Analogues of Ramanujan s 24 squares formula

Transcription

1 International Journal of Number Theory Vol., No. 5 (24) 99 9 c World Scientific Publishing Company DOI:.42/S Analogues of Ramanujan s 24 squares formula Faruk Uygul Department of Mathematics American University of Sharjah Sharjah, UAE fuygul@aus.edu Kenneth S. Williams School of Mathematics and Statistics Carleton University, Ottawa, ON, Canada KS 5B6 kwilliam@connect.carleton.ca Received 29 July 23 Accepted 7 November 23 Published 2 February 24 Ramanujan s formula for the number r 24 (n) of representations of a positive integer n as a sum of 24 squares asserts that r 24 (n) = 6 σ (n) 32 σ (n/2) σ (n/4) where ( )n τ(n) τ(n/2), 8X d >< if k N, d N σ (k) = d k >: if k Q\N, and Ramanujan s function τ(n) isgivenby Y X q ( q n ) 24 = τ(n)q n, q C, q <. n= n= In this paper, we determine a class of formulae analogous to Ramanujan s formula for r 24 (n). Keywords: Ramanujan s 24 squares formula; sums of squares and triangular numbers; Ramanujan s tau function; sum of divisors function; Eisenstein series; infinite products; Dedekind eta function; modular forms. Mathematics Subject Classification 2: E25, F27 99

2 F. Uygul & K. S. Williams. Notation Let N, N = N {}, Z and C denote the sets of positive integers, nonnegative integers, integers and complex numbers, respectively. Throughout this paper, q denotes a complex variable with q <. We now define the functions that we need. For k N the infinite product E k is defined by E k := n N( q kn ). (.) Ramanujan s theta functions ϕ and ψ are defined by ϕ(q) :=+2 q n2 = q n2 (.2) n N n Z and ψ(q) := q n(n+)/2, (.3) n N see [2, p. 6]. By Jacobi s triple product identity [2, p. ] we have and ϕ(q) =E 2 E5 2 E 2 4 (.4) ψ(q) =E E2 2. (.5) Ramanujan s tau function τ(n) (n N) isgivenby qe 24 = n N τ(n)q n, (.6) see [5, Eq. (92), p. 5]. The first five values of τ are τ() =,τ(2) = 24,τ(3) = 252,τ(4) = 472 and τ(5) = 483. The Bernoulli numbers B n (n N )are defined by t e t = t n B n, t C, t < 2π, (.7) n! n= so that B =,B = 2, B 2 = 6, B 4 = 3, B 6 = 42, B 8 = 3, B = 5 66, B 2 = 273, B 4 = 7 6,... and B 2n+ = (n N). The classical normalized Eisenstein series E 2k (q) (k N) is defined by E 2k (q) := 4k B 2k n= n 2k q n q n. (.8) The sum of the powers of divisors function σ k (n) is defined for k N and n N by σ k (n) := d k. (.9) d N d n

3 Analogues of Ramanujan s 24 squares formula For m N we set σ k (m) :=. By taking k =6in(.8), we obtain (as B 2 = E 2 (q) = ) σ (n)q n. (.) n= Finally, if f(q) = n N f n q n, we define [f(q)] n := f n,n N.. Introduction Ramanujan s famous formula [5, p. 62] for the number r 24 (n) ofrepresentations of a positive integer n as a sum of 24 squares asserts that r 24 (n) = 6 σ (n) 32 σ (n/2) σ (n/4) ( )n τ(n) τ(n/2), (.) where the sum of divisors function σ (n) was defined in (.9) and Ramanujan s tau function in (.6). Ramanujan [3; 5, p. 53] conjectured and Mordell [] proved that τ satisfies τ(kn) =τ(k)τ(n) k τ(n/k) (.2) for all positive integers n and all primes k, seealso[6, p. 297]. Taking k =2in(.2) we have so that τ(2n) = 24τ(n) 248τ(n/2) (.3) ( ) n τ(n) =τ(n)+48τ(n/2) + 496τ(n/4). (.4) Using (.4) in(.) we obtain Ramanujan s formula in the equivalent form r 24 (n) = 6 σ (n) 32 σ (n/2) σ (n/4) τ(n) τ(n/2) τ(n/4). (.5) Multiplying both sides of (.5) byq n, and summing over n N, we obtain (as r 24 () = ) r 24 (n)q n = E 2(q) 2 E 2(q 2 )+ 496 E 2(q 4 ) where n= (q) (q 2 ) (q 4 ), (.6) (q) := n= τ(n)q n (= qe 24 ). (.7)

4 2 F. Uygul & K. S. Williams Hence, as n= by (.4), we have deduced the identity r 24 (n)q n = ϕ 24 (q) =E 48 E 2 2 E 48 4 (.8) ( q n ) 48 ( q 2n ) 2 ( q 4n ) 48 n= = E 2(q) 2 E 2(q 2 )+ 496 E 2(q 4 ) (q) (q 2 ) (q 4 ) (.9) from Ramanujan s 24 squares formula. Conversely, using (.8), (.) and(.7) in(.9), and then equating coefficients of q n (n N), we recover Ramanujan s 24 squares theorem in the form (.5). It is our aim in this paper to determine all identities like (.9). In fact, for reasons that will be made clear in Sec. 2, we determine all identities of a slightly more general type than (.9). We define Ω(q) :=( (q) 2 (q 4 )) /3 = q 3 n= ( q n ) 8 ( q 4n ) 6 = ω(n)q n (.) (so that ω() = ω(2) =, ω(3) = ) and prove that there are precisely 28 identities of the type q r n= ( q n ) a ( q 2n ) a2 ( q 4n ) a4 = A E 2 (q)+a 2 E 2 (q 2 )+A 4 E 2 (q 4 ) n= + B (q)+b 2 (q 2 )+B 4 (q 4 )+CΩ(q), (.) where r, a,a 2,a 4 are integers and A,A 2,A 4,B,B 2,B 4,C are rational numbers. We prove in Sec. 2 the following result. Theorem.. The only integers r, a,a 2,a 4 and rational numbers A,A 2,A 4, B,B 2,B 4,C satisfying (.) are given in Table. We now begin our examination of the identities in Theorem.. Eachidentity is identified by the number in the left-hand column of Table. Identity is precisely formula (.9) and so is equivalent to Ramanujan s 24 squares formula. Thus the identities of Theorem. are analogues of Ramanujan s 24 squares formula. Theorem. shows that there are exactly formulae like (.9), namely those with C =, that is, identities, 5, 6, 3, 4, 5, 6, 23, 24, 28.

5 Analogues of Ramanujan s 24 squares formula 3 Table. 28 identities of Theorem.. No. a a2 a4 r A A2 A4 B B2 B4 C (Continued)

6 4 F. Uygul & K. S. Williams Table. (Continued) No. a a2 a4 r A A2 A4 B B2 B4 C

7 Analogues of Ramanujan s 24 squares formula 5 Since ( ( q) n )= ( q n ) ( q 2n ) 3 ( q 4n ), (.2) n= n= we see that the mapping q q transforms the left-hand side of (.) as follows: q r n= ( q n ) a ( q 2n ) a2 ( q 4n ) a4 ( ) r q r n= ( q n ) a ( q 2n ) 3a+a2 ( q 4n ) a+a4. (.3) To determine the effect of q q on the right-hand side of (.), we must determine E 2 ( q), ( q) andω( q). Proposition.. (i) E 2 ( q) = E 2 (q) + 498E 2 (q 2 ) 496E 2 (q 4 ), (ii) ( q) = (q) 48 (q 2 ) 496 (q 4 ), (iii) Ω( q) = Ω(q) 6 (q 4 ). Proof. (i) By (.) wehave The identity E 2 (q)+e 2 ( q) = σ (2n)q 2n. (.4) n= σ (2n) = 249σ (n) 248σ (n/2), n N, (.5) is easily proved. Using (.5) in(.4), and appealing to(.), we obtain σ (2n)q 2n = n= Putting (.6) into(.4), we deduce E 2(q ) 6552 E 2(q 4 ) (.6) E 2 (q)+e 2 ( q) = 498E 2 (q 2 ) 496E 2 (q 4 ) from which (i) follows. (ii) From (.7) and(.4) weobtain ( q) = ( ) n τ(n)q n = (τ(n)+48τ(n/2) + 496τ(n/4))q n n= n= from which (ii) follows. (iii) By [, Eq. (3.), p. 242] we have E 24 2 = E6 E8 4 +6qE8 E6 4. (.7)

8 6 F. Uygul & K. S. Williams Multiplying (.7) by q 3 E 8 E8 4, we obtain the identity q 3 E 8 E24 2 E8 4 = q3 E 8 E6 4 6q4 E4 24. (.8) By (.) and(.7) the right-hand side of (.8) is Ω(q) 6 (q 4 ). By (.2) and (.) the left-hand side of (.8) is q 3 (E E3 2 E 4 )8 E4 6 = Ω( q). This completes the proof of (iii). From (.3) and Proposition. we see that the identity (.) with parameters r, a,a 2,a 4,A,A 2,A 4,B,B 2,B 4,C transforms under q q into the identity (.) with parameters where r,a,a 2,a 4,A,A 2,A 4,B,B 2,B 4,C, r = r, a = a,a 2 =3a + a 2,a 4 = a + a 4, A =( ) r+ A,A 2 =( ) r ( 498A + A 2 ), A 4 =( ) r ( 496A + A 4 ),B =( ) r+ B, B 2 =( ) r ( 48B + B 2 ), B 4 =( )r ( 496B + B 4 6C),C =( ) r+ C. Thus the identities 7, 8, 9, 2, 2, 22, 23, 24, 25, 26, 27, 28 in Theorem. are obtained from the identities,, 2, 7, 8, 9, 5, 6, 3, 4, 2, respectively by changing q to q, and thus are not really new. Hence, we need only to consider the identities 6 in Theorem.. First we observe that the identities 6,, 4 and 5 are trivial. Identity 6 is qe 24 E2 72 E 24 4 = (q) + 48 (q 2 ) (q 4 ). This follows by mapping q q in (q) =qe 24 andappealingto(.2) and Proposition.(ii). Identity is q 3 E 8 E24 2 E4 8 = 6 (q 4 )+Ω(q). This follows by mapping q q in Ω(q) =q 3 EE and appealing to (.2) and Proposition.(iii). Identities 4 and 5 are q 4 E 24 4 = (q 4 ), q 2 E 24 2 = (q 2 ), respectively, which follow from (q) = qe 24 by mapping q q 4 and q q 2 respectively. Further, identities 3 and 6 result from mapping q q 2 in identities 5 and 28 respectively. Thus we are left to consider the identities 5, 7 and 2. Of these identities we find that two are known (identities and 5) and the remaining eight are new (see Theorems.2.9).

9 Analogues of Ramanujan s 24 squares formula 7 By considering the first few terms in the expansion of E b Eb2 2 Eb4 4 Eb8 8 Eb6 (b,b 2,b 4,b 8,b 6 Z) inpowersofq, it is easy to check that E b Eb2 2 Eb4 4 Eb8 8 Eb6 6 = ifandonlyifb = b 2 = b 4 = b 8 = b 6 =. Using this result, together with (.4) and(.5), it is easy to prove that E a Ea2 2 Ea4 4 = ϕ t (q)ϕ u (q 2 )ϕ v (q 4 )ψ w (q)ψ x (q 2 )ψ y (q 4 )ψ z (q 8 ) for some integers t, u, v, w, x, y, z if and only if 4a +2a 2 + a 4 =. (.9) The condition (.9) is satisfied by identities, 2, 3, 5, 7 and but not by identities 4, 8, 9 and 2. If (.9) is satisfied, then Thus, as ϕ(q)ψ 2 (q)ψ(q 2 )=, we have for any integer t. E a Ea2 2 Ea4 4 = ψ a (q)ψ 2a a2 (q 2 ). (.2) E a Ea2 2 Ea4 4 = ϕ t (q)ψ 2t a (q)ψ t 2a a2 (q 2 ) (.2) Identity. As we have already mentioned this identity is formula (.9). Equating the coefficients of q n (n N), we recover Ramanujan s 24 squares formula (.5). Identity 2. Taking a = 4,a 2 =96,a 4 = 32 and t =6in(.2), identity 2 becomes qϕ 6 (q)ψ 8 (q) =qe 4 E2 96 E 32 4 = 6552 E 2(q) 6552 E 2(q 2 )+ 69 (q) (q2 ) (q 4 ) + 256Ω(q). (.22) Equating the coefficients of q n (n N) in(.22), we obtain the following new result analogous to Ramanujan s 24 squares formula. Theorem.2. Let n N. Then the number of representations of n as a sum of 6 squares and 8 triangular numbers is r(n =6 +8 ) = σ (n) σ (n/2) + 69 τ(n) τ(n/2) + 892τ(n/4) + 256ω(n).

10 8 F. Uygul & K. S. Williams Identity 3. Taking a = 32,a 2 =72,a 4 = 6 and t =8in(.2), identity 3 becomes q 2 ϕ 8 (q)ψ 6 (q) =q 2 E 32 E2 72 E4 6 = 4832 E 2(q) 4832 E 2(q 2 ) 56 (q) (q2 ) (q 4 ) + 6Ω(q). (.23) Equating the coefficients of q n (n N,n 2) in (.23), we obtain another new formula analogous to Ramanujan s 24 squares formula. Theorem.3. Let n N satisfy n 2. Then the number of representations of n 2 as a sum of 8 squares and 6 triangular numbers is r(n 2=8 +6 ) = 56 σ (n) 56 σ (n/2) 56 τ(n) + 23 τ(n/2) + 256τ(n/4) + 6ω(n). 382 Identity 4. Equating the coefficients of q n (n N) in identity 4, we obtain the following new result. Theorem.4. Let n N. Then [E 32 E2 96 E 4 4 ] n = 6 σ (n/2) σ (n/4) + 32τ(n) τ(n/2) + τ(n/4) 496ω(n). Identity 5. Taking a = 24, a 2 = 48, a 4 =andt =in(.2), identity 5 becomes E 2(q) 259 (q) (q2 ). q 3 ψ 24 (q) =q 3 E 24 E 48 2 = Equating the coefficients of q n (n N,n 3), we deduce t 24 (n 3) = σ (n) σ (n/2) E 2(q 2 ) 259 τ(n) τ(n/2), where t 24 (n 3) denotes the number of representations of the nonnegative integer n 3 as a sum of 24 triangular numbers. A general formula for the number t 2s (n) of representations of n as a sum of 2s triangular numbers was first given by Ramanujan [4, Sec. 2; 5, pp. 9 9]. Cooper [4, Theorems 3.5 and 3.6] has given an elementary proof of Ramanujan s formula. The special case 2s =24

11 Analogues of Ramanujan s 24 squares formula 9 of Ramanujan s formula gives the above formula for t 24 (n 3). The formula for t 24 (n 3) has been stated explicitly by Ono, Robins and Wahl [2, Theorem 8, p. 86] and Cooper [4, p. 37]. Identity 7. Taking a = 6,a 2 =24,a 4 =6andt =in(.2), identity 7 becomes q 4 ψ 6 (q)ψ 8 (q 2 )=q 4 E 6 E2 24 E4 6 = E 2(q) E 2(q 2 ) 259 (q) (q2 ) (q 4 ) 6 Ω(q). Equating the coefficients of q n (n N,n 4), we obtain the following result. Theorem.5. Let n N satisfy n 4. Then the number of representations of n 4 as a sum of 6 triangular numbers plus twice the sum of 8 triangular numbers is r(n 4=6 +2(8 )) = σ (n) σ (n/2) τ(n) 259 τ(n/2) τ(n/4) ω(n). Identity 8. Equating the coefficients of q n (n N,n 2) in identity 8 gives the following result. Theorem.6. Let n N satisfy n 2. Then [q 2 E 6 E 48 2 E 8 4 ] n = τ(n/2) + 256τ(n/4) + 6ω(n). Identity 9. Equating the coefficients of q n (n N) in identity 9, we obtain the following result. Theorem.7. Let n N. Then [E 6 E 72 2 E 32 4 ] n = 6 σ (n/2) σ (n/4) + 6τ(n) τ(n/2) τ(n/4) 496ω(n). Identity. Taking a = 8,a 2 =,a 4 =32andt =in(.2), identity becomes q 5 ψ 8 (q)ψ 6 (q 2 )=q 5 E 8 E32 4 = E 2(q) E 2(q 2 ) 259 (q) (q2 ) 8 (q4 ) 256 Ω(q).

12 F. Uygul & K. S. Williams Equating the coefficients of q n (n N,n 5), we obtain the following result. Theorem.8. Let n N satisfy n 5. Then the number of representations of n 5 as a sum of 8 triangular numbers plus twice the sum of 6 triangular numbers is r(n 5=8 +2(6 )) = σ (n) σ (n/2) 259 τ(n) τ(n/2) 8 τ(n/4) 256 ω(n). Identity 2. Equating the coefficients of q n (n N) in identity 2 gives the following result. Theorem.9. Let n N. Then [qe 8 E48 2 E4 6 ] n = τ(n)+32τ(n/2) 256ω(n). 2. Proof of Theorem.. Let H denote the Poincaré upper half-plane, that is H = {z C Im(z) > }. For z Hthe Dedekind eta function η(z) is defined by η(z) :=e 2πiz/24 n= ( e 2πinz ), see [5, p. 9] for example. Let Γ denote the modular group, that is {( ) a b Γ:=SL 2 (Z) = a, b, c, d Z,ad bc =}. c d The Dedekind eta function η(z) is a holomorphic modular form of weight /2 for Γ with a certain multiplier system, which we denote by. For all M Γ, (M) is a 24th root of unity depending only on M. The determination of (M) was first addressed by Dedekind in the nineteenth century, then by Rademacher in 93 and later by Petersson in 954. As we need the value of (M) explicitly, we now state Petersson s formula in the form given by Knopp [7, p. 5]. In order to do this we introduce the notation ( c d ) and ( d c ) for ( a b c d ) Γ, so that a, b, c, d Z and ad bc =.Ifc =,thend = ± and we define ( ) ( ) :=, :=. If c is even and c,thend is odd and we define ( ( ) c c := ( ) d) (sgn(c) )(sgn(d) )/4, d

13 Analogues of Ramanujan s 24 squares formula where ( c x d ) is the Jacobi symbol of quadratic reciprocity and sgn(x) = x nonzero real number x. Ifc is odd and d =,thenc = ± and we define ) ) :=, :=. ( Finally, if c is odd and d, we define ( ) d := c We can now state Petersson s formula. ( ( ) d. c for any Theorem 2.. For M =( a c function η(z) is given by b d ) Γ the multiplier system of the Dedekind eta and η ν (M) = ( c η ν (M) = d) ( d c ) e 2πi((a+d)c bd(c2 ) 3c)/24 e 2πi((a+d)c bd(c2 )+3d 3 3cd)/24 if c is odd if c is even. For a positive integer N, we define the Hecke congruence group of level N by {( ) } a b Γ (N) := Γ c d c modn. Clearly Γ () = Γ, Γ (N) ΓandΓ (kn) Γ (N) for any positive integer k. We denote the complex vector space of holomorphic modular forms of weight k for Γ (N) with trivial multiplier system by M k (Γ (N)). We now address the question: Under what conditions does the eta quotient η a (z)η a2 (2z)η a4 (4z) (a,a 2,a 4 Z) belongtom 2 (Γ (4))? This kind of question was probably first addressed by Newman []. Theorem 2.2. Let a,a 2,a 4 Z. Then if and only if and η a (z)η a2 (2z)η a4 (4z) M 2 (Γ (4)) a + a 2 + a 4 =24, (2.) a +2a 2 +4a 4, (2.2) a +2a 2 + a 4, (2.3) 4a +2a 2 + a 4, (2.4) 8 a, 24 a 2, 8 a 4. (2.5)

14 2 F. Uygul & K. S. Williams Proof. Suppose first that a,a 2,a 4 Z satisfy (2.) (2.5). We show that g(z) := η a (z)η a2 (2z)η a4 (4z) is a holomorphic modular form of weight 2 for Γ (4) with trivial multiplier system. From the infinite product representation of η(z) we see that η(z) is a nonzero holomorphic function on H. Thus g(z) is a holomorphic function on H (for any a,a 2,a 4 Z). By (2.2) (2.4) wehave (c, m) 2 m a m m N m 4 for all c N with c 4. Thus by [9, Corollary 2.3, p. 37], g(z) is holomorphic at all cusps (including ). Since η(z) is a modular form of weight /2 for Γ with multiplier system, g(z) transforms like a modular form of weight 2 (a + a 2 + a 4 )=2 for Γ (4) with multiplier system ν g given by ( a ν g c ) b d ( a b = c d ) a ( a 2b c/2 d ) a2 ( a 4b c/4 d ) a4, for all ( a b c d) Γ (4). Thus to complete the proof that g(z) belongs to M 2 (Γ (4)) we have only to prove that ( ) ( ) a b a b ν g = forall Γ (4). c d c d Now by (2.) wehave ( a ν g c ) b d ( a b = c d ) a ( a 2b c/2 d ( ) a b c d = ( ) a 4b c/4 d a ) a2 ( a 4b c/4 d ) 24 a a 2 as is a 24th root of unity and by (2.5) wehave24 a 2.Since8 a (by (2.5)) we complete the proof by showing that ( ) 8 a b c d ( ) a b ( ) = forall Γ (4). (2.6) a 4b c d c/4 d

15 Analogues of Ramanujan s 24 squares formula 3 Suppose first that c (mod8)sothatc/4 is even. Then, by Petersson s theorem (Theorem 2.), we have with ω = e 2πi/24 ( ) a b c d ( ) = a 4b c/4 d ( ) c/4 d ( c d) ω (a+d)c bd(c2 )+3d 3 3cd ω (a+d)(c/4) 4bd((c/4)2 )+3d 3 3(c/4)d = ±ω 3(a+d)c 4 3bdc2 4 3bd 9cd 4 so that ( ) a b c d ( ) a 4b c/4 d 8 = ω 6(a+d)c 6bdc2 24bd 8cd. Now c (mod 8) implies 6(a + d)c 6bdc 2 24bd 8cd (mod 24) so that (2.6) holds. Now suppose that c 4(mod8)sothatc/4 is odd. Then, by Theorem 2., we have ( ) a b ( ν c η d) c d ( ) = a 4b c/4 d ( d c/4 ω (a+d)c bd(c2 )+3d 3 3cd ) ω (a+d)(c/4) 4bd((c/4)2 ) 3c/4 = ±ω 3(a+d)c 4 3bdc2 4 3bd+3d 3 3cd+ 3c 4 so that ( ) a b c d ( ) a 4b c/4 d 8 = ω 6(a+d)c 6bdc2 24bd+24d 24 24cd+6c. Now c (mod 4) implies 6(a + d)c 6bdc 2 24bd +24d 24 24cd +6c (mod 24) so that (2.6) holds. This completes the proof that g(z) M 2 (Γ (4)). Conversely, suppose that a,a 2,a 4 Z are such that g(z) :=η a (z)η a2 (2z)η a4 (4z) M 2 (Γ (4)).

16 4 F. Uygul & K. S. Williams We prove that (2.) (2.5) hold.asg(z) is a modular form of weight 2, Eq. (2.) holds. As g(z) is holomorphic at all cusps, inequalities (2.2) (2.4) holdby[9, Corollary 2.3, p. 37]. Thus we are left to prove (2.5). As g has a trivial multiplier system for Γ (4) we have ( ) a ( ) a2 ( ) 24 a a 2 a b a 2b a 4b = c d c/2 d c/4 d for all ( a b) c d Γ (4). Choose ( a b) ( c d = 8 9) Γ (4). Then by Theorem 2., we have with ω = e 2πı/24 ( ) ( ) ( ) 2 4 = ω 7, = ω 22, = ω 2, so that 7a +22a 2 +2(24 a a 2 ) (mod 24), and thus 3a +4a 2 (mod 24). (2.7) Next choose ( a b) ( c d = 4 ) Γ (4). Here ( ) ( ) ( ) = ω 2, = ω 22, = ω 23, 4 2 so that 2a +22a (24 a a 2 ) (mod 24), and hence 3a + a 2 (mod 24). (2.8) From (2.8) we deduce a 2 (mod 3). Subtracting (2.8) from(2.7), we obtain 3a 2 (mod 24) so a 2 (mod8).hencea 2 (mod 24) and (by (2.8)) a (mod 8). Finally a 4 =24 a a 2 (mod 8). This completes the proof of (2.5). The proof of Theorem 2.2 is now complete. We now determine the integers a,a 2,a 4 satisfying (2.) (2.5). Theorem 2.3. The only integers a,a 2,a 4 Table 2. satisfying (2.) (2.5) are given in Proof. Let a,a 2,a 4 Z satisfy (2.) (2.5). By (2.5) wehavea = 8A and a 2 =24A 2 for some integers A and A 2.Then(2.) (2.4) give a 4 =24 8A 24A 2, (2.9) A +2A 2 4, (2.) A 2, (2.) A + A 2. (2.2)

17 Analogues of Ramanujan s 24 squares formula 5 Table 2. Values of a,a 2,a 4 satisfying (2.) (2.5). No. a a 2 a 4 no a a 2 a Thus 6 A 6and A 2 5. A simple computer search through the lattice points of the box {(x, y) Z 2 6 x 6, y 5} yielded 28 lattice points (A,A 2 ) satisfying (2.9) (2.2). The corresponding 28 values of (a,a 2,a 4 ) are listed in Table 2. For z Hand k N we define the Eisenstein series ξ 2k (z) by ξ 2k (z) = 4k B 2k σ 2k (n)e 2πinz n= so that with q = e 2πiz we have E 2k (q) = 4k B 2k n= σ 2k (n)q n = ξ 2k (z). It is well known that ξ 2k (z) is a modular form of weight 2k for the full modular group Γ with trivial multiplier system if k 2[9, p. 9]. Thus in particular we have ξ 2 (z), ξ 2 (2z), ξ 2 (4z) M 2 (Γ (4)). Theorem 2.4. A basis for the complex vector space M 2 (Γ (4)) comprises ξ 2 (z), ξ 2 (2z), ξ 2 (4z), η 24 (z), η 24 (2z), η 24 (4z), η 8 (z)η 6 (4z). Proof. By the corollary to Proposition 4 in [8] wehave We have already noted that and, by Theorem 2.2, wehave dim(m 2 (Γ (4))) = 7. ξ 2 (z), ξ 2 (2z), ξ 2 (4z) M 2 (Γ (4)), (2.3) η 24 (z), η 24 (2z), η 24 (4z), η 8 (z)η 6 (4z) M 2 (Γ (4)). (2.4)

18 6 F. Uygul & K. S. Williams The seven modular forms listed in (2.3) and(2.4) are linearly independent over C. Hence they form a basis for M 2 (Γ (4)). By [9, Corollary 2.3, p. 37] η 24 (z),η 24 (2z),η 24 (4z) andη 8 (z)η 6 (4z) arecuspidal eta products. Proof of Theorem.. Let r, a,a 2,a 4 be integers and A,A 2,A 4,B,B 2,B 4,C be rational numbers such that (.)holds.setq = e 2πiz.As q < wehavez H. Now E 2 (q k )=E 2 (e 2πikz )=ξ 2 (kz), k =, 2, 4, and q r n= (q k )=η 24 (kz), k =, 2, 4, Ω(q) =η 8 (z)η 6 (4z), ( q n ) a ( q 2n ) a2 ( q 4n ) a4 = q r a 24 a 2 2 a 4 6 η a (z)η a2 (2z)η a4 (4z), so that (.) becomes q r a 24 a 2 2 a 4 6 η a (z)η a2 (2z)η a4 (4z) = A ξ 2 (z)+a 2 ξ 2 (2z)+A 4 ξ 2 (4z) + B η 24 (z)+b 2 η 24 (2z)+B 4 η 24 (4z)+Cη 8 (z)η 6 (4z). The right-hand side belongs to M 2 (Γ (4)). Hence q r a 24 a 2 2 a 4 6 η a (z)η a2 (2z)η a4 (4z) M 2 (Γ (4)). Since η a (z)η a2 (2z)η a4 (4z) is a modular form of weight 2 (a + a 2 + a 4 ) with respect to some multiplier system, it must be the case that q r a 24 a 2 2 a 4 6 transforms like a modular form with respect to some multiplier system. However this is clearly not possible if r a 24 a2 2 a4 a 6. Hence we must have r = 24 + a2 2 + a4 6 and thus η a (z)η a2 (2z)η a4 (4z) M 2 (Γ (4)). Then, by Theorem 2.2, we see that a,a 2,a 4 must satisfy (2.) (2.5). Hence (a,a 2,a 4 ) is one of the 28 triples given in Theorem 2.3. For each of the 28 triples (a,a 2,a 4 ) in Theorem 2.3, Theorem 2.4 ensures that A,A 2,A 4,B,B 2,B 4,C are uniquely determined by η a (z)η a2 (2z)η a4 (4z) =A ξ 2 (z)+a 2 ξ 2 (2z)+A 4 ξ 2 (4z) + B η 24 (z)+b 2 η 24 (2z)+B 4 η 24 (4z)+Cη 8 (z)η 6 (4z). Expanding each of η a (z)η a2 (2z)η a4 (4z),ξ 2 (z),...,η 8 (z)η 6 (4z) inpowersofe 2πiz and equating the coefficients of e 2πinz (n =,, 2, 3, 4, 5, 6), we obtain seven linear equations for the seven quantities A,A 2,...,C. From these we obtain the values of A,A 2,...,C givenintheorem..

19 Analogues of Ramanujan s 24 squares formula 7 3. Final Remarks The second author has recently given the following arithmetic formula for τ(n), namely, τ(n) = x 2 4 x2 2 (x2 3x2 3 )(x2 2 3x2 4 ), (x,...,x 8) Z 8 x 2 + +x2 8 =2n as a special case of a general product-to-sum formula [6, Theorem 4.]. As another application of this product-to-sum formula we obtain a formula for ω(n). In Theorem. of [6] wetaker =,s=2,t=,u=,v =2,w =,x =,y =sothat k =2,l=3,m= 8. We obtain the following formula for ω(n). Theorem 3.. Let n N satisfy n 3. Then ω(n) = (x 2 8 2x2 7 )(x2 2 2x2 8 )(x4 3 3x2 3 x2 4 ). (x,...,x 8) Z 8 x 2 + +x2 4 +2x x2 8 =n We conclude with the following property of ω(n). Theorem 3.2. Let n N. Then ω(n) = for n 2 (mod 4). Proof. We recall the parameters x and z defined in [2, p. 2], namely, x = ϕ4 ( q) ϕ 4 (q), z = ϕ2 (q). From Berndt s catalogue of formulas [2, pp ] we deduce Hence E =2 /6 q /24 x /24 ( x) /6 z /2, E 2 =2 /3 q /2 x /2 ( x) /2 z /2, E 4 =2 2/3 q /6 x /6 ( x) /24 z /2. A(q) :=q 3 E 8 E24 2 E8 4 =2 2 x 3 ( x)z 2. Jacobi s change of sign principle [3, p. 565] tells us that x( q) = x x, z( q) =( x)/2 z, so Thus A( q) = 2 2 x 3 ( x) 2 z 2. A(q)+A( q) =2 2 x 4 ( x)z 2. (3.)

20 8 F. Uygul & K. S. Williams Cheng s rotation principle [3, p. 565] asserts x(iq) = 8i( x)/4 ( ( x) /2 ), z(iq) = i ( i( x) /4 ) 4 2 ( i( x)/4 ) 2 z. (3.2) Mapping q to iq in (3.), and appealing to (3.2), we find A(iq)+A( iq) =2 2 x 4 ( x)z 2. Thus A(q) A(iq)+A( q) A( iq) =.Seta n =[A(q)] n.then a n q n = a n ( i n +( ) n ( i) n )q n =. 4 Hence n= n 2 (mod4) n= ω(n) =[q 3 E 8 2 E4] 8 n =[A(q)] n = a n = forn 2 (mod 4), as claimed. Theorems.2,.3,.5,.8, and3.2 show that the number of representations (x,...,x 6,x 7,...,x 24 ) Z 6 N 8 of n as x x2 6 + x 7(x 7 +) x 24(x 24 +), 2 the number of representations (x,...,x 8,x 9,...,x 24 ) Z 8 N 6 x x2 8 + x 9(x 9 +) 2 the number of representations (x,...,x 24 ) N 24 x (x +) x 6(x 6 +) x 24(x 24 +), 2 of n 4as of n 2as + x 7 (x 7 +)+ + x 24 (x 24 +), and the number of representations (x,...,x 24 ) N 24 of n 5as x (x +) x 8(x 8 +) 2 + x 9 (x 9 +)+ + x 24 (x 24 +), depend only upon σ (n),σ (n/2),τ(n) andτ(n/2) when n 2(mod4). In this paper we have considered certain eta quotients in the space M 2 (Γ (4)). In [7] the second author has essentially considered similar eta quotients in the space M 2 (Γ (4)) but in an entirely different manner. References [] A. Alaca, Ş. Alaca and K. S. Williams, An infinite class of identities, Bull. Austral. Math. Soc. 75 (27) [2] B.C.Berndt,Number Theory in the Spirit of Ramanujan (American Mathematical Society, Providence, RI, 26). [3] N. Cheng and K. S. Williams, Convolution sums involving the divisor function, Proc. Edinb. Math. Soc. 47 (24)

21 Analogues of Ramanujan s 24 squares formula 9 [4] S. Cooper, On sums of an even number of squares, and an even number of triangular numbers: An elementary approach based on Ramanujan s ψ summation formula, in q-series with Applications to Combinatorics, Number Theory, and Physics, Contemporary Mathematics, Vol. 29 (American Mathematical Society, Providence, RI, 2), pp [5] F. Diamond and J. Shurman, A First Course in Modular Forms (Springer, New York, 25). [6] H.M.FarkasandI.Kra,Theta Constants, Riemann Surfaces and the Modular Group (American Mathematical Society, Providence, RI, 2). [7] M. I. Knopp, Modular Functions in Analytic Number Theory (Chelsea Publishing Company, New York, 993). [8] N. Koblitz, Introduction to Elliptic Curves and Modular Forms (Springer, New York, 993). [9] G. Köhler, Eta Products and Theta Series Identities (Springer, Berlin, 2). [] L. J. Mordell, On Mr. Ramanujan s empirical expansions of modular functions, Proc. Cambridge Philos. Soc. 9 (97) [] M. Newman, Construction and application of a class of modular functions (II), Proc. London Math. Soc. (3 ) 9 (959) [2] K. Ono, S. Robins and P. T. Wahl, On the representation of integers as sums of triangular numbers, Aequationes Math. 5 (995) [3] S. Ramanujan, On certain arithmetical functions, Trans. Cambridge Philos. Soc. 22 (96) [4], On certain trigonometrical sums and their applications in the theory of numbers, Trans. Cambridge Philos. Soc. 22 (98) [5], Collected Papers, eds. G. H. Hardy, P. V. SeshuAiyarandB. M. Wilson (American Mathematical Society, Providence, RI, 2). [6] K. S. Williams, Quadratic forms and a product-to-sum formula, Acta Arith. 58 (23) [7], The parents of Jacobi s four squares theorem are unique, Amer. Math. Monthly 2(4) (23)