent the generating functions for p 1 (n), p (n) p 3 (n) respectively. Hence the above identity is equivalent to p 1 (n)q n + p (n)q n = p 3 (n)q n, where we set p 1 (0) = p (0) = p 3 (0) = 1. Equating the coefficients of q n in the above, we lead to the desired result. The following table verifies the case for n = in the above theorem. p 1 () = 36 : 1 r + 1 r, 1 w + 1 w, 1 b + 1 b, 1 g + 1 g, 1 y + 1 y, 1 o + 1 o, 1 m + 1 m, 1 p + 1 p, 1 r + 1 w, 1 r + 1 b, 1 r + 1 g, 1 r + 1 y, 1 r + 1 o, 1 r + 1 m, 1 r + 1 p, 1 w + 1 b, 1 w + 1 g, 1 w + 1 y, 1 w + 1 o, 1 w + 1 m, 1 w + 1 p, 1 b + 1 g, 1 b + 1 y, 1 b + 1 o, 1 b + 1 m, 1 b + 1 p, 1 g + 1 y, 1 g + 1 o, 1 g + 1 m, 1 g + 1 p, 1 y + 1 o, 1 y + 1 m, 1 y + 1 p, 1 o + 1 m, 1 o + 1 p, 1 m + 1 p. p () = 4 : r, w, b, g. p 3 () = 0 : 1 r + 1 r, 1 w + 1 w, 1 b + 1 b, 1 g + 1 g, 1 r + 1 w, 1 r + 1 b, 1 r + 1 g, 1 w + 1 b, 1 w + 1 g, 1 b + 1 g, r, w, g, b, y, o, m, bl, p, i Acknowledgements The authors are extremely grateful to the referee for numerous suggestions which have significantly improved the presentation of this work. The authors would also like to thank Dr. K R Vasuki, Department of Studies in Mathematics, Manasa Gangothri, University of Mysore, Mysore, India for his guidance during the preparation of this paper. The research of the first author is partially supported under Extra Mural Research Funding by SERB, a statutory body of DST, Government of India (File No. EMR/016/001601). References [1] Adiga C, Vasuki K R Naika M S M, Some new evaluations of Ramanujan s cubic continued fractions, New Zeal J. Math., 31 (00) [] Adiga C, Kim T, Naika M S M Madhusudan H S, On Ramanujan s cubic continued fractions explicit evaluations of theta functions, Indian J. ure Appl. Math., 35 (004) [3] Baruah N D, On some of Ramanujans identities for eta functions, Indian J. Math., 43 (000) [4] Baruah N D Saikia N, Some general theorems on the explicit evaluations of Ramanujans cubic continued fraction, J. Computational Appl. Math., 160 (003) [5] Berndt B C, Ramanujan s Notebooks, art III (1991) (New York: Springer) [6] Berndt B C Zhang L C, Ramanujan s identities for eta functions, Math. Ann., 9 (199) [7] Berndt B C, Ramanujan s Notebooks, art IV (1994) (New York: Springer) [8] Berndt B C, Chan H H, Zhang L-C, Ramanujan s class invariants cubic continued fraction, Acta Arith., 73 (1995) 67 85

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