EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION

Transcription

1 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION MADELINE LOCUS DAWSEY* AND RIAD MASRI Abstract. In this paper, we establish an asymptotic formula with an effective bound on the error term for the Andrews smallest parts function sptn). We use this formula to prove recent conjectures of Chen concerning inequalities which involve the partition function pn) and sptn). Further, we strengthen one of the conjectures, and prove that for every ɛ > 0 there is an effectively computable constant Nɛ) > 0 such that for all n Nɛ), we have 6 n pn) < sptn) < 6 + ɛ ) n pn). Due to the conditional convergence of the Rademacher-type formula for sptn), we must employ methods which are completely different from those used by Lehmer to give effective error bounds for pn). Instead, our approach relies on the fact that pn) and sptn) can be expressed as traces of singular moduli. 1. Introduction and Statement of Results The smallest parts function sptn) of Andrews is defined for any integer n 1 as the number of smallest parts among the integer partitions of size n. For example, the partitions of n = 4 are with the smallest parts underlined): 4, + 1, 2 + 2, , , and so spt4) = 10. The spt-function has many remarkable properties. For example, Andrews [] proved the following analogues of the well-known Ramanujan congruences for the partition function pn): spt5n + 4) 0 mod 5), spt7n + 5) 0 mod 7), spt1n + 6) 0 mod 1). One can compute sptn) by making use of the generating function sptn)q n = n=1 n=1 q n 1 q n ) 2 q n+1 ; q), *This author was previously known as Madeline Locus. 1

2 2 LOCUS AND MASRI a; q) = 1 aq n ). n=0 We use this generating function to compute the values of sptn) required for this paper. In this paper, we will prove the following asymptotic formula for sptn) with an effective bound on the error term. Theorem 1.1. Let µn) := 24n 1/6. Then for all n 1, we have sptn) = 24n 1 eµn) + E s n), E s n) < )2 qn) 24n 1) 2 e µn)/2 with log24n 1) qn) := loglog24n 1)) Our interest in proving the effective bounds of Theorem 1.1 was motivated in part by the following recent conjectures of Chen [10] concerning inequalities which involve pn) and sptn). Conjecture Chen). 1) For n 5, we have 6 n pn) < sptn) < n pn). 2) For a, b) 2, 2) or, ), we have ) For n 6, we have 4) For n > m > 1, we have 5) For n 1, we have 6) For n 7, we have spta) sptb) > spta + b). sptn) 2 > sptn 1) sptn + 1). sptn) 2 > sptn m) sptn + m). sptn 1) sptn) sptn 1) sptn) ) > sptn) n sptn + 1). ) 1 + > sptn) 24n /2 sptn + 1). Remark. Conjectures 1) and 2) are slight modifications of Chen s original claims. By combining Theorem 1.1 with classical work of Lehmer [20] which gives effective bounds for pn), we will prove the following result.

3 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION Theorem 1.2. All of Chen s conjectures are true. We will also use Theorem 1.1 to prove the following more precise version of Theorem 1.2 regarding Conjecture 1). Theorem 1. Refined Theorem 1.2 1)). For each ɛ > 0, there is an effectively computable constant Nɛ) > 0 such that for all n Nɛ), we have ) 6 6 npn). npn) < sptn) < + ɛ Remark. The constant Nɛ) of Theorem 1. can be computed in practice. For example, by letting ɛ = 1 6/ in Theorem 1., we get Theorem 1.2 1) for n N1 6/). Then, by computing N1 6/) = 12788, we use a computer to verify Theorem 1.2 1) in the exceptional range 5 n < Remark. In analogy with the Hardy-Ramanujan asymptotic for pn), Bringmann [6] used the circle method to establish the asymptotic sptn) 1 8n e 2n as n. Bringmann s asymptotic for sptn) implies that 6 1) sptn) npn) as n. Theorem 1. refines the asymptotic 1). We now describe our approach to Theorem 1.1. In particular, we explain some of the difficulties involved in proving effective bounds for sptn). In [25], Rademacher established an exact formula for pn) as the absolutely convergent infinite sum 2) pn) = 2 24n 1) /4 c=1 ) A c n) 24n 1 I /2, c 6c I ν is the I-Bessel function, A c n) is the Kloosterman sum A c n) := e isd,c) e 2idn c, and sd, c) is the classical Dedekind sum sd, c) := c 1 r=1 d mod c) d,c)=1 r dr c c dr 1 ). c 2 Recently, Ahlgren and Andersen [1] gave a Rademacher-type exact formula for sptn) as the conditionally convergent infinite sum sptn) = 6 24n 1) 1 A c n) ) ) 24n 1 ) 4 I1/2 I /2. c 6c c=1

4 4 LOCUS AND MASRI In order to give an effective bound on the error term for pn), Lehmer [20] truncated the absolutely convergent sum 2) and applied bounds for the Kloosterman sum A c n). On the other hand, since the formula ) is only conditionally convergent, bounding sptn) is a much more delicate matter. In fact, to resolve the difficult problem of proving that ) converges, Ahlgren and Andersen used advanced methods from the spectral theory of automorphic forms. To give an effective bound on the error term for sptn), we will use different types of formulas for pn) and sptn) which express these functions as traces of singular moduli. To state these formulas, consider the weight 2 weakly holomorphic modular form for Γ 0 6) defined by gz) := 1 E 2 z) 2E 2 2z) E 2 z) + 6E 2 6z) ) 2 2, z = x + iy H. ηz)η2z)ηz)η6z) By applying the Maass weight-raising operator to gz), one gets the following weight zero weak Maass form for Γ 0 6): 1 d P z) := 2i dz + 1 ) gz). 2y Bruinier and Ono [8] proved the following formula for pn). Theorem Bruinier Ono). For all n 1, we have 1 4) pn) = P τ Q ), 24n 1 the sum is over the Γ 0 6) equivalence classes of discriminant 24n+1 positive definite, integral binary quadratic forms Q = [a, b, c] such that 6 a and b 1 mod 12), and τ Q is the Heegner point given by the root Q τ Q, 1) = 0 in the complex upper half-plane H. 5) Similarly, consider the weight zero weakly holomorphic modular form for Γ 0 6) defined by fz) := 1 E 4 z) 4E 4 2z) 9E 4 z) + 6E 4 6z) ) ηz)η2z)ηz)η6z) Ahlgren and Andersen [1] proved the following analogue of 4) for sptn). Theorem Ahlgren Andersen). For all n 1, we have 6) sptn) = 1 fτ Q ) P τ Q ) ). 12 [Q] Identities which express Fourier coefficients of harmonic weak Maass forms as traces of singular moduli have been used in many contexts to give strong asymptotic formulas. For example, Bringmann and Ono [7] expressed pn) as a twisted trace of singular moduli by arithmetically reformulating Rademacher s exact formula 2) for pn). Folsom and the second author [16] combined the Bringmann-Ono formula with spectral methods and subconvexity bounds for quadratic twists of modular L functions to give an asymptotic formula for pn) with a power-saving error term. In particular, by calculating the main term in this asymptotic formula in terms of the truncated main term in Rademacher s exact formula for pn), these authors improved the exponent in Lehmer s bound [20]. This exponent was further improved by Ahlgren and Andersen [2]. [Q]

5 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 5 In the works [4, 22, 2], spectral methods and subconvexity bounds were again used to give asymptotic formulas with power-saving error terms for twisted traces of singular moduli. These results were applied to study a variety of arithmetic problems, including the distribution of pn) and sptn), and the distribution of partition ranks. Note that although the constants in the error terms of these results are effective, it would be very difficult to actually give explicit numerical values for these constants because of the methods involved in the proofs. There is an alternative approach, which we now describe. Using 4), the formula 6) can be written as 7) sptn) = n 1 Sn) pn), 12 Sn) is the trace of singular moduli for fz) given by Sn) := [Q] fτ Q ). Then by applying Lehmer s effective bounds for pn) in 7), we will reduce the proof of Theorem 1.1 to the following asymptotic formula for Sn) with an effective bound on the error term. Theorem 1.4. For all n 1, we have Sn) = 224n 1) 1/4 I 1/2 24n 1/6 ) + En), En) < )2 qn)+1 24n 1) 2 e µn)/2 with qn) := log24n 1) loglog24n 1)) In [14], Dewar and Murty used the formula 4) to derive the Hardy-Ramanujan asymptotic formula for pn) without using the circle method. We will use a related approach to prove Theorem 1.4. In order to give effective bounds, additional care must be taken. For instance, we must give effective bounds for the Fourier coefficients of fz), which requires a careful study of the asymptotic properties of Bessel functions in transitionary ranges. Organization. The paper is organized as follows. In Section 2, we review some facts regarding quadratic forms and Heegner points. In Section, we prove Theorem 1.4. In Section 4, we study the asymptotic properties of pn) and sptn), and prove Theorem 1.1. In Section 5, we prove Theorem 1.. Finally, in Section 6, we prove the remaining conjectures. Acknowledgments. The authors would like to thank Ken Ono for many valuable suggestions regarding this work. They would also like to thank Adrian Barquero-Sanchez, Sheng-Chi Liu, Wei-Lun Tsai, and Matt Young for very helpful discussions, Michael Griffin and Lea Beneish for computing the values of sptn) which were required to prove the conjectures, and Karl Mahlburg for comments which helped to improve the exposition of the paper.

6 6 LOCUS AND MASRI 2. Quadratic forms and Heegner points Let N 1 be a positive integer and D < 0 be a negative discriminant coprime to N. Let Q D,N be the set of positive definite, integral binary quadratic forms QX, Y ) = [a Q, b Q, c Q ]X, Y ) = a Q X 2 + b Q XY + c Q Y 2 of discriminant b 2 Q 4a Qc Q = D < 0 with a Q 0 mod N). There is a right) action of Γ 0 N) on Q D,N defined by Q = [a Q, b Q, c Q ] Q σ = [ a σ Q, b σ Q, cq] σ, ) α β for σ = Γ γ δ 0 N) we have a σ Q = a Q α 2 + b Q αγ + c Q γ 2, b σ Q = 2a Q αβ + b Q αδ + βγ) + 2c Q γδ, c σ Q = a Q β 2 + b Q βδ + c Q δ 2. Given a solution r mod 2N) of r 2 D mod 4N), we define the subset of forms Q D,N,r := {Q = [a Q, b Q, c Q ] Q D,N : b Q r mod 2N)}. Then the group Γ 0 N) also acts on Q D,N,r. The number of Γ 0 N) equivalence classes in Q D,N,r is given by the Hurwitz-Kronecker class number HD). The preceding facts remain true if we restrict to the subset Q prim D,N of primitive forms in Q D,N ; i.e., those forms with a Q, b Q, c Q ) = 1. In this case, the number of Γ 0 N) equivalence classes in Q prim D,N,r is given by the class number hd). To each form Q Q D,N, we associate a Heegner point τ Q which is the root of Q X, 1) given by τ Q = b Q + D 2a Q H. The Heegner points τ Q are compatible with the action of Γ 0 N) in the sense that if σ Γ 0 N), then 8) στ Q ) = τ Q σ 1.. Proof of Theorem 1.4 In this section we prove Theorem 1.4, which will be used crucially in the proof of Theorem 1.1. Let D n := 24n + 1 for n Z + and define the trace of fz) by Sn) := fτ Q ). [Q] Q Dn,6,1 /Γ 0 6)

7 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 7 First, we decompose Sn) as a linear combination of traces involving primitive forms. Let < 0 be any discriminant with 1 mod 24) and define the class polynomials H n X) := X fτ Q )) and Ĥ X) := [Q] Q Dn,6,1 /Γ 0 6) [Q] Q prim,6,1 /Γ 06) X fτ Q )). Let {W m } m 6 be the group of Atkin-Lehner involutions for Γ 0 6). Since f 0 W m = f for m = 1, 6 and f 0 W m = f for m = 2,, then arguing exactly as in the proof of [9, Lemma 7] we get the identity 9) H n X) = hd εu) n/u 2 ) Ĥ 2εu)X), Dn/u u>0 u 2 D n εu) = 1 if u ±1 mod 12) and εu) = 1 otherwise. Comparing terms on both sides of 9) yields the class number relation and the decomposition 10) HD n ) = hd n /u 2 ) u>0 u 2 D n Sn) = hd εu) n/u 2 )+1 S u n), u>0 u 2 D n S u n) := [Q] Q prim Dn/u 2,6,1 /Γ 06) fτ Q ). Next, following [14] we express S u n) as a trace involving primitive forms of level 1. The group Γ 0 6) has index 12 in SL 2 Z). We choose the following 12 right coset representatives: ) 1 0 γ :=, 0 1 ) ) r γ 1/,r :=, r = 0, 1; ) ) s γ 1/2,s :=, s = 0, 1, 2; ) ) t γ 0,t :=, t = 0, 1, 2,, 4, We denote this set of coset representatives by C 6. The matrices γ C 6 are scaling matrices for the four cusps {, 1/, 1/2, 0} of the modular curve X 0 6), which have widths 1, 2,, and 6, respectively. In particular, we have γ ) =, γ 1/,r ) = 1/, γ 1/2,s ) = 1/2, and γ 0,t ) = 0.

8 8 LOCUS AND MASRI Now, let Q denote a set of primitive, reduced forms representing the equivalence classes in Q prim,1 /SL 2Z). For each Q Q, there is a unique choice of coset representative γ Q C 6 such that This induces a bijection 11) [Q γ 1 Q ] Qprim,6,1 /Γ 06). Q Q prim,6,1 /Γ 06) Q [Q γ 1 Q ]; see the proposition on page 505 in [17], or more concretely, [14, Lemma ], an explicit list of the matrices γ Q C 6 is given. Using the bijection 11) and the compatibility relation 8) for Heegner points, the trace S u n) can be expressed as 12) S u n) = fτ Q ) = f γ Q τ Q )). [Q] Q prim Dn/u 2,6,1 /Γ 06) Q Q Dn/u 2 Therefore, to study the asymptotic distribution of S u n), we need the Fourier expansion of fz) with respect to the scaling matrices γ, γ 1/,r, γ 1/2,s, and γ 0,t. Observe that the Fourier expansion of fz) at the cusp is given explicitly by see [1, equation 4.6)]) f 0 γ z) = e z) e z) + a0) + a0) = 4 2 l 6 am) = 2 l 6 am) = 2 l 6 µl) µl) µl) m=1 0<c 0 mod6/l) c,l)=1 0<c 0 mod6/l) c,l)=1 0<c 0 mod6/l) c,l)=1 1 m {am)emz) + a m)e mz)}, S l, 0; c ) lc) 2, S l, m; c ) ) 4 m I 1, m > 0, lc lc S l, m; c ) 4 ) m J 1, m < 0. lc lc Here µl) is the Möbius function, Sa, b; c) is the Kloosterman sum Sa, b; c) := ) ad + bd e, c d mod c) c,d)=1 ez) := e 2iz and I 1, J 1 are the usual Bessel functions of order 1 note that d is the multiplicative inverse of d mod c)). Since fz) is an eigenfunction for the Atkin-Lehner involutions of level 6, the Fourier expansion of fz) with respect to the scaling matrices γ 1/,r, γ 1/2,s, γ 0,t can be determined

9 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 9 from the Fourier expansion at. In particular, if ζ 6 := e1/6) is a primitive sixth root of unity, we have see for example [14,.2)]) f 0 γ 1/,r z) = 1) r [e z/2) e z/2)] + a0) + a m,r z), f 0 γ 1/2,s z) = ζ 2s 6 [e z/) e z/)] a0) + f 0 γ 0,t z) = ζ t 6 [e z/6) e z/6)] + a0) + m=1 b m,s z), m=1 c m,t z), m=1 a m,r z) := 1)r m [am)emz/2) + a m)e mz/2)], b m,s z) := 1 m [ ζ +2ms 6 am)emz/) + ζ 2ms 6 a m)e mz/) ], c m,t z) := 1 m [ ζ mt 6 am)emz/6) + ζ mt 6 a m)e mz/6) ]. More generally, given a form Q Q and corresponding coset representative γ Q C 6, we let h Q {1, 2,, 6} be the width of the cusp γ Q ) and ζ Q be the sixth root of unity such that 1) f 0 γ Q z) = ζ Q [e z/h Q ) e z/h Q )] ± a0) + C m,q z), m=1 C m,q z) := 1 m [ ζ 1 Q am)emz/h Q) + ζ Q a m)e mz/h Q ) ]. In the following lemma we give effective bounds for the Fourier coefficients am). Lemma.1. The following bounds hold. 1) For m = 0 we have 2) For m < 0 we have ) For m > 0 we have C 1 := 16 24) /2 a0) 8 4. am) C 1 m, 4 6 e 2 log2) log) ζ 2 /2). am) C 2 m exp4 m), C 2 := 2 24) /2 4 6 e 2 log2) log) eζ 2 /2).

10 10 LOCUS AND MASRI Proof. We first estimate a0). Since l, c) = 1, we can evaluate the Ramanujan sum as S l, 0; c ) = ) ld e = ) ld e = µc), c c d mod c) c,d)=1 d mod c) c,d)=1 the last equality follows from [19, equation.4)]. Therefore, S l, 0; c ) = µc) 1, and we get a0) 4 2 l l c=1 µl) 0<c 0 mod6/l) c,l)=1 0<c 0 mod6/l) c,l)=1 1 c 2 = 16 2 ζ2) = c 2 S l, 0; c ) lc 2 We next estimate am) for m < 0. Using the series J 1 x) = x 1) k x 2k 2 2 2k k!k + 1)!, x > 0 we find that k=1 14) J 1 x) x for 0 < x < 1. 2 On the other hand, by [24, page ] we have the uniform asymptotic formula { 2 [ ] [ ]} J 1 x) = cosx /4) 1 + R J) 2 x, 1) sinx /4) x 8x RJ) x, 1), the error terms satisfy the bounds see [24, equation 1.0)]) R J) 2 x, 1) < 1, R J) x, 1) < 1 for x 1. Hence 15) 16) 2 1 J 1 x) 6 for x 1. x Let M = 4 m / l. Then using the Weil bound and the estimates 14) and 15), we get am) 2 l 6 Sa, b; c) τc)a, b, c) 1/2 c 1/2 µl) 0<c 0 mod6/l) c,l)=1 ) S l, m; c J 1 M/c) lc

11 and EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION m 1/4 S m 1/2 S 2, S 1 := l 6 S 2 := l 6 l 1/4 l 1 0<c<M c,l)=1 c M c,l)=1 τc) l, m, c ) 1/2 τc) l, m, c ) 1/2 c /2. For all ɛ > 0 we have the following effective bound for the divisor function, 17) For future reference, we note that Then using 17) with ɛ = 1/2 we get Also, we have S 1 C1/2) m 1/2 l 6 τn) Cɛ)n ɛ Cɛ) := C1/2) = l 1/4 S 2 4 m 1/2 Then after combining estimates, we get C 1 := 16 24) /2 c=1 max α 0 p<2 1/ɛ α + 1 p αɛ. 4 6 e 2 log2) log). 0<c<M c,l)=1 c 1/2 8 4)/2 C1/2) m 5/4. τc) c /2 = 4 m 1/2 ζ 2 /2). am) C 1 m, m < e 2 log2) log) ζ 2 /2). Finally, we estimate am) for m > 0. Using the inequality see [21, 6.25)]) I 1 x) < exp 2x)) x expx) for x > 0 2 we find that 18) I 1 x) e x for 0 < x < 1. 2 On the other hand, by [24, equation A.4)] we have I 1 x) = i K 1xe i ) ± i e±i K 1 x), x > 0

12 12 LOCUS AND MASRI K 1 is the usual Bessel function. Then I 1 x) 1 ) K1 xe i 1 + ) K1 xe i 2 + K 1x). Now, by [24, page 26] we have the uniform asymptotic formulas K ) 1 xe i [ = ±i 2x expx) 1 + R K) 1 xe i, 1 )] and [ ] K 1 x) = 2x exp x) 1 + R K) 1 x, 1), the error terms satisfy the bounds see [24, equations 1.26) and B.1)]) R K) 1 xe i, 1 ) 8 sup Λ 1 2xe i r ) 1 for x 1 r 1 and see [24, equation 1.0)]) R K) 1 x, 1) < 1 for x 1. Here for p > 0 and ω 0, we have Λ p ω) := ω p e ω Γ1 p, ω) Γ1 p, ω) is the incomplete Gamma function. Hence we get ) I 1 x) 2 expx) + exp x)) 4 expx) for x 1. x x Again, let M = 4 m/ l. Then using the estimates 16), 18) and 19), we get am) 2 ) S l, m; c µl) I 1 M/c) l 6 lc and S := l 6 0<c 0 mod6/l) c,l)=1 4 2m 1/4 S em 1/2 S 4, l 1/4 S 4 := l 6 0<c<M c,l)=1 l 1 τc) l, m, c ) 1/2 exp4 m/ lc) c M c,l)=1 τc) l, m, c ) 1/2 As before, using 17) with ɛ = 1/2 we get S C1/2)m 1/2 exp 4 m ) c /2. l 6 l 1/4 8 4)/2 C1/2)m 5/4 exp 4 m ), 0<c<M c,l)=1 c 1/2

13 and Then combining estimates yields EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 1 S 4 = S 2 4m 1/2 c=1 τc) c /2 = 4m1/2 ζ 2 /2). am) C 2 mexp 4 m ), m > 0 C 2 := 2 24) /2 4 6 e 2 log2) log) eζ 2 /2). We are now in position to prove Theorem 1.4, which we restate for the convenience of the reader. Theorem.2. For all n 1, we have with Sn) = 224n 1) 1/4 I 1/2 24n 1/6 ) + En), En) < )2 qn)+1 24n 1) 2 exp 24n 1/12) qn) := Proof. By 10), 12) and 1), we have Sn) = hd εu) n/u 2 )+1 S u n) u>0 u 2 D n = hd εu) n/u 2 )+1 u>0 u 2 D n = hd εu) n/u 2 )+1 u>0 u 2 D n log24n 1) loglog24n 1)) Q Q Dn/u 2 Q Q Dn/u 2 E 1 n) := ±a0) hd εu) n/u 2 )+1 hd n /u 2 ) + u>0 u 2 D n f 0 γ Q τ Q ) ζ Q [e τ Q /h Q ) e τ Q /h Q )] + E 1 n), hd εu) n/u 2 )+1 m=1 u>0 Q Q u 2 Dn/u 2 D n We first give an effective upper bound for E 1 n). We have hd 20) εu) n/u 2 )+1 hd n /u 2 ) hd n /u 2 ) = HD n ), u>0 u>0 u 2 D n u 2 D n C m,q τ Q ).

14 14 LOCUS AND MASRI and thus Lemma.1 gives ±a0) hd εu) n/u 2 )+1 hd n /u 2 ) 8 u>0 4 HD n ). u 2 D n Next, using the definition of C m,q z) we get hd εu) n/u 2 )+1 C m,q τ Q ) m=1 u>0 Q Q u 2 Dn/u 2 D n am) m ζ 1 Q emτ Q/h Q ) + m=1 Observe that u>0 Q Q u 2 Dn/u 2 D n m=1 e mτ Q /h Q ) = ζ b Qm 2a Q h Q exp a m) m m D n /u 2 a Q h Q u>0 Q Q u 2 Dn/u 2 D n ), ζ Q e mτ Q /h Q ). ζ 2aQ h Q is the primitive 2a Q h Q -th root of unity ) 1 ζ 2aQ h Q := e. 2a Q h Q Since Q Q Dn/u 2 is reduced, the corresponding Heegner point τ Q lies in the standard fundamental domain F for SL 2 Z). In particular, we have Dn /u Imτ Q ) = 2 /2, 2a Q which implies that Since h Q 6, we have 21) a Q m D n /u 2 a Q h Q Then using 20) and 21), we get ζ 1 Q emτ Q/h Q ) Q Q Dn/u 2 Dn /u 2. Q Q Dn/u 2 m 4. exp m ) Dn /u 2 a Q h Q Similarly, since hd n /u 2 )exp m/4 ). ) e mτ Q /h Q ) = ζ b Qm Dn /u 2a Q h Q exp m 2, a Q h Q

15 the same argument gives EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 15 Q Q Dn/u 2 Hence by Lemma.1, we have hd εu) n/u 2 )+1 m=1 u>0 u 2 D n { HD n ) C 1 m=1 ζ Q e mτ Q /h Q ) hd n/u 2 )exp m/4 ). Q Q Dn/u 2 C m,q τ Q ) m 1/2 exp m/4 ) + C 2 Combining the preceding estimates yields E 1 n) { 8 4 HD n ) + HD n ) C 1 m=1 m=1 m 1/2 exp m/4 ) + C 2 We now estimate the infinite sums. First, we have m 1/2 exp m/4 ) m=1 Next, write and observe that Then if m 769, we have so that 0 x 1/2 exp x/4 )dx = m 1/2 exp 4 m m/4 )}. m=1 4 m m/4 = m 4 4 ), m 4 4 > 0 m > 768. m 4 4 m 4 4 > m 4 4 ) m m 4 4 ). 769 m 1/2 exp 4 m m/4 )}. 4 ) /2 4 ) /2 Γ/2) = 2. Now, we split the sum into appropriate ranges and use the preceding bound to get m 1/2 exp 4 m m/4 ) m=1 768 = m=1 m 1/2 exp 4 m m/4 ) + m=769 m 1/2 exp m 4 4 )) m

16 16 LOCUS AND MASRI 768 m=1 m 1/2 exp 4 m m/4 ) + A calculation shows that 768 m 1/2 exp 4 m m ) 4 m=1 Also, if we let then m= β := m 1/2 exp m 4 4 )) 769 Combining things, we have shown that m=769 m 1/2 exp m 4 4 )). 769 x 1/2 exp 4 x x ) 4 dx < E 1 n) Cβ)HD n ), x 1/2 exp βx) dx = β /2 Γ/2, 769 β). Cβ) is the explicit positive constant defined by Cβ) := 8 4 ) /2 4 + C C [ ) + β /2 Γ/2, 769 β) ]. We now find an explicit upper bound for Cβ). We have < 260. Also, since and ζ/2) 2.61 <, we have so that C1/2) = 4 6 e 2 log2) log) 1.74 < 2 C 1 = 16 24) /2 C1/2) ζ 2 /2) < 2 24) / < 48, C 1 4 ) /2 2 < 48 4 ) / < Similarly, we have C 2 = 2 24) /2 C1/2) eζ 2 /2) < 64 24) / e < 5208, and Γ/2, 769 β) 0.82 < 1, so that β / < C 2 [ ) + β /2 Γ/2, 769 β) ] < 5208 [ ) ] < After combining estimates, we get Cβ) < ) <

17 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 17 Summarizing, we have shown that E 1 n) < )HD n ). It remains to analyze the main term. We write the main term as hd εu) n/u 2 )+1 ζ Q [e τ Q /h Q ) e τ Q /h Q )] u>0 Q Q u 2 Dn/u 2 D n = ζ Q [e τ Q /h Q ) e τ Q /h Q )] + E 2 n), Q Q Dn 22) and 2) E 2 n) := hd εu) n/u 2 )+1 u 2 u 2 D n Q Q Dn/u 2 Observe that for any form Q = [a Q, b Q, c Q ] Q Dn/u 2, we have a Q h Q 0 mod 6) ζ Q [e τ Q /h Q ) e τ Q /h Q )]. e τ Q /h Q ) e τ Q /h Q ) = ζ b Q 2a Q h Q [exp ) )] D n /u 2 exp D n /u 2. a Q h Q a Q h Q Now, by [14, 4.2)] there are exactly 4 forms Q Q Dn with a Q h Q = 6, and these are given by Q 1 = [1, 1, 6n], Q 2 = [2, 1, n], Q = [, 1, 2n], Q 4 = [6, 1, n]. Moreover, the corresponding coset representatives γ Qi C 6 such that are given by Write [Q i γ 1 Q i ] Q prim D n,6,1 /Γ 06) γ Q1 = γ 0,1, γ Q2 = γ 1/2, 1, γ Q = γ 1/,0, γ Q4 = γ. Q Q Dn ζ Q [e τ Q /h Q ) e τ Q /h Q )] = E n) := 4 ζ Qi [e τ Qi /h Qi ) e τ Qi /h Qi )] + E n), i=1 Q Q Dn Q Q i ζ Q [e τ Q /h Q ) e τ Q /h Q )]. By 22) we have a Q h Q 12 for all Q Q i, hence using 2) we get E n) [ exp D n /a Q h Q ) exp ] D n /a Q h Q ) Q Q Dn Q Q i Q Q Dn Q Q i exp Dn /a Q h Q )

18 18 LOCUS AND MASRI hd n )exp D n /12). Similarly, by 22) we have a Q h Q 6, so that Dn /u 2 a Q h Q D n /12 for u 2. Hence by 2) we have 24) E 2 n) u 2 u 2 D n u 2 u 2 D n Next, using the identity we have Q Q Dn/u 2 Q Q Dn/u 2 [ ) )] D n /u exp 2 exp D n /u 2 a Q h Q a Q h Q ) D n /u exp 2 a Q h Q HD n )exp D n /12). I 1/2 z) = 2 expz) exp z)) z 2 e z) e z) = 2 yi 1/2 2y)e x), z = x + iy. Therefore, since a Qi h Qi = 6 for i = 1, 2,, 4, we get 4 ζ Qi [e τ Qi /h Qi ) e τ Qi /h Qi )] = [2 D n 1/4 I 1/2 ] D n /6) expi/6) 12 i=1 4 ζ Qi. Also, from the Fourier expansion of fz) with respect to γ 0,1, γ 1/2, 1, γ 1/,0, and γ given previously, we have Hence expi/6) ζ Q1 = ζ 1 6, ζ Q2 = ζ 2 1) 6, ζ Q = 1) 0, ζ Q4 = 1. 4 ζ Qi = expi/6) i=1 i=1 ) ζ6 1 + ζ 2 1) 6 + 1) = 4 cos/6) = 2. Then simplifying yields 4 ζ Qi [e τ Qi /h Qi ) e τ Qi /h Qi )] = 2 D n 1/4 I 1/2 D n /6). i=1 By combining the preceding results, we get En) := E 1 n) + E 2 n) + E n) with Sn) = 2 D n 1/4 I 1/2 D n /6) + En), En) E 1 n) + E 2 n) + E n) < 2HD n )exp D n /12) )HD n ) < )HD n )exp D n /12).

19 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 19 To complete the proof, we require only a crude effective upper bound for the Hurwitz- Kronecker class number HD n ). Write D n = d n f 2 n with d n < 0 a fundamental discriminant and f n Z +. Then we have the class number relation 25) Inserting the formula see e.g. [12, p. 2]) into 25) yields Now, a simple estimate yields HD n ) = hd n /u 2 ) = hu 2 d n ). u>0 u>0 u 2 D n u f n hu 2 d n ) = u p u 1 χ ) d n p) hd n ) p HD n ) = δn)hd n ), δn) := u u>0 u f n p u 1 χ ) d n p). p δn) D n τ D n )2 ω Dn ) ω D n ) is the number of prime divisors of D n. We have τ D n ) < 2 D n, and by [26, Théorème 1] we have { } log D n ) ω D n ) max 1, loglog D n )) Hence Using the class number formula and the evaluation another simple estimate yields δn) 2 qn)+1 D n. hd n ) = Lχ dn, 1) = d n /2 dn Lχ d n, 1) hd n ) d n. Then combining the preceding estimates gives d n 1 t=1 HD n ) 2 qn)+1 D n 2. Finally, using the class number bound we get log D n ) loglog D n )) χ dn t)t, En) < )2 qn)+1 D n 2 exp D n /12). =: qn).

20 20 LOCUS AND MASRI This completes the proof. 4. Effective asymptotics for sptn) To prove our results, we will require an effective asymptotic formula for pn) due to Lehmer [20]. For convenience, define µn) := 6 24n 1. Inspired by the Hardy-Ramanujan asymptotic for pn), Rademacher [25] obtained the following exact formula: 2 ) A c n) µn) pn) = I 24n 1) /4 /2, c c A c n) is the Kloosterman sum and sd, c) is the Dedekind sum A c n) := sd, c) := c 1 r=1 c=1 d mod c) d,c)=1 r dr c c e isd,c) e 2idn c dr 1 ). c 2 Using Rademacher s formula, Lehmer [20] proved the following result. Theorem 4.1 Lehmer). For all n 1, we have 12 N { A c n) pn) = 1 c 24n 1 c µn) c=1 R 2 n, N) < N 2/ 2 { N ) e µn)/c c ) } e µn)/c + R 2 n, N), µn) 2µn) e µn)/n e µn)/n) N 2 µn) 2 We first use Theorem 4.1 to quickly deduce the following effective bound. Lemma 4.2. For all n 1, we have Proof. Using the identity 26) pn) = 2 24n 1 I /2 x) = ) e µn) + E p n), µn) E p n) 124)e µn)/2. [ ) e x ) ] e x, x x x we may write Theorem 4.1 with the choice N = 2) as 2 2 ) A c n) µn) 27) pn) = I 24n 1) /4 /2 c c c=1 + R 2 n, 2), }.

21 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 21 [ ) { } R 2 n, 2) < 2 2 e µn)/2 e µn)/2) + 1/6 2 2/ µn) 2 ) ] 2 2. µn) Now, using 27) and 26) we get E p n) := pn) = 2 24n 1) I 2 A 2 n) /4 /2 µn)) + 24n 1) / ) e µn) + E p n), µn) = 2 24n 1 [ n 1) /4 2 2 µn) ) µn) ) µn) I /2 + R 2 n, 2) 2 e µn) + A ) ] 2n) µn) I /2 + R 2 n, 2). 2 2 By [21, 6.24)], we have the bound x ) / ) I /2 x) < 2 Γ5/2) e x )e x x /2 e x, x > 0. Then an estimate using the trivial bound A c n) < c and the bound 28) yields ) 2 A 2 n) µn) I 24n 1) /4 /2 12e µn)/ Similarly, two straightforward estimates yield n 1) /4 2 µn) and Hence µn) R 2 n, 2) < 1296)e µn)/2. E p n) 124)e µn)/2. ) e µn) 16e µn)/ Proof of Theorem 1.1. Using 4), the formula 6) can be written as 29) sptn) = 1 [Sn) 24n 1)pn)]. 12 Now, a straightforward calculation using 24) shows that 224n 1) 1/4 I 1/2 µn)) = 2 e µn) + E 4 n), E 4 n) 24n 1)e µn)/2.

22 22 LOCUS AND MASRI Hence the asymptotic formula in Theorem 1.4 can be written as 0) Sn) = 2 e µn) + Ẽn), Ẽn) := En) + E 4n) with Ẽn) En) + E 4n) < )2 qn)+1 24n 1) 2 e µn)/2. Then using 29), 0), and Lemma 4.2, another straightforward calculation yields the error term satisfies the bound sptn) = 24n 1 eµn) + E s n), E s n) := Ẽn) 12 24n 1 E p n) 12 E s n) < )2 qn) 24n 1) 2 e µn)/2. As pointed out by Bessenrodt and Ono [5], it is straightforward to obtain from Theorem 4.1 that 1 1 ) e µn) < pn) < ) e µn) 12n n 12n n for all n 1. We will use Theorem 1.1 to prove the following analogous statement for sptn), n is replaced by any positive integral power of n. Theorem 4.. For each λ Z + and k Z +, there is an effective positive integer B k λ) such that for all n B k λ), we have 1 1 ) e µn) < sptn) < 24n 1 λn k ) e µn). 24n 1 λn k Proof. By Theorem 1.1 we have the bounds 24n 1 eµn) E s n) < sptn) < 24n 1 eµn) + E s n), E s n) < )2 qn) 24n 1) 2 e µn)/2. Clearly, there is an effective positive integer B k λ) such that the inequality )2 qn) 24n 1) 2 e µn)/2 < 24n 1 1 λn k eµn) holds for all n B k λ). For instance, if λ = k = 1 then B 1 1) = This completes the proof.

23 1) and 2) EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 2 5. Proof of Theorem 1. By Theorem 1.1 and Lemma 4.2, we may write αn) := Also, for ɛ > 0 we define 24n 1 γn) := sptn) = αn)e µn) + E s n) pn) = βn)e µn) + E p n), 6 2, βn) := 1 24n 1 ) 6 n. n, γn, ɛ) := + ɛ ) 6. 24n 1 We must prove that there exists an effective positive constant Nɛ) > 0 such that for all n Nɛ), we have ) 4) γn)pn) < sptn) < γn, ɛ)pn). First, using 1) and 2) we find that the lower bound in ) is equivalent to c 1 n)e µn) > γn)e p n) E s n), c 1 n) := αn) βn)γn). Now, the error bounds in Lemma 4.2 and Theorem 1.1 imply that γn)e p n) E s n) c 2 n)e µn)/2, c 2 n) := 124)γn) )2 qn) 24n 1) 2. Then noting that c 1 n) > 0 for all n 1, we find that 4) is implied by the bound or equivalently, the bound 5) e µn)/2 > c n) := c 2n) c 1 n), [ 12 n > 1 ) 2 24 logc n)) + 1]. A calculation shows that 5) holds for all n N := Similarly, using 1) and 2) we find that the upper bound in ) is equivalent to 6) c 4 n, ɛ)e µn) > E s n) γn, ɛ)e p n), c 4 n, ɛ) := βn)γn, ɛ) αn). The error bounds in Lemma 4.2 and Theorem 1.1 imply that E s n) γn, ɛ)e p n) c 5 n, ɛ)e µn)/2, c 5 n, ɛ) := 124)γn, ɛ) )2 qn) 24n 1) 2.

24 24 LOCUS AND MASRI Moreover, there exists an effective positive constant N 1 ɛ) > 0 such that c 4 n, ɛ) > 0 for all n N 1 ɛ). Then arguing as above, we find that if n N 1 ɛ), the bound 6) is implied by the bound [ 12 n > 1 ) 2 7) 24 logc 6n, ɛ)) + 1], c 6 n, ɛ) := c 5 n, ɛ)/c 4 n, ɛ). Clearly, there exists an effective positive constant N 2 ɛ) N 1 ɛ) such that 7) holds for all n N 2 ɛ). Let Nɛ) := max{n, N 2 ɛ)}. Then the inequalities ) hold for all n Nɛ). 6. Proof of Theorem Proof of Conjecture 1). Let ɛ = 1 6/ in Theorem 1.. We need to determine the constant N1 6/). A calculation shows that the inequality c 4 n, 1 6/) > 0 holds if n N 1 1 6/) N 1 1 6/) = 4. Next, we need to find the smallest positive integer N 2 1 6/) 4 such that the bound [ 12 n > 1 24 logc 6n, 1 ) 2 6/)) + 1] holds for all n N 2 1 6/). A calculation shows that this constant is given by N 2 1 6/) = We now have N1 6/) := max{n, N 2 1 6/)} = max{12788, 11997} = Therefore, the inequalities 6 npn) < sptn) < npn) hold for all n Finally, one can verify with a computer that these inequalities also hold for 5 n < Proof of Conjecture 2). We follow closely the proof of [5, Theorem 2.1]. By taking λ = k = 1 in Theorem 4. recall that B 1 1) = 1500), we find that 8) 1 1 ) e µn) < sptn) < 24n 1 n ) e µn) 24n 1 n holds for all n One can verify with a computer that 8) also holds for 1 n < Now, assume that 1 < a b, and let b = Ca C 1. From 8) we get the inequalities spta)sptca) > 2 24a ) 1 1 ) e µa)+µca) 24Ca 1 a Ca and spta + Ca) < 24a + Ca) 1 ) e µa+ca). a + Ca

25 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 25 Hence, for all but finitely many cases, it suffices to find conditions on a > 1 and C 1 such that 9) e µa)+µca) µa+ca) > 24a 1 ) 24Ca a+ca ) ). 24a + Ca) a 1 1 Ca For convenience, define ) a+ca T a C) := µa) + µca) µa + Ca) and S a C) := ) ). 1 1 a 1 1 Ca Then by taking logarithms, we find that 9) is equivalent to 24a 1 ) 24Ca 1 40) T a C) > log + log S a C). 24a + Ca) 1 As functions of C, it can be shown that T a C) is increasing and S a C) is decreasing for C 1, and thus T a C) T a 1) and logs a 1)) logs a C)). Hence it suffices to show that T a 1) > log 24a 1 24Ca 1 24a + Ca) 1 ) + logs a 1)). Moreover, since 24Ca 1 24a + Ca) 1 1 for all C 1 and all a > 1, it suffices to show that ) 24a 1 41) T a 1) > log + logs a 1)). By computing the values T a 1) and S a 1), we find that 41) holds for all a 5. To complete the proof, assume that 2 a 4. For each such integer a, we calculate the real number C a for which ) 24a 1 T a C a ) = log + logs a C a )). The values C a are listed in the table below. a C a By the discussion above, if b = Ca a is an integer for which C > C a, then 40) holds, which in turn gives the theorem in these cases. Only finitely many cases remain, namely the pairs of integers 2 a 4 and 1 b/a C a. We compute spta), sptb), and spta + b) in these cases to complete the proof.

26 26 LOCUS AND MASRI 6.. Proof of Conjecture ). We require some lemmas and a proposition analogous to those of Desalvo and Pak [1] in order to prove the remaining conjectures. The following is [1, Lemma 2.1]. Lemma 6.1. Suppose hx) is a positive, increasing function with two continuous derivatives for all x > 0, and that h x) > 0 is decreasing, and h x) < 0 is increasing for all x > 0. Then for all x > 0, we have By Theorem 1.1, we may write h x 1) < hx + 1) 2hx) + hx 1) < h x + 1). 42) fn) := and Lemma 6.2. Let Then for all n 4, we have sptn) = fn) + E s n) 24n 1 eµn) E s n) )2 qn) 24n 1) 2 e µn)/2. F n) := 2 logfn)) logfn + 1)) logfn 1)). 24 ) /2 1 24n + 1) 1 n < F n) < ) /2 ) 2. 24n 1) 1 24n + 1) 1 Proof. We can write fn) from 42) as so that Then we have fn) = 1 2 µn) eµn), logfn)) = µn) logµn)) log2 ). F n) = 2µn) µn + 1) µn 1) 2 logµn)) + logµn + 1)) + logµn 1)). Since the functions µx) and µx) := logµx)) satisfy the hypotheses of Lemma 6.1, we get Computing derivatives gives 24 24n + 1) 1 ) /2 µ n + 1) + µ n 1) < F n) < µ n 1) + µ n + 1) n 1) 1 ) 2 < F n) < for all n 2, from which we deduce that 24 24n 1) 1 ) / n + 1) 1 ) 2 24 ) /2 1 24n + 1) 1 n < F n) < ) /2 ) 2 24n 1) 1 24n + 1) 1 for all n 4.

27 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 27 Lemma 6.. Define the functions y n := E s n) /fn), and Mn) := )µn)2 qn) 24n 1) 2 e µn)/2, gn) := Mn) 1 Mn). Then for all n 2, we have [ ] 1 y n ) 2 log > 2gn) Mn + 1) Mn 1) 1 + y n+1 ) 1 + y n 1 ) and [ ] 1 + y n ) 2 log < 2Mn) + gn + 1) + gn 1). 1 y n+1 ) 1 y n 1 ) Proof. First observe that for all n 1, we have 4) 0 < y n := E sn) fn) < )2 qn) 24n 1) 2 e µn)/2 1 2 µn) eµn) = Mn). Note also that y n < 1 for all n 1. Then using 4) and the inequalities log1 x) x 1 x for 0 < x < 1 and we get log1 + x) < x for x > 0, [ ] 1 y n ) 2 log = 2 log 1 y n ) log 1 + y n+1 ) log 1 + y n 1 ) 1 + y n+1 ) 1 + y n 1 ) > 2 y n 1 y n y n+1 y n 1 > 2gn) Mn + 1) Mn 1) for all n 2. Similarly, we get [ ] 1 + y n ) 2 log = 2 log1 + y n ) log1 y n+1 ) log1 y n 1 ) 1 y n+1 ) 1 y n 1 ) < 2y n + y n+1 + y n 1 1 y n+1 1 y n 1 < 2Mn) + gn + 1) + gn 1) for all n 2. Proposition 6.4. Let spt 2 n) := 2 logsptn)) logsptn + 1)) logsptn 1)).

28 28 LOCUS AND MASRI Then we have 1 24n) /2 < spt 2n) < 2 n /2, the lower bound holds for all n and the upper bound holds for all n Proof. We first bound sptn) by fn) 1 E ) sn) fn) Then recalling that < sptn) < fn) 1 + E ) sn). fn) F n) := 2 logfn)) logfn + 1)) logfn 1)) and y n := E s n) /fn), we take logarithms in the preceding inequalities to get [ ] [ ] 1 y n ) y n ) 2 F n) + log < spt 1 + y n+1 ) 1 + y n 1 ) 2 n) < F n) + log. 1 y n+1 ) 1 y n 1 ) It follows immediately from Lemmas 6.2 and 6. that for all n 4, we have spt 2 n) > 24 ) /2 1 2gn) Mn + 1) Mn 1) 24n + 1) 1 n2 and 44) spt 2 n) < ) /2 ) 2 + 2Mn) + gn + 1) + gn 1). 24n 1) 1 24n + 1) 1 Finally, a calculation shows that for all n and 24 24n + 1) 1 ) /2 1 n 2 2gn) Mn + 1) Mn 1) > 1 24n) / ) /2 ) 2 + 2Mn) + gn + 1) + gn 1) < 2 24n 1) 1 24n + 1) 1 n /2 for all n This completes the proof. To prove Conjecture ), we must show that sptn) 2 > sptn 1)sptn + 1) for n 6. Taking logarithms, we see that this is equivalent to spt 2 n) > 0. By the lower bound in Proposition 6.4, we have spt 2 n) > 0 for all n Finally, one can verify with a computer that spt 2 n) > 0 for all 6 n < This completes the proof.

29 EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION Proof of Conjecture 4). We follow closely the proof of [1, Theorem 5.1]. Recall that a sequence {ak)} k=0 of non-negative integers is log-concave if ak) 2 ak 1)ak + 1) for all k 1. Moreover, it is known that log-concavity implies strong log-concavity al i)ak + i) ak)al), for all 0 k l n and 0 i l k see e.g. [27]). Now, we have proved that sptn) 2 > sptn 1)sptn + 1) for all n 6. Therefore, if we take k = n m, l = n + m, and i = m, then sptn) 2 > sptn m)sptn + m) for all n > m > 1 with n m > 6. We next consider the case n > m > 1 with 1 n m 6. We will prove that 45) sptn) 2 sptm + 1) 2 > spt6)spt6 + 2m) sptn m)sptn + m) for all 1 n m 6 with m On the other hand, one can verify with a computer that sptn) 2 > sptn m)sptn + m) for all 1 n m 6 with m < This completes the proof of Conjecture 4), subject to verifying the inequalities 45). Since n m + 1, we have Moreover, since n m 6 we have 46) and thus sptn) 2 sptm + 1) 2. sptn m) < spt6) < 90000, spt6)spt6 + 2m) sptn m)sptn + m). This verifies the first and third inequalities in 45). It remains to prove that 47) sptm + 1) 2 > spt6)spt6 + 2m) for all m Taking logarithms in 47), we see that it suffices to prove 48) 2 logsptm + 1)) logspt6)) logspt6 + 2m)) > 0 for all m By [18] and [15], respectively, we have the lower and upper bounds e 2 m 2me 1/6m < pm) < e 2/)m. Then by Theorem 1. with the choice ɛ = 2 2) /, we get 6m e 2 m 12m 49) < sptm) < e 2/)m. 2me1/6m

30 0 LOCUS AND MASRI Using the inequalities 46) and 49), we see that 48) is implied by the lower bound ) 6m + 1) 2 log 2 2 m + 1)e 1/6m+1) +4 ) m) m + 1 log90000) log m) > 0 for all m A calculation shows this is true for the given range of m Proof of Conjecture 5). Taking logarithms, we find that Conjecture 5) is equivalent to spt 2 n) < log ) n for all n 1. By the upper bound in Proposition 6.4 and some straightforward estimates, we have spt 2 n) < 2 n < 1 /2 n + 1 < log ) n for all n Finally, one can verify with a computer that the conjectured inequality holds for all 1 n < This completes the proof Proof of Conjecture 6). We follow closely the proof of [11, Conjecture 1.]. Taking logarithms, we find that Conjecture 6) is equivalent to ) spt 2 n) < log n /2 for all n 7. By 44) we have spt 2 n) < ) /2 ) 2 + 2Mn) + gn + 1) + gn 1) 24n 1) 1 24n + 1) 1 for all n 4. On the other hand, by [11, 2.)] we have 24 ) /2 < 24 24n + 1) 1 24n) / ) /2 ) 2 + 2n 5/2 for all n 50, and by the second inequality following [11, 2.2)] we have 288 ) 2 < 1 24n + 1) 1 6n 1 5/2 2n 2 for all n 50. Therefore, for all n 50 we have spt 2 n) < 24 ) n) + 5 /2 24) /2 n 1 + 2Mn) + gn + 1) + gn 1). 5/2 2n2 Now, a calculation shows that 5 n 1 + 2Mn) + gn + 1) + gn 1) < 0 5/2 2n2 for all n Hence spt 2 n) < 24 24n) /2 for all n Then using the inequality ) 2 24 = ) 24) /2 24n) /2 24n) /2 x1 x) < log1 + x) for x > 0,

31 we get EFFECTIVE BOUNDS FOR THE ANDREWS SPT-FUNCTION 1 spt 2 n) < log ) ) = log n) /2 24n /2 for all n Finally, one can verify with a computer that this inequality also holds for all 7 n < This completes the proof. References [1] S. Ahlgren and N. Andersen, Algebraic and transcendental formulas for the smallest parts function. Adv. Math ), [2] S. Ahlgren and N. Andersen, Kloosterman sums and Maass cusp forms of half integral weight for the modular group. International Mathematics Research Notices, Vol. 2016, [] G. E. Andrews, The number of smallest parts in the partitions of n. J. Reine Angew. Math ), [4] J. Banks, A. Barquero-Sanchez, R. Masri, and Y. Sheng, The asymptotic distribution of Andrews smallest parts function. Arch. Math., ), [5] C. Bessenrodt and K. Ono, Maximal multiplicative properties of partitions. Ann. Comb ), [6] K. Bringmann, On the explicit construction of higher deformations of partition statistics. Duke Math. J ), [7] K. Bringmann and K. Ono, An arithmetic formula for the partition function. Proc. Amer. Math. Soc ), [8] J. H. Bruinier and K. Ono, Algebraic formulas for the coefficients of half-integral weight harmonic weak Maass forms. Adv. Math ), [9] J. H. Bruinier, K. Ono, and A. V. Sutherland, Class polynomials for nonholomorphic modular functions. J. Number Theory ), [10] W. Y. Chen, The spt-function of Andrews. arxiv: [11] W. Y. Chen, L. X. Wang, and G. Y. Xie, Finite differences of the logarithm of the partition function. Math. Comp ), [12] H. Cohen, A course in computational algebraic number theory. Graduate Texts in Mathematics, 18. Springer-Verlag, Berlin, [1] S. Desalvo and I. Pak, Log-concavity of the partition function. Ramanujan J ), [14] M. Dewar and R. Murty, A derivation of the Hardy-Ramanujan formula from an arithmetic formula. Proc. Amer. Math. Soc ), [15] P. Erdös, On an elementary proof of some asymptotic formulas in the theory of partitions. Ann. Math ), [16] A. Folsom and R. Masri, Equidistribution of Heegner points and the partition function. Math. Ann ), [17] B. Gross, W. Kohnen, and D. Zagier, Heegner points and derivatives of L-series. II. Math. Ann ), [18] G. H. Hardy and S. Ramanujan, Asymptotic formulae in combinatory analysis. Proc. London Math. Soc ), [19] H. Iwaniec and E. Kowalski, Analytic number theory. American Mathematical Society Colloquium Publications, 5. American Mathematical Society, Providence, RI, xii+615 pp. [20] D. H. Lehmer, On the remainders and convergence of the series for the partition function. Trans. Amer. Math. Soc ), [21] Y. L. Luke, Inequalities for generalized hypergeometric functions. J. Approx. Theory ), [22] R. Masri, Fourier coefficients of harmonic weak Maass forms and the partition function. Amer. J. Math ), [2] R. Masri, Singular moduli and the distribution of partition ranks modulo 2. Math. Proc. Cambridge Philos. Soc ), [24] G. Nemes, Error bounds for the large-argument asymptotic expansions of the Hankel and Bessel functions. arxiv: v2

32 2 LOCUS AND MASRI [25] H. Rademacher, On the expansion of the partition function in a series. Ann. of Math ), [26] G. Robin, Estimation de la fonction de Tchebychef θ sur le k-ième nombre premier et grandes valeurs de la fonction ωn) nombre de diviseurs premiers de n. French) [Estimate of the Chebyshev function θ on the k-th prime number and large values of the number of prime divisors function ωn) of n] Acta Arith ), [27] B. E. Sagan, Inductive and injective proofs of log concavity results. Discret. Math ), Department of Mathematics and Computer Science, Emory University, Atlanta, GA 022 address: madeline.locus@emory.edu Department of Mathematics, Mailstop 68, Texas A&M University, College Station, TX address: masri@math.tamu.edu