Ternary Quadratic Forms and Eta Quotients

Transcription

1 Canad. Math. Bull. Vol. 58 (4), 015 pp Canadian Mathematical Society 015 Ternary Quadratic Forms and Eta Quotients Kenneth S. Williams Abstract. Let η(z) (z C, Im(z) > 0) denote the Dedekind eta function. We use a recent productto-sum formula in conjunction with conditions for the non-representability of integers by certain ternary quadratic forms to give explicitly ten eta quotients f (z) = η a(m1) (m 1 z) η a(mr) (m rz) = c(n)e πinz, z C, Im(z) > 0, such that the Fourier coefficients c(n) vanish for all positive integers n in each of infinitely many non-overlapping arithmetic progressions. For example, we show that if f (z) = η 4 (z)η 9 (4z)η (8z) we have c(n) for all n in each of the arithmetic progressions {16k + 14} k 0, {64k + 56} k 0, {56k + 4} k 0, {104k + 896} k 0, Introduction Let N denote the set of positive integers and N 0 = N {0}. Let H denote the Poincaré upper half-plane {z C Im(z) > 0}. The Dedekind eta function is defined by η(z) = e πiz/1 (1 e πinz ), z H. An eta quotient f (z) is a holomorphic function of the form f (z) = η a(m1) (m 1 z) η a(mr) (m r z), z H, where r N, m 1,..., m r N satisfy m 1 < < m r, and a(m 1 ),..., a(m r ) are nonzero integers. We suppose that m 1 a(m 1 ) + + m r a(m r ) = 4 so that the eta quotient f (z) has a Fourier expansion of the form f (z) = c(n)e πinz, c(1) = 1, where the Fourier coefficients c(n) are integers. We adopt the notation [ f (z)] n = c(n), n N. Many questions concerning the vanishing or non-vanishing of the Fourier coefficients of eta quotients have been addressed; see, for example, [], [3, p. 133], [4], [5], and [6]. In this note we are interested in determining explicit eta quotients f (z) such that [ f (z)] n for all n in infinitely many non-overlapping arithmetic progressions. We Received by the editors July 6, 014; revised May 1, 015. Published electronically July 13, 015. AMS subject classification: 11F0, 11E0, 11E5. Keywords: Dedekind eta function, eta quotient, ternary quadratic form, vanishing of Fourier coefficients, product-to-sum formula. 858

2 Ternary Quadratic Forms and Eta Quotients 859 do this by using the author s recent product-to-sum formula [7, Theorem 1.1, p. 80] to express the Fourier coefficients [ f (z)] n of certain eta quotients f (z) in the form (1.1) [ f (z)] n = (x 1,x,x 3) Z 3 Q(x 1,x,x 3)=n P(x 1, x, x 3 ), n N, where P Q[x 1, x, x 3 ] and Q = x 1 + ax + bx 3 for some a, b N with 1 a b 4. Classical results on the representability of n N by the ternary quadratic form Q [1, pp ] give infinitely many arithmetic progressions such that if n belongs to any one of them, then n is not represented by Q, and so by (1.1), [ f (z)] n for these n. The 10 eta quotients constructed in this manner are given in Theorem 1.1(i) (x). Theorem 1.1 is proved in Section. Theorem 1.1 For all e N 0 we have (i) [η (z)η(z)η (3z)η(4z)η 5 (6z)η (8z)η (1z)] n for n 5 e+1 (mod e+4 ), (ii) [η (z)η 8 (z)η (4z)η 3 (6z)] n for n 3 e+1 (mod 3 e+ ), (iii) [η 3 (z)η (3z)η 8 (6z)η (1z)] n for n 3 e (mod 3 e+1 ), (iv) [η 3 (z)η (4z)η 3 (6z)η 5 (8z)η (16z)] n for n 5 3 e+1 (mod 3 e+ ), (v) [η (3z)η 6 (4z)η 5 (6z)η (1z)] n for n 8 3 e+1 (mod 4 3 e+ ), (vi) [η (3z)η 6 (4z)η 5 (6z)η (1z)] n for n 11 3 e+1 (mod 4 3 e+ ), (vii) [η (z)η 7 (z)η (4z)] n for n 7 e (mod e+3 ), (viii) [η 4 (z)η 9 (4z)η (8z)] n for n 7 e+1 (mod e+4 ), (ix) [η 4 (z)η (z)η (3z)η 4 (4z)η 5 (6z)η (1z)] n for n 3 e+1 (mod 3 e+ ), (x) [η 4 (z)η (z)η (4z)η 5 (8z)η (16z)] n for n 7 e (mod e+3 ). Proof of Theorem 1.1 For k N and q C with q < 1, we define E k = E k (q) = (1 q kn ), so that E k = E 1 (q k ). We begin with a lemma that enables us to eliminate some uninteresting cases that arise in the proof of Theorem 1.1 as well as assisting in the formulation of some conjectures in Section 3. Lemma.1 Let n N. (i) If n 0, 3 (mod 4), then [qe1 E 1 E4 6] n. (ii) If n 0, 6 (mod 8), then [qe1 E E4 1 E8 7E 16 ] n. (iii) If n 0, 3 (mod 4), then [qe1 E 5E4 4] n. (iv) If n, 3 (mod 4), then [qe3 E4 6E5 6 E 1 ] n. (v) If n 3 (mod 4), then [qe1 4 EE 4E 8 5E 16 ] n.

3 860 K. S. Williams Proof (i) By a classical theorem of Jacobi, we have E1 E 1 = 1 + ( 1) m q m. m=1 Thus for n N, we have 1 if n = 1, [qe1 E 1 ] n = ( 1) m if n = m + 1, m N, 0 if n m + 1, m N 0. If n 0, 3 (mod 4), then we have n m + 1 (m N 0 ) so that [qe 1 E 1 ] n. Hence, [qe 1 E 1 E 6 4] n if n 0, 3 (mod 4). (ii) Replacing q by q in Jacobi s identity, we obtain EE 4 1 = 1 + ( 1) m q m. m=1 Multiplying the series for E 1 E 1 and E E 1 4 together, we deduce qe1 E E4 1 = q + ( 1) n q n +1 + ( 1) n q n ( ( 1) l+m ) q n. l,m 1 l +m +1=n As n + 1 1,, 5 (mod 8), n + 1 1, 3 (mod 8) and l + m + 1 1,, 3, 4, 5, 7 (mod 8), we have [qe 1 E E 1 4 ] n for n 0, 6 (mod 8). Thus, [qe 1 E E 1 4 E 7 8E 16 ] n if n 0, 6 (mod 8). (iii) By another classical identity of Jacobi, we have Thus for n N we have [qe1 EE 5 4 ] n = E1 EE 5 4 = 1 + q m. m=1 1 if n = 1, if n = m + 1, m N, 0 if n m + 1, m N 0. If n 0, 3 (mod 4) then n m + 1 (m N 0 ) so [qe 1 E 5 E 4 ] n. Thus, [qe 1 E 5 E 4 4] n = [qe 1 E 5 E 4 E 6 4] n if n 0, 3 (mod 4). (iv) Mapping q to q 3 in the identity of Jacobi given in the proof of (iii), we have Hence for n N, we have [qe3 E6E 5 1 ] n = E3 E6E 5 1 = 1 + q 3m. m=1 1 if n = 1, if n = 3m + 1, m N, 0 if n 3m + 1, m N 0.

4 Ternary Quadratic Forms and Eta Quotients 861 If n, 3 (mod 4), then n 3m + 1 (m N 0 ) so [qe 3 E 5 6 E 1 ] n. Thus [qe 3 E 6 4E 5 6E 1 ] n if n, 3 (mod 4). (v) Ramanujan defined the theta function ϕ by and proved that Hence, 3 j=0 ϕ(q) = q n = 1 + q n n= ϕ (q) + ϕ ( q) = ϕ (q ), ϕ(q)ϕ( q) = ϕ ( q ). ( 1) j ϕ(i j q)ϕ 3 ( i j q) = ϕ(q)ϕ( q)( ϕ (q) + ϕ ( q)) ϕ(iq)ϕ( iq)( ϕ (iq) + ϕ ( iq)) = ϕ ( q )ϕ (q ) ϕ (q )ϕ ( q ). Thus, [ϕ(q)ϕ 3 ( q)] n for n (mod 4). Now ϕ(q) = E1 E 5E 4 and ϕ( q) = E1 E 1 so ϕ(q)ϕ 3 ( q) = E1 4 EE 4. Hence, Then [E 4 1 E E 4 ] n for n (mod 4). [E 4 1 E E 4E 5 8E 16 ] n = [E 4 1 E E 4 E 4 4E 5 8E 16 ] n = [E 4 1 E E 4 ] n for n (mod 4), and finally [qe1 4 EE 4E 8E 5 16 ] n for n 3 (mod 4). This completes the proof of the lemma. Taking q = e πiz (z H) so that q < 1, we have η(kz) = e πikz/1 E k = q k/4 E k. We now state the product-to-sum formula proved by the author in [7, Theorem 1.1, p. 80], which we will use in the proof of Theorem 1.1. Product-to-Sum Formula Let k N 0 and l N. Let r, s, t, u N 0 be such that r + s + t + u = k. Let v, w, x, y N 0 be such that v + w + x + y = l. Set m = k + l so that m N and m. Let and P(x 1,..., x m ) = 1 l r+v g=r+1 (xg xg+s+l+y) r+v+w (xg 3xg+s+t+l+y) g=r+v+1 r+v+w+x (xg 4xg+s+t+l+y+u) g=r+v+w+1 r+l g=r+v+w+x+1 Q(x 1,..., x m ) = x x r+l+y + x r+l+y x r+s+l+v+y (x 4 g 3x gx g+y) + 3x r+s+l+v+y x r+s+t+l+v+w+y + 4x r+s+t+l+v+w+y x m.

5 86 K. S. Williams Let a 1 = r + v + 4y, a 6 = 5t + 3w, a = 5r s + v + 3w + y, a 8 = s + 5u + v, a 3 = t, a 4 = r + 5s u + v + 6x + 4y, Then for n N with n l, we have and a 1 = t, a 16 = u. [ q l E a1 1 E a Ea3 3 Ea4 4 Ea6 6 Ea8 8 Ea1 1 Ea16 16 ] n = (x 1,...,x m) Z m Q(x 1,...,x m)=n P(x 1,..., x m ) a 1 + a + 3a 3 + 4a 4 + 6a 6 + 8a 8 + 1a a 16 = 4l. Proof of Theorem 1.1 We use the special case of the product-to-sum formula when k = l = 1, so that m = 3 and Q is a positive diagonal ternary quadratic form all of whose coefficients are 1,, 3, or 4 with at least one of the coefficients equal to 1. Let r, s, t, u, v, w, x, y N 0 satisfy (.1) r + s + t + u = 1, v + w + x + y = 1. Define A N and B, C, D N 0 by (.) A = r + y + 1, B = s + v, C = t + w, D = u + x, so that (.3) A + B + C + D = 3. Then and P(x 1, x, x 3 ) = 1 r+v g=r+1 (.4) Q(x 1, x, x 3 ) = A i=1 (xg xg+s+y+1) r+v+w (xg 3xg+s+t+y+1) g=r+v+1 r+v+w+x (xg 4xg+s+t+y+u+1) g=r+v+w+1 x i + A+B i=a+1 x i + 3 A+B+C i=a+b+1 r+1 g=r+v+w+x+1 (x 4 g 3x gx g+y) xi + 4 A+B+C+D xi. i=a+b+c+1 Define the integers a 1, a, a 3, a 4, a 6, a 8, a 1 and a 16 as in the product-to-sum formula. Then, as l = 1, by the product-to-sum formula we have and (.5) a 1 + a + 3a 3 + 4a 4 + 6a 6 + 8a 8 + 1a a 16 = 4 [ η a1 (z)η a (z)η a3 (3z)η a4 (4z)η a6 (6z)η a8 (8z)η a1 (1z)η a16 (16z)] n = [ qe a1 1 E a Ea3 3 Ea4 4 Ea6 6 Ea8 8 Ea1 1 Ea16 16 ] n = (x 1,x,x 3) Z 3 Q(x 1,x,x 3)=n P(x 1, x, x 3 ).

6 Ternary Quadratic Forms and Eta Quotients 863 We next examine the 16 possible values of the vector (r, s, t, u, v, w, x, y) N 8 0 subject to the restrictions in (.1). The ternary form Q corresponding to each such vector is determined from (.), (.3), and (.4) and each eta quotient η a1 (z)η a (z)η a3 (3z)η a4 (4z)η a6 (6z)η a8 (8z)η a1 (1z)η a16 (16z) from the formulae for a 1, a, a 3, a 4, a 6, a 8, a 1, a 16 in terms of r, s, t, u, v, w, x, y given in the product-to-sum formula. The values are given in Table 1. It is known from the work of Dickson and Jones (see, for example, [1, pp ]) that the positive integers n for which Q(x 1, x, x 3 ) = n is not solvable in integers x 1, x, and x 3 are as given in Table, where k, l N 0. Appealing to (.5), Table 1 and Table, we obtain the following results. Case 1 gives (.6) [ η 6 (z)η 1 (4z)η (8z)] n for n 7 e+1 (mod e+4 ) for all e N 0, but this is not interesting, as clearly (.6) holds for all n N with n 0 (mod ). Case gives (.7) [ η (z)η 1 (z)η 6 (4z)] n for n 7 e (mod e+3 ) for all e N 0, which is again not interesting, as by Lemma.1(i) (.7) holds for all n 0, 3 (mod 4). Case 3 gives [ η (z)η(z)η (3z)η(4z)η 5 (6z)η (8z)η (1z)] n for n 5 e+1 (mod e+4 ) for all e N 0, which is Theorem 1.1(i). Case 4 gives (.8) [ η (z)η(z)η 1 (4z)η 7 (8z)η (16z)] n for n 7 e+1 (mod e+4 ) for all e N 0, but this is not interesting, as (.8) holds for all n 0, 6 (mod 8) by Lemma.1(ii). Case 5 gives [ η (z)η 8 (z)η (4z)η 3 (6z)] n for n 3 e+1 (mod 3 e+ ) for all e N 0, which is Theorem 1.1(ii). Case 6 gives (.9) [ η(z)η 5 (4z)η 3 (6z)η (8z)] n for n 5 e+1 (mod e+4 ) for all e N 0, which is not interesting, as (.9) holds for all n 0 (mod ). Case 7 gives [ η 3 (z)η (3z)η 8 (6z)η (1z)] n for n 3 e (mod 3 e+1 ) for all e N 0, which is Theorem 1.1(iii).

7 864 K. S. Williams case (r, s, t, u, v, w, x, y) (a 1, a, a 3, a 4, a 6, a 8, a 1, a 16 ) Q(x 1, x, x 3 ) 1 (1, 0, 0, 0, 1, 0, 0, 0) (0, 6, 0, 1, 0,, 0, 0) x1 + x + x3 (0, 1, 0, 0, 1, 0, 0, 0) (, 1, 0, 6, 0, 0, 0, 0) x1 + x + x3 3 (0, 0, 1, 0, 1, 0, 0, 0) (, 1,, 1, 5,,, 0) x1 + x + 3x3 4 (0, 0, 0, 1, 1, 0, 0, 0) (, 1, 0, 1, 0, 7, 0, ) x1 + x + 4x3 5 (1, 0, 0, 0, 0, 1, 0, 0) (, 8, 0,, 3, 0, 0, 0) x1 + x + 3x3 6 (0, 1, 0, 0, 0, 1, 0, 0) (0, 1, 0, 5, 3,, 0, 0) x1 + x + 3x3 7 (0, 0, 1, 0, 0, 1, 0, 0) (0, 3,, 0, 8, 0,, 0) x1 + 3x + 3x3 8 (0, 0, 0, 1, 0, 1, 0, 0) (0, 3, 0,, 3, 5, 0, ) x1 + 3x + 4x3 9 (1, 0, 0, 0, 0, 0, 1, 0) (, 5, 0, 4, 0, 0, 0, 0) x1 + x + 4x3 10 (0, 1, 0, 0, 0, 0, 1, 0) (0,, 0, 11, 0,, 0, 0) x1 + x + 4x3 11 (0, 0, 1, 0, 0, 0, 1, 0) (0, 0,, 6, 5, 0,, 0) x1 + 3x + 4x3 1 (0, 0, 0, 1, 0, 0, 1, 0) (0, 0, 0, 4, 0, 5, 0, ) x1 + 4x + 4x3 13 (1, 0, 0, 0, 0, 0, 0, 1) (, 7, 0,, 0, 0, 0, 0) x1 + x + x3 14 (0, 1, 0, 0, 0, 0, 0, 1) (4, 0, 0, 9, 0,, 0, 0) x1 + x + x3 15 (0, 0, 1, 0, 0, 0, 0, 1) (4,,, 4, 5, 0,, 0) x1 + x + 3x3 16 (0, 0, 0, 1, 0, 0, 0, 1) (4,, 0,, 0, 5, 0, ) x1 + x + 4x3 Table 1: Eta quotients and ternary forms corresponding to (r, s, t, u, v, w, x, y) integers not integers not Q(x 1, x, x 3 ) Q(x represented by Q 1, x, x 3 ) represented by Q x1 + x + x3 4 k (8l + 7) x1 + x + 3x3 4 k (16l + 10) x 1 + x + x 3 4 k (16l + 14) x 1 + x + 4x 3 4 k (16l + 14) x 1 + x + 3x 3 9 k (9l + 6) x 1 + 3x + 3x 3 9 k (3l + ) x 1 + x + 4x 3 8l + 3, 4 k (8l + 7) x 1 + 3x + 4x 3 4l +, 9 k (9l + 6) x 1 + x + x 3 4 k (8l + 7) x 1 + 4x + 4x 3 4l +, 4l + 3, 4 k (8l + 7) Table : Integers not represented by ternary quadratic forms Case 8 gives (.10) [ η 3 (z)η (4z)η 3 (6z)η 5 (8z)η (16z)] n for n (mod 4) and n 3 e+1 (mod 3 e+ ) for all e N 0. Clearly (.10) holds for all n 0 (mod ) so the former is not interesting while the latter is interesting only

8 Ternary Quadratic Forms and Eta Quotients 865 when n 1 (mod ), that is, when n 5 3 e+1 (mod 3 e+ ), which is Theorem 1.1(iv). Case 9 gives (.11) [ η (z)η 5 (z)η 4 (4z)] n for n 3 (mod 8) and n 7 e (mod e+3 ) for all e N 0. But this is not interesting, as (.11) holds for all n 0, 3 (mod 4) by Lemma.1(iii). Case 10 gives (.1) [ η (z)η 11 (4z)η (8z)] n for n 7 e+1 (mod e+4 ) for all e N 0. But this is not interesting as (.1) holds for all n 0 (mod ). Case 11 gives (.13) [ η (3z)η 6 (4z)η 5 (6z)η (1z)] n for n (mod 4) and n 3 e+1 (mod 3 e+ ) for all e N 0. By Lemma.1(iv) (.13) holds for all n, 3 (mod 4). Thus (.13) is only interesting for those n satisfying n 3 e+1 (mod 3 e+ ) and n 0, 1 (mod 4), that is, for n 8 3 e+1 (mod 4 3 e+ ) and n 11 3 e+1 (mod 4 3 e+ ). These are parts (v) and (vi) of Theorem 1.1, respectively. Case 1 gives (.14) [ η 4 (4z)η 5 (8z)η (16z)] n for n (mod 4), n 3 (mod 4) and n 7 e (mod e+3 ) for all e N 0, but again this is not interesting as (.14) clearly holds for all n / 1 (mod 4). Case 13 gives [η (z)η 7 (z)η (4z)] n for n 7 e (mod e+3 ) for all e N 0, which is Theorem 1.1(vii). Case 14 gives [ η 4 (z)η 9 (4z)η (8z)] n for n 7 e+1 (mod e+4 ) for all e N 0, which is Theorem 1.1(viii). Case 15 gives [ η 4 (z)η (z)η (3z)η 4 (4z)η 5 (6z)η (1z)] n for n 3 e+1 (mod 3 e+ ) for all e N 0, which is Theorem 1.1(ix). Finally, Case 16 gives [ η 4 (z)η (z)η (4z)η 5 (8z)η (16z)] n for n 3 (mod 8) and n 7 e (mod e+3 ) for all e N 0. By Lemma.1(v) the equality holds for all n 3 (mod 4). The latter congruence is Theorem 1.1(x).

9 866 K. S. Williams 3 Final Remarks We briefly discuss whether or not the criteria in each of the ten parts of Theorem 1.1 form a complete description of the set of vanishing coefficients for the corresponding eta quotient. (i) As [ η (z)η(z)η (3z)η(4z)η 5 (6z)η (8z)η (1z)] n for n = 6, 4, 9, 39, 54, 60, 78,..., the congruences n 5 e+1 (mod e+4 ) do not form a complete description of when the coefficients of the eta quotient vanish and it does not seem easy to formulate such a description. We note that the data up to n = 000 suggest the following conjecture. Conjecture 3.1 [ η (z)η(z)η (3z)η(4z)η 5 (6z)η (8z)η (1z)] n for n 3 e+1 (mod 3 e+ ). (ii) The data up to n = 1000 suggest the following conjecture. Conjecture 3. If [ η (z)η 8 (z)η (4z)η 3 (6z)] n then n 3 e+1 (mod 3 e+ ) for some e N 0. If Conjecture 3. is true, then by Theorem 1.1(ii) we have [ η (z)η 8 (z)η (4z)η 3 (6z)] n if and only if n 3 e+1 (mod 3 e+ ) for some e N 0. (iii) The data up to n = 1000 suggest the following conjecture. Conjecture 3.3 If [ η 3 (z)η (3z)η 8 (6z)η (1z)] n, then n 3 e (mod 3 e+1 ) for some e N 0. If Conjecture 3.3 is true then by Theorem 1.1(iii) we have if and only if [ η 3 (z)η (3z)η 8 (6z)η (1z)] n n 3 e (mod 3 e+1 ) for some e N 0. (iv) The data up to n = 1000 suggest the following conjecture. Conjecture 3.4 If [ η 3 (z)η (4z)η 3 (6z)η 5 (8z)η (16z)] n and n is odd, then n 5 3 e+1 (mod 3 e+ ) for some e N 0.

10 Ternary Quadratic Forms and Eta Quotients 867 If Conjecture 3.4 is true, then by Theorem 1.1(iv) we have [ η 3 (z)η (4z)η 3 (6z)η 5 (8z)η (16z)] n if and only if n 0 (mod ) or n 5 3 e+1 (mod 3 e+ ) for some e N 0. (v)(vi) The data up to n = 1000 suggest the following conjecture. Conjecture 3.5 If n 0, 1 (mod 4) and [ η (3z)η 6 (4z)η 5 (6z)η (1z)] n, then n 8 3 e+1 (mod 4 3 e+ ) or n 11 3 e+1 (mod 4 3 e+ ) for some e N 0. If Conjecture 3.5 is true, then by Theorem 1.1(v)(vi) and Lemma.1(iv) we have [ η (3z)η 6 (4z)η 5 (6z)η (1z)] n if and only if n (mod 4), n 3 (mod 4), n 8 3 e+1 (mod 4 3 e+ ) or n 11 3 e+1 (mod 4 3 e+ ) for some e N 0. (vii) The data up to n = 1000 suggest the following conjecture. Conjecture 3.6 If [ η (z)η 7 (z)η (4z)] n, then n 7 e (mod e+3 ) for some e N 0. If Conjecture 3.6 is true, then by Theorem 1.1(vii) we have [ η (z)η 7 (z)η (4z)] n if and only if n 7 e (mod e+3 ) for some e N 0. (viii) The data up to n = 1000 suggest the following conjecture. Conjecture 3.7 If [ η 4 (z)η 9 (4z)η (8z)] n, then n 7 e+1 (mod e+4 ) for some e N 0. If Conjecture 3.7 is true, then by Theorem 1.1(viii) we have [ η 4 (z)η 9 (4z)η (8z)] n if and only if n 7 e+1 (mod e+4 ) for some e N 0. (ix) Theorem 1.1(ix) and the data up to n = 1000 suggest the following conjecture. Conjecture 3.8 [ η 4 (z)η (z)η (3z)η 4 (z)η 5 (6z)η (1z)] n if and only if n 3 e+1 (mod 3 e+ ) or n = 3 e+1 for some e N 0. (x) The data up to n = 1000 suggest the following conjecture.

11 868 K. S. Williams Conjecture 3.9 If n / 3 (mod 4) and [ η 4 (z)η (z)η (4z)η 5 (8z)η (16z)] n then n 7 e (mod e+3 ) for some e N. If Conjecture 3.9 is true, then Lemma.1(v) and Theorem 1.1(x) give [ η 4 (z)η (z)η (4z)η 5 (8z)η (16z)] n if and only if n 3 (mod 4) or n 7 e (mod e+3 ) for some e N. Over the past twenty years or so, many new results concerning the representability and non-representability of positive integers by ternary quadratic forms have been proved by a number of authors, for example, Berkovich, Bhargava, Duke, Jagy, Kaplansky, and Oh, as well as many others. However, as the product-to-sum formula used in this paper applies only to the ten ternaries x 1 + ax + bx 3 with 1 a b 4, we cannot use these new results in conjunction with the product-to-sum formula to obtain further results like those in Theorem 1.1. Acknowledgment The author thanks the referee for his/her comments, which led to improvements in this paper. References [1] L. E. Dickson, Modern elementary theory of numbers. University of Chicago Press, Chicago, IL, [] B. Gordon and S. Robins, Lacunarity of Dedekind η-products. Glasgow Math. J. 37(1995), no. 1, [3] L. J. P. Kilford, Modular forms: a classical and computational introduction. Imperial College Press, London, 008. [4] D. H. Lehmer, The vanishing of Ramanujan s function τ(n). Duke Math. J. 14(1947), [5] V. K. Murty, Lacunarity of modular forms. J. Indian Math. Soc. 5(1987), [6] J.-P. Serre, Sur la lacunarité des puissances de η. Glasgow Math. J. 7(1985), [7] K. S. Williams, Quadratic forms and a product-to-sum formula. Acta Arith. 158(013), no. 1, Centre for Research in Algebra and Number Theory, School of Mathematics and Statistics, Carleton University, Ottawa, ON K1S 5B6 williams@math.carleton.ca