PROOF OF THE ALDER-ANDREWS CONJECTURE

Transcription

1 PROOF OF THE ALDER-ANDREWS CONJECTURE CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER Abstract Motivated by classical identities of Euler Schur Rogers Ramanujan Alder investigated q d n Q d n the number of partitions of n into d-distinct parts into parts which are ± mod d + 3 respectively He conjectured that q d n Q d n Andrews Yee proved the conjecture for d = s also for d 3 We complete the proof of Andrews s refinement of Alder s conjecture by determining effective asymptotic estimates for these partition functions correcting refining earlier work of Meinardus thereby reducing the conjecture to a finite computation Introduction Statement of Results The famous identity of Euler states that the number of partitions into odd parts equals the number of partitions into distinct parts the first Rogers-Ramanujan identity tells us that the number of partitions into parts which are ± mod 5 equals the number of partitions into parts which are -distinct a d-distinct partition is one the difference between any two parts is at least d Another related identity is a theorem of Schur which states that the partitions of n into parts which are ± mod 6 are in bijection with the partitions of n into 3-distinct parts no consecutive multiples of 3 appear In 956 these three facts encouraged HL Alder to consider the partition functions q d n := pn d-distinct parts Q d n := pn parts ± mod d + 3 Conjecture Alder If d n = q d n Q d n then for any d n we have that d n 0 By the above discussion the conjecture is true for d 3 the inequality can be replaced by an equality for d = Large tables of values seem to suggest however that q d n Q d n are rarely equal Andrews [] refined Alder s conjecture see [3] for more information on this conjecture: Conjecture Alder-Andrews For 4 d 7 n d + 9 or d 8 n d + 6 d n > 0 Remark For any given d there are only finitely many n not covered by the Alder-Andrews conjecture a simple argument shows that d n 0 for these n In essence Alder s conjecture asks us to relate the coefficients of Q d nq n = q n d+ q n n=0 n= q d nq n = n=0 n=0 q dn +n q q q n

2 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER Although the first generating function is essentially a weight 0 modular form the second is generally not modular except in the cases d = This is the root of the difficulty in proving Alder s conjecture since the task is to relate Fourier coefficients of two functions which have different analytic properties However there have been several significant advances toward proving Alder s conjecture Using combinatorial methods Andrews [] proved that Alder s conjecture holds for all values of d which are of the form s s 4 In addition Yee [9] [0] proved that the conjecture holds for d = 7 for all d 3 These results are of great importance because they resolve the conjecture except for 4 d 30 d 7 5 In addition Andrews [] deduced that lim n d n = + using powerful results of Meinardus [6] [7] which give asymptotic ressions for the coefficients q d n Q d n Unfortunately a mistake in [7] implies that one must argue further to establish this limit We correct the proof of Meinardus s main theorem see the discussion after 3 show that the statement of the theorem remains unchanged We first prove the following result which can be made licit: Theorem Let d 4 let α [0 ] be the root of α d + α = 0 If A := d log α + α rd r= then for every positive integer n we have r A /4 d n = πα d dα d + n 3/4 na + E d n E d n = O n 5 6 na Remark The main term of d n is the same as the main term for q d n cf Theorem 3 In the course of proving Theorem we derive licit approximations for Q d n q d n see Theorems 3 respectively Using these results we obtain the following: Theorem The Alder-Andrews Conjecture is true In order to prove Theorems we consider q d n Q d n independently then compare the resulting effective estimates Accordingly in Section we give licit asymptotics for Q d n culminating in Theorem Next in Section 3 we laboriously make Meinardus s argument effective correct in order to give an licit asymptotic formula for q d n in Theorem 3 In Section 4 we use the results from Sections 3 to prove Theorems Acknowledgements The authors thank Ken Ono for advising us on this project during the student number theory seminar at UW-Madison The authors would also like to thank Kathrin Bringmann Kimmo Eriksson for their helpful comments Estimate of Q d n with licit error bound As before let Q d n denote the number of partitions of n whose parts are ± mod d + 3 From the work of Meinardus we have that Q d n 3d sin n 3 4 π n π 3d + 3

3 PROOF OF THE ALDER-ANDREWS CONJECTURE 3 In this formula only the order of the error is known We will bound the error licitly following closely the method of Meinardus [6] as it is presented by Andrews in Chapter 6 of [] This allows us to prove the following theorem: Theorem If d 4 n is a positive integer then Q d n = 3d sin n 3 4 π n π + Rn 3d + 3 Rn is an licitly bounded function see 0 at the end of this section Remark An exact formula for Q d n is known due to the work of Subrahmanyasastri [8] In addition by using Maass-Poincaré series Bringmann Ono [5] obtained exact formulas in a much more general setting However we do not employ these results since the formulas are extremely complicated Theorems do not require this level of precision Preliminary Facts Consider the generating function f associated to Q d n fτ := q n = + Q d nq n n ± n 0 q = e τ Rτ > 0 Let τ = y + πix We can then obtain a formula for Q d n by integrating fτ against e nτ Consequently we require an approximation of fτ so that we may make use of this integral formula To do this we need an additional function gτ := q n n ± n 0 Lemma If arg τ > π x then Rgτ gy c 4 y c is an licitly given constant depending only on d Proof For notational convenience we will consider the ression y Rgτ gy Exping we find that y Rgτ gy = S + S + S 3 cosπx e 3d+8y e d+5y e d+4y + e y S := e y e y d+6y e y cosπd + 3x + cosπd + x e d+7y e d+5y e d+4y + e d+y S := e y e y d+6y e y cosπd + 3x + S 3 := n= cosπd + 3x e d+5y + e d+4y e y e y d+6y e y cosπd + 3x + When y = 0 each of S S S 3 is defined Namely S = 0 S = 0 S 3 = Since these functions are even in x we may assume x 0 Further the condition arg τ > π 4 implies that y < πx To find c we note that each S i 0 so it suffices to bound one

4 4 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER away from 0 We do this in three different cases Case : Suppose that y Since > x > y it follows that cosπx > cos π that S is bounded away from 0 In particular π cos e 3d+8 e d+5 e d+4 + e S e π e π + Case : Suppose that y < x k < y for some positive integer k Although less obvious than in Case S will again be bounded away from 0: S π cos π e 3d+8y e d+5y e d+4y + e y e π e y + 8π y e y so S π 3 cos π d + d + 3 e π e π + + 8π 4 e π Case 3 : Suppose that y < x k y for some non-negative integer k We additionally assume x k This is permitted since every x is covered as we vary k It will be S 3 that is bounded away from 0 Let u := πd + 3 x Now we have that S 3 k note that 0 u π y u π 4π e π e u π cos u u + e π cos u cos u = cos πd + 3x a tedious analysis of the derivative of this function implies for d 4 that 8π 3 S 3 e π e + 4e Obviously we may take c to be the minimum of the bounds 3 Using Lemma we now obtain an approximation for fτ Lemma 3 If arg τ π x then 4 fτ = f τ = O y π 3d + 3 τ + log sin π + f τ is an licitly bounded function Furthermore if y y max is sufficiently small 0 < δ < 0 < ε 3 < δ β := 3 δ 4 yβ x then there is a

5 PROOF OF THE ALDER-ANDREWS CONJECTURE 5 constant c 3 depending on d ε δ such that π fy + πix 3d + 3 y c 3 y ε Remark The discussion of the size of y max will follow 4 Proof From page 9 of Andrews [] we have that log fτ = τ π 3d log sin π + πi Ds is the Dirichlet series which converges for Rs > Writing Ds := Ds = d + 3 s ζ n ± n 0 n s +i i τ s + ζ s d + d + 3 d + 3 s Γsζs + Dsds ζs a is the Hurwitz zeta function we see that Ds can be analytically continued to the entire complex plane except for a pole of order residue at s = see for example page 55 of Apostol s book [4] We bound the integral by noting that Ds ζs obtaining +i s Γsζs + Dsds ξ y πi i τ ξ := π ζ + it ζ The first statement of the lemma follows Remark Numerical estimates show that ξ < 4 + it Γ + it dt To prove the second statement we again follow the method of Andrews [] We consider two cases: y β x y y x In the first we see that arg τ π so π π 4 we apply the first statement of the lemma getting log fy + πix π y 3d π y + 4π x y + log 3d + 3 sin π + ξ y c 4 := π 3d + 3 y c 4 y δ π 4 3 δ 3d + 3 y max log sin π y δ max ξy +δ max

6 6 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER In the second case we have that using Lemma we obtain log fy + πix = log fy + R gτ gy log fy + πix π 3d + 3 y c 5 y 4 c 5 := c y max log sin π ξy 3 max We let c 3 :=min c 4 y max ε δ c5 y max ε take y max to be small enough so that c 3 >0 Remark In the proof of Theorem we need only bound Q d n since it is of lower order than q d n We shall ignore the restriction on y max for convenience Proof of Theorem From the Cauchy integral theorem we have Q d n = πi τ0 +πi τ 0 fτ nτ dτ = fy + πix ny + nπix dx Applying the saddle point method we take y = n π/ 3d + 3 we let m := ny for notational simplicity Assuming the notation in Lemma 3 for n 6 we have y max π so that both cases in the proof of the second statement of Lemma 3 are nonvacuous We have that y β Q d n = e m fy + πix πinx dx + e m R y β y β By Lemma 3 so R := [ R 5 e m R Using Lemma 3 write Q d n = m log + y β fy + πix πinx dx π 3d + 3 m n c3 m n m c 3 m ε π 3d + 3 ε ] ε sin π m/n β ϕ x dx + mr d + 3 m/n β [ ϕ x := m + πixn ] + πinx + g x m β

7 g x ξ π 3m PROOF OF THE ALDER-ANDREWS CONJECTURE 7 After making the change of variables πx = m/nω we obtain Q d n = m + log mn + log sin π log π I := c0 m β c 0 m β ϕ ω dω I + mr π β c 0 := π 3d + 3 ϕ ω := m + iω + iω + g ω We must now find an asymptotic ression for I Write 6 I = c0 m β c0 m β c 0 m β mω dω + R R := mω ϕ 3 ω dω c 0 m β with ϕ 3 ω := m + iω + ω + g ω + iω Simplifying we find that 7 ϕ 3 ω c 3 0m 3δ 4 + ξ π 3md + 3 Substituting m min = δ 4 π 0 δ 4 3d + 3 for m in 7 it follows that Thus letting c 6 := ϕ 3max ϕ 3max ϕ 3 ω 44+8δ 3δ 6 π 76+8δ 3δ 6 3d + 3 we have ϕ 3 ω m + 3δ 4 Hence we conclude that R c 0 c 7 m δ Computing the integral in 6 we see that 8 g m m c0 m β c 6 c ξc 6 m 3δ 4 min mω π dω = c 0 m m β c 0 m δ 4 + ξπ δ 8 =: ϕ 3max π 3d g m Thus we find that I = π m these results we obtain the desired ression for Q d n n 3 4 3d Q d n = 4 sin n π =: m + 3δ 4 c7 +g m+r Combining π 3d Rn

8 8 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER 0 Rn n 4 n + δ π sin π c 7 π + δ 3 sin π n π n π n δ 8 π δ 3δ + 4 3d n π c 3 n ε π Estimate of q d n with licit error bound Theorem of [7] with k = m = l = d gives q d n A /4 πα d dα d + n 3/4 na 3ε α A depend only on d see their definitions below in Theorem 3 We will bound the error licitly following closely the paper of Meinardus [7] We make his calculations effective we obtain the following theorem Theorem 3 Let α be the unique real number in [0 ] satisfying α d + α = 0 let A := d log α + α rd r= If n is a positive integer then r A /4 q d n = πα d dα d + n 3/4 na + r d n r d n can be bounded licitly see the end of this section 3 Preliminary Facts For fixed d 4 we have the generating function 3 fz := q d ne nz n=0 with z = x + iy Hence we obviously have that 3 q d n = π fze nz dy π Therefore to estimate q d n we require strong approximations for fz Lemma 3 Assuming the notation above for y x +ε x < β β := min π α d ξ log ρ πd d + ρ π ε 4 0 < ξ < is a constant we have that π fz = α d dα d + e A z + ferr z f err z = o is an licitly bounded function Lemma 33 Assuming the notation above for x < β x +ε < y π we have that π A fx + iy e ηρxε + f ρ x dx x + d log α + f ρ x f f are functions given in Lemma 34 η is an licitly given constant Remark Although ε = 4 in [7] we will benefit by varying ε in our work

9 PROOF OF THE ALDER-ANDREWS CONJECTURE 9 To prove Lemmas 3 33 we follow [7] write by the Cauchy Integral Theorem 33 fz = Hw zθw z dw πi C w C is a circle of radius ρ := α centered at the origin 34 Hw z := we nz Θw z := e d nn z w n n= Therefore we estimate Hw z Θw z n= Lemma 34 Let ρ = α d = α suppose w = ρe iϕ with π ϕ < π Then for y x +ε x < β w r 35 Hw z = z r + log w + f w z r= π log 36 Θw z = dz w dz log w + f w z as x 0 f w z = O x f w z = O x + [ c 0 x π ϕ + c x ε ] are licitly bounded functions Proof First 34 the inverse Mellin transform yield 37 log Hw z = πi +i i z s ΓsζsDs + w ds ζs is the Riemann zeta function Γs is the Gamma function Ds w := w r r which is defined as a function of s for all fixed w with w < r s Note that if θ 0 := arctan x ε then z / it z / e θ 0 t + x ε 4 x / e θ 0 t By changing the curve of integration accounting for the poles at s = 0 we have log Hw z = w r log w + + f z r w z f w z = πi r z / it Γ + it ζ + x ε 4 5 π 3 ζ 3 ρ ρ + it D + it w 4 π θ 0 x =: f x This proves the first statement as x ε θ 0 = arctan x ε both tend toward 0 as x 0 The transformation properties of theta functions functions give π 38 Θw z = dz e log w dz/ dz e π µ πiµ dz dz log w dz/ µ= dt

10 0 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER The argument on page 95 of [7] completes the proof of the lemma with 39 [ ] f w z e d z 8 e dx +x ε 8 + 4π ξ dx+x ε + π ξ dx+x ε π ϕ =: f ϕ z = f 0 z + f ϕ z dx + x ε We now prove Lemmas 3 33 using Lemma 34 Proof of Lemma 3 Recall from 33 that fz = Hw zθw z dw πi w Let ϕ 0 = x c with 3 8 < c < Then 30 fz = πi ρe iϕ 0 ρe iϕ 0 C Hw zθw z dw w + πi B π ϕ π dx+x ε π log ρ d Hw zθw z dw w x ε + d z 8 B is the circle C without the arc ρe iϕ 0 to ρe iϕ 0 We first estimate the second integral in 30 We note the error of Meinardus [7] in the bound of Θw z provided between 5 6 From Lemma 34 we have that 3 π x log ρ Θw z d z ρ dx + y yϕ log ρ dx +y ϕ x dx + y + yϕ log ρ + f dx + y w z The term does not appear in [7] tends to infinity if yϕ < 0 This term arises from the main term of Θw z so its contribution cannot be ignored Furthermore it is O x ε hence cannot be combined into the negative O x c term arising from ϕ x However the bound Meinardus claims on the product Hw zθw z is correct dx +y To see this we need more than the bound Hw z Hρ x that was originally thought to be sufficient From Lemma 34 we have that 3 Hw z f w z + ρ R R z r= w r r = = x x + y x x + y + y x + y r= ρ r cosrϕ r + ρ r r + r= r= x x + y y x + y r= ρ r sinrϕ rϕ r z r= r= w r r ρ r sinrϕ r ρ r yϕ log ρ cosrϕ r x + y

11 PROOF OF THE ALDER-ANDREWS CONJECTURE Since ρ = α d ρ = α combining this with 3 3 we see that π + ρ Hw zθw z f w z + Ax d z ρ x + y ϕ x x + y [ ] d + ρ r cosrϕ y ρ r sinrϕ rϕ ϕ r ϕ x r r= + f w z Hence as x 0 we recover Meinardus s bound on Hw zθw z Using the notation of Lemma Hw zθw z dw B w 34 ψϕ z := π d z + ρ ρ e ψϕz dϕ +f ϕ z ϕ x dx + y + x x + y B r= r= f x + Ax B x + y ψϕ z + ρ r cosrϕ y r x + y [ + f 0 z π ϕ dx + x ε r= ] dϕ ρ r sinrϕ rϕ r We evaluate the two integrals of 33 separately For the integral ψϕ z d ϕ we consider two parts based on the sign of yϕ We B assume that y > 0 When ϕ > 0 sinrϕ rϕ < 0 for all r so ψϕ z > 0 Then 35 π ϕ 0 ψϕ z d ϕ ν > is a constant When ϕ < 0 we note that sinrϕ rϕ > 0 so 36 ψϕ z = ϕ x dx + y + x x + y ϕ x dx + y + x x + y =: ˆψϕ z ψ ϕ ϕ 0 z [ ψϕ 0 z ψνϕ 0 z] + ψ ϕ νϕ 0 z [ ψνϕ 0 z ψπ z] r= r= ρ r cosrϕ x+ε r x + y r= ρ r r cosrϕ + ϕ3 x +ε α d 6x + y whence using that π αd x ε π d 37 π [ ψ ϕ zdϕ ϕ 0 ˆψ ˆψ ϕ 0 z ϕ ϕ 0 z [ + ˆψ ˆψ νϕ 0 z ϕ νϕ 0 z ρ r r r 3 ϕ 3 6 ] ˆψ νϕ 0 z + ] ˆψ π z

12 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER We now consider the second integral in 33 A weaker bound on ψϕ z suffices In particular we have ψϕ z kϕ x k := x + y d παd y + ρ 6 x π 4 which is positive since x < β Hence we have that 38 B ψϕ z + π ϕ dϕ dx + x ε dx + x ε π kdx + x ε kϕ 0 + [ kπ + πϕ 0 dx + x ε π ] dx + x ε Using in 33 gives an licit bound for the second integral of 30 say E B z Following page 97 of [7] the first integral of 30 is given by I := πi ρe iϕ 0 ρe iϕ 0 39 I main := 30 I error := f 3 w z Hw zθw z dw w = πdz ϕ0 ϕ 0 ρe ϕ0 ϕ 0 A z + d log α ϕ dα d + dϕ dz I main + I error ρe iϕ log + f 3 w z + f w z + f w z ρ ϕ dα d + dϕ dz 6 ρe ϕ 3 Then we have 3 I error ϕ 0 ρ cos ϕ0 ρ πzd 3 I main = dα d + Hence it follows that 33 I = ρe f x + 6 ρe ϕ3 0 + f ϕ 0 z ϕ 0 ϕ dα d + dϕ dz α d A dα d + + z Êϕ 0 w z 34 Êϕ 0 w z [ α d Ax d z πd z x + y ρ cos ϕ0 + ϕ 0 ρ =: E ϕ0 z ϕ 0 dα d + f x + ρe 6 ρe ϕ3 0 ϕ 0x dα d + d x + y + f ϕ 0 z ]

13 PROOF OF THE ALDER-ANDREWS CONJECTURE 3 Hence we finally see that f err z E ϕ0 z + E B z α d dα d + Ax x + y Proof of Lemma 33 In order to bound f for x +ε < y π note that Θw z Θρ x by 34 which also yields that { log Hw z = R{log Hw z} log Hρ x + R w } e nz ρ e nx n n On the other h we have that { R w } e nz ρ e nx ρx ε β ε e β e β n n e β cos β +ε + e β To see this note that { R w } e nz ρ n n This then gives R { w n e nz} ρ n e nx ρx ε x ε e x e nx ρe x e x e x cos x +ε + e x e x e x cos x +ε + e x β ε e β e β e β cos β +ε + e β =: η The statement of Lemma 33 now follows from 33 Lemma 34 3 Proof of Theorem 3 From 3 it follows that q d n = I + I I := x +ε fze nz dy I := x +ε π + fze nz dy π x π +ε π x +ε In this proof we let x = A 35 I = γe A x γ := π α d dα d + n Following the idea of page 9 of [7] we split I as E 3 := γe A x x +ε x +ε e E := γe A x y A x 3 dy + E + E 3 x +ε x +ε e xy +iy 3 x +ε x +ε e y A x 3 e A x x +y f err z dy y 4 +ixy 3 x 3 x +y dy

14 4 CLAUDIA ALFES MARIE JAMESON AND ROBERT J LEMKE OLIVER The first integral in 35 can be written 36 γe A x x +ε x +ε e y A x 3 dy = γe A x πx 3 37 E γ A x ε e A x Axε For E we further split the integral: E = γe A x with ε > ε ε > Then 3 A 38 γe x 39 A γe x y x +ε x +ε y x +ε e Finally for E 3 we have +γe A x y x +ε e y A x 3 e A y A x 3 e A e y A x 3 e A x +ε y x +ε e y 4 +ixy 3 x 3 x +y y 4 +ixy 3 x 3 x +y y 4 +ixy 3 x 3 x +y y A x 3 e A dy dy A + E y 4 +ixy 3 x 3 x +y A γe x dy dy Ax 3ε πx 3 A A γ Axε x 3 + x ε x +x ε + γx3 A 330 E 3 γe A x f max err πx 3 + x ε A x Axε To bound I we apply Lemma 33 to find that π 33 I e ηρxε + f ρ x nx + A dx x + d log α + f ρ x Finally we obtain q d n = A /4 πα d dα d + n 3/4 na + E + E + E 3 + I E + E + E 3 + I is bounded using the ressions in The result follows with r d n E + E + E 3 + I 4 Proof of Alder s Conjecture Using Theorems 3 we are now able to prove our main results Proof of Theorem Applying the results of Sections 3 we have that d n = q d n Q d n = A /4 πα d dα d + n 3/4 na + E d n

15 PROOF OF THE ALDER-ANDREWS CONJECTURE 5 E d n = r d n Q d n In the proof of Lemma 3 we relax the restriction on y max Thus Theorem implies Q d n = O π 3d+9 n / + c 0 n 6 c 0 is some positive constant By Theorem 3 r d n E + E + E 3 + I a careful examination of each of these terms shows that E = O n 5 6 e An E = O n 3 ε 4 e An E 3 = O n 5 6 e An I = O n 4 e An ηρx ε Hence by choosing ε 7 the result follows 8 Proof of Theorem The works of Yee [9][0] Andrews [] show that d n 0 when d 3 can be easily modified to show that the inequality is strict when n d + 6 For each remaining 4 d 30 we use Theorems 3 to compute the smallest n such that our bounds imply d n > 0 We denote this n by Ωd a C++ program computed the values of d n Ω d n which then confirmed the remaining cases of the Alder-Andrews Conjecture As an example we find that when d = 30 Ω To get this we take δ = 0 0 ε = 5 0 in Theorem in Theorem 3 ε = 6906 ε = ξ = 99 c = ν = Other d are similar all satisfy Ωd Ω30 References [] G E Andrews On a partition problem of H L Alder Pacific J Math 36: [] G E Andrews The theory of partitions Addison-Wesley Publishing Co Reading Mass-London- Amsterdam 976 Encyclopedia of Mathematics its Applications Vol [3] G E Andrews K Eriksson Integer partitions Cambridge University Press Cambridge 004 [4] T M Apostol Introduction to analytic number theory Springer-Verlag New York 976 Undergraduate Texts in Mathematics [5] K Bringmann K Ono Coefficients of harmonic maass forms accepted for publication [6] G Meinardus Asymptotische Aussagen über Partitionen Math Z 59: [7] G Meinardus Über Partitionen mit Differenzenbedingungen Math Z 6: [8] V V Subrahmanyasastri Partitions with congruence conditions J Indian Math Soc NS 36: [9] A J Yee Partitions with difference conditions Alder s conjecture Proc Natl Acad Sci USA 047: electronic 004 [0] A J Yee Alder s conjecture J Reine Angew Math 66: Lehrstuhl A für Mathematik RWTH Aachen Templergraben 64 D-506 Aachen Germany address: claudiaalfes@rwth-aachende Department of Mathematics University of Wisconsin Madison Wisconsin address: jameson@mathwiscedu Department of Mathematics University of Wisconsin Madison Wisconsin address: lemkeoli@mathwiscedu