Yang Y - Partitions Into Primes - Trans. Amer. Math. Soc. 352 (2000) No.6 2581-2600

8/9/2019 Yang Y - Partitions Into Primes - Trans. Amer. Math. Soc. 352 (2000) No.6 2581-2600

http://slidepdf.com/reader/full/yang-y-partitions-into-primes-trans-amer-math-soc-352-2000-no6-2581-2600 1/20

TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 352, Number 6, Pages 2581–2600S 0002-9947(00)02386-2Article electronically published on February 14, 2000

PARTITIONS INTO PRIMES

YIFAN YANG

Abstract. We investigate the asymptotic behavior of the partition function

pΛ(n) defined by∞n=0 pΛ(n)xn =

∞m=1(1−xm)−Λ(m), where Λ(n) denotes

the von Mangoldt function. Improving a result of Richmond, we show that

log pΛ(n) = 2 ζ (2)n+O(

√n exp−c(log n)(log2 n)−2/3(log3 n)−1/3), where

c is a positive constant and logk denotes the k times iterated logarithm. We

also show that the error term can be improved to O(n1/4) if and only if the

Riemann Hypothesis holds.

1. Introduction

The asymptotic behavior of partition functions has been extensively studied inthe literature. The most famous result is the asymptotic formula

p(n) ∼ 1

(4√

3)neπ

√2n/3 (n → ∞)

for the ordinary partition function p(n), proved in 1918 by Hardy and Ramanujan[3]. The asymptotic behavior of more general partition functions has been studiedby many authors, including Ingham [4], Kohlbecker [6], Meinardus [7], [8], Rothand Szekeres [12], Schwarz [13], [14], and Richmond [10], [11].

Of particular interest are functions related to partitions into primes. As anapplication of an asymptotic formula for general partition functions, Roth and

Szekeres [12] showed that the number qP(n) of partitions of n into distinct primessatisfies

log qP(n) = π

2

3

n

log n

1/2 1 + O

loglog n

log n

.

A similar, but in some ways more natural, partition function is the function pΛ(n) defined by

∞n=0

pΛ(n)xn =∞

m=1

(1 − xm)−Λ(m),

where

Λ(m) = log p, m= pr,

0, else,is the von Mangoldt function. The function pΛ(n) represents a weighted count of the number of partitions of n into prime powers. This weighted partition functionwas first introduced and studied in 1950 by Brigham [1], who proved a conditional

Received by the editors March 3, 1998.

2000 Mathematics Subject Classification. Primary 11P82; Secondary 11M26, 11N05.

c2000 American Mathematical Society

2581



2582 YIFAN YANG

result under the assumption of the Riemann Hypothesis. More recently, Richmond[11], using an asymptotic formula for general partition functions (see [10]) andVinogradov’s zero-free region for the Riemann zeta function, obtained the following

unconditional result.

Theorem A (Richmond). There exists a positive constant c such that for all sufficiently large n

log pΛ(n) = 2

ζ (2)n1/2

1 + O

exp(−c(log n)4/7(log log n)−3/7)

.(1.1)

Richmond also proved a conditional result.

Theorem B (Richmond). Let θ be the least upper bound for the real parts of thenon-trivial zeros of the Riemann zeta function. Then

log pΛ(n) = 2

ζ (2)n1/2 + O(nθ/2).(1.2)

In particular, if the Riemann Hypothesis is true, then log pΛ(n) = 2

ζ (2)n1/2 + O(n1/4).(1.3)

In our first result we show that the error term in Richmond’s unconditional result(1.1) can be substantially improved.

Theorem 1. There exists a positive constant c such that for all sufficiently largen

log pΛ(n) = 2

ζ (2)n1/2

1 + O

exp

− c log n

(log2 n)2/3(log3 n)1/3

,(1.4)

where logk denotes the k times iterated logarithm.

Note that the exponential factor in (1.4) is much smaller than the corresponding

factor in the error term of the prime number theorem obtained by using the Vino-gradov zero-free region, namely exp(−(log n)3/5−). The reason for this unexpect-edly small error term lies in the fact that log pΛ(n) behaves in many respects morelike the power series f (x) =

∞n=1 Λ(n)xn than the partial sum Ψ(u) =

n≤u Λ(n);

indeed, it would not be hard to show that, as x → 1−, f (x) differs from its approx-imation 1/(1 − x) by a similarly small error term.

While the estimate of Theorem A can be substantially improved, our next resultshows that Theorem B is best possible.

Theorem 2. We have

log pΛ(n) − 2

ζ (2)n1/2 = Ω±(n1/4).

Our final result gives a converse to Theorem B.

Theorem 3. Let θ be the least upper bound for the real parts of the zeros of theRiemann zeta function, and let θ be the greatest lower bound for all α for which

log pΛ(n) = 2

ζ (2)n1/2 + O(nα/2)

as n → ∞. Then θ ≤ θ.

Combining this result with Theorem B yields the following corollary.



PARTITIONS INTO PRIMES 2583

Corollary. Let θ and θ be defined as in Theorem 3. Then θ = θ. In particular,the Riemann Hypothesis is true if and only if

log pΛ(n) = 2

ζ (2)n1/2

+ O(n1/4+

) for all > 0.

Let

P Λ(u) =n≤u

pΛ(n)

be the summatory function of pΛ(n). In Section 4 we show that Theorems 1-3 aretrue if and only if the corresponding statements with n replaced by u and pΛ(n)replaced by P Λ(u) are true. Hence it suffices to prove Theorems 1-3 with P Λ(u)instead of pΛ(n).

Our main tools for proving these results are Abelian and Tauberian theoremsthat relate estimates for P Λ(u) to estimates for the Laplace-Stieltjes transform of P Λ(u), defined by

F Λ(x) =

∞0

e−xu dP Λ(u).

We will state and prove these results in Section 2. In Section 3 we establish somelemmas relating the behavior of general weighted partition functions pw(n) definedby

∞n=0

pw(n)xn =∞

m=1

(1 − xm)−w(m)

to analytic properties of the Dirichlet series

f w(s) =∞

n=1

w(n)

ns,

where w(m) is a non-negative function defined on the set of positive integers. Ourmethods here are, to some extent, similar to those in Meinardus [7]. However,Meinardus made much stronger assumptions on the analytic properties of f w(s).These assumptions are not satisfied in the case w(n) = Λ(n), so that Meinardus’results are not applicable to pΛ(n). In Section 4 we complete the proof of Theorems1-3. In Section 5 we use one of our Tauberian results to give a new proof of TheoremB that is simpler and more elementary than the original proof of Richmond.

2. Some Abelian and Tauberian Results

Our first result is an elementary Abelian result which generalizes and extends aresult of Freiman [2, p. 276].

Proposition 1. Suppose that P (u) is a non-negative and non-decreasing function satisfying P (t) = O(et) for every > 0. For x > 0 let

F (x) =

∞0

e−xu dP (u)

be the Laplace-Stieltjes transform of P (u). Suppose that for some constants A > 0and 0 < a < 1 the inequality

log P (u) ≥ Aua − r(u)(2.1a)



2584 YIFAN YANG

holds for all sufficiently large u, where r(u) is a positive differentiable function satisfying

r(u)u

−a

→ 0 monotonically as u → ∞,(R1)and

r(u)/(log u) → ∞ monotonically as u → ∞.(R2)

Then

log F (x) ≥ Bx−b − r(Bx−b−1)(2.2a)

for all sufficiently small x, where

B = (Aa)1/(1−a)(1 − a)/a, b = a/(1 − a), B = Bb.(2.3)

Similarly, if

log P (u) ≤ Aua + r(u)(2.1b)

for all sufficiently large u, then with the same notation

log F (x) ≤ Bx−b + 2r(Bx−b−1)(2.2b)

for all sufficiently small x.

As a simple consequence of Proposition 1, we have the following corollary.

Corollary. Let P (u), F (x), A, a, B, b and B be given as in Proposition 1. Sup-pose that r(u) satisfies conditions (R1) and (R2). Then

log F (x) − Bx−b = Ω+(r(Bx−b−1))

as x → 0 implies that

log P (u) − Aua = Ω+(r(u))

as u

→ ∞. The same statement holds if Ω+ is replaced by Ω−.

Proof of Proposition 1. Suppose first that (2.1a) holds for sufficiently large u. Bythe monotonicity of P (t) and the assumption that P (t) is non-negative, we havefor all x > 0 and all u > 0,

F (x) =

∞0

e−xt dP (t) = x

∞0

e−xtP (t) dt(2.4)

≥ x

∞u

e−xtP (t) dt ≥ x

∞u

e−xtP (u) dt = e−xuP (u).

(Note that the convergence of the integral defining F (x) is ensured by the assump-tion that P (t) = O(et) for every > 0.)

By (2.1a) it follows that

F (x)

≥exp

−xu + Aua

−r(u)

(2.5)

for all sufficiently large u and all x > 0. We choose u = ux to maximize −xu + Aua,i.e., we let ux be defined by

x = Aau−(1−a)x .(2.6)

If x is sufficiently small, then ux will be large enough for (2.5) to hold. We notethat

xux = x(Aax−1)1/(1−a) = Bx−a/(1−a) = Bx−b(2.7)




and

Auax = A(Aax−1)a/(1−a) = (Aa)1/(1−a)a−1x−b = Ba−1x−b.(2.8)

Substituting these expressions into the right-hand side of (2.5), we obtain

log F (x) ≥ B(−1 + a−1)x−b − r(Bx−b−1) = Bx−b − r(Bx−b−1),

which proves (2.2a). It remains to prove that (2.1b) implies (2.2b).Let u0 be a positive constant such that (2.1b) and conditions (R1)-(R2) are

satisfied for u ≥ u0. Given a positive number x, we let ux be defined by (2.6) andassume that x is small enough so that ux ≥ 2u0. We write

F (x) = x

u0

0

+

(1−µ)ux

u0

+

(1+µ)ux

(1−µ)ux

+

∞(1+µ)ux

e−xtP (t) dt(2.9)

= xI 1 + I 2 + I 3 + I 4,

where µ is a parameter to be chosen so that xI 3 has the same order of magnitude

as F (x). It turns out that the optimal choice for µ is

µ = µ(ux) = K

u−ax r(ux),(2.10)

where K is a positive constant to be chosen later in terms of A and a. We notethat condition (R1) implies that µ(ux) → 0 as ux → ∞. In particular, we have0 < µ < 1/2 if x is sufficiently small in terms of K , which we will henceforthassume.

The integral I 1 is bounded by u0

0P (t) dt and thus of order O(1). To estimate

I 4, we define φ(t) for t ≥ u0 by

φ(t) = −xt + Ata + r(t).

Since, by condition (R1), r(t)t−a is monotonically decreasing to 0, we have r(t)t−a−ar(t)t−a−1

≤0 and r(t)t−1

≤r(ux)u−a

x u−(1−a)x = K −2µ2u−(1−a)

x for all t≥

ux.Thus, for all t ≥ (1 + µ)ux we obtain

φ(t) = −x + Aat−(1−a) + r(t) ≤ −x + Aat−(1−a) + ar(t)t−1

≤ −x + Aa(1 + µ)−(1−a)u−(1−a)x + K −2aµ2u−(1−a)

x .

By (2.6), the last expression is equal to (−1 + (1 + µ)−(1−a) + O(µ2))x. Since(1+µ)−(1−a) = 1−(1−a)µ+O(µ2), we have −1+(1+µ)−(1−a)+O(µ2) ≤ −(1−a)µ/2for all sufficiently small x. Hence for all sufficiently small x and t ≥ (1 + µ)ux weobtain

φ(t) ≤ −1 − a

2xµ.

It follows that

xI 4 ≤ x ∞

(1+µ)uxeφ(t) dt ≤ −

2

(1 − a)µ ∞

(1+µ)uxeφ(t)φ(t) dt =

2

(1 − a)µ eφ((1+µ)ux)

=2

(1 − a)µexp −x(1 + µ)ux + A(1 + µ)aua

x + r((1 + µ)ux) .

We substitute (2.7) and (2.8) into the last expression and note that condition (R1)implies that r((1 + µ)ux)((1 + µ)ux)−a ≤ r(ux)u−a

x , i.e.,

r((1 + µ)ux) ≤ (1 + µ)ar(ux)(2.11)



2586 YIFAN YANG

for all sufficiently small x. This yields

xI 4

≤

2

(1 − a)µ

expBx−b −(1 + µ) + (1 + µ)aa−1 + (1 + µ)a(Aa)−1r(ux)u−a

x ≤ exp

Bx−b

(a−1 − 1) +

1

2(a − 1)µ2 + O(µ3) + (1 + µ)a(Aa)−1r(ux)u−a

x

+ O(log µ−1)

= exp

Bx−b

1 − 1

2aµ2 + O(µ3) + (1 + O(µ))A−1(1 − a)−1r(ux)u−a

x

+ O(log µ−1)

,

where we have used the identity B(a−1 − 1) = B which follows from (2.3). Ex-pressing r(ux)u−a

x in terms of µ using (2.10), the last expression becomes

exp

Bx−b

1 − µ2a

2− K −2A−1(1 − a)−1

+ O(µ3) + O(xb log µ−1)

.

By (2.8), (2.10) and condition (R2) we have

xb log µ−1 u−ax log ux = o(u−a

x r(ux)) = o(µ2).

Choosing now K sufficiently large in terms of A and a yields

xI 4 ≤ exp

Bx−b(1 − aµ2/4)

(2.12)

for all sufficiently small x. By a similar argument we have

xI 2 ≤ exp

Bx−b(1 − aµ2/4)

(2.13)

for all sufficiently small x.We now estimate I 3. By (2.1b) we have

xI 3 = x

(1+µ)ux

(1−µ)ux

e−xtP (t) dt ≤ x

(1+µ)ux

(1−µ)ux

e−xt+Ata+r(t) dt

We observe that t = ux maximizes −xt + Ata and, by condition (R2), r(t) ≤r((1 + µ)ux) for all t ≤ (1 + µ)ux. Hence, by (2.7), (2.8) and (2.11),

xI 3 ≤ 2xµux exp −xux + Auax + r((1 + µ)ux)(2.14)

≤ 2xµux exp

Bx−b + (1 + µ)ar(Bx−b−1)

for all sufficiently small x. Combining (2.9), (2.12), (2.13) and (2.14), and notingthat, by (2.7) and (2.10),

log(xµux) = log

Bx−bK

u−a

x r(ux)

= O

log(ua

xr(ux))

= O(log x−1),

we finally obtain

F (x) ≤ O(x) + O(expBx−b − aµ2/4)

+ exp

Bx−b + (1 + µ)ar(Bx−b−1) + O(log x−1)

= exp

Bx−b + (1 + µ)ar(Bx−b−1) + O(log x−1)

≤ exp

Bx−b + 2r(Bx−b−1)

for all sufficiently small x, where in the last step we have used the assumption (R2).This yields the desired estimate (2.2b).

Our next result is a Tauberian counterpart to Proposition 1.




Proposition 2. Let P (u) and F (x) be defined as in Proposition 1. Suppose that for some constants B,b > 0 and B = Bb the inequality

log F (x) − Bx

−b ≤ r(B

x

−b−1

)(2.15)holds for all sufficiently small x, where r(u) is a positive differentiable function satisfying conditions (R1) and (R2) in Proposition 1 with a = b/(1 + b). Then

−C

uar(u) ≤ log P (u) − Aua ≤ r(u)(2.16)

for all sufficiently large u, where A and a are determined by (2.3) and C is a constant depending on B and b.

Proof. Suppose that (2.15) holds for all sufficiently small x. We first prove theupper bound for log P (u).

Given a positive number u, by (2.4) and assumption (2.15) we have

P (u) ≤ exuF (x) ≤ exp

xu + Bx−b + r(Bx−b−1)

(2.17)

for all sufficiently small x. We choosex = xu = Aau−(1−a),(2.18)

where A and a are determined by (2.3), and note that with this choice of x we haveu = ux, where ux is given by (2.6). We assume that u is large enough so that (2.17)holds for x = xu. Using (2.7) and (2.8) with xu and u in place of x and ux, we seethat

xuu + Bx−bu = Bx−b

u (1 + b−1) = Aauax

1 +

1 − a

a

= Aua

x(2.19)

and

Bx−b−1u = u.(2.20)

Thus, we obtain the upper bound

P (u) ≤ expxuu + Bx−bu + r(Bx−b−1

u ) = expAua + r(u)(2.21)

for all sufficiently large u. This implies the upper bound in (2.16). It remains toprove the lower bound.

Let u0 be a large positive constant such that (2.21) holds for u ≥ u0. Assumingthat u ≥ 2u0, we split the integral defining F (xu) into four parts as before. LetI 1, I 2, I 3 and I 4 be defined by (2.9) with x = xu and ux replaced by u. We

let µ = µ(u) = K

u−ar(u) be defined as in (2.10), where K is a large positiveconstant to be chosen later. Using the upper bound (2.21) and arguing exactly asin the proof of Proposition 1, we see that the upper bounds (2.12) and (2.13) forI 4 and I 2 remain valid for sufficiently large u with xu and u in place of x and ux.We note that, by (2.3), (2.8) and the definition of µ,

14 Baµ2x−bu = 14 BaK 2r(u)u−ax−bu = 14 BaK 2r(u) AaB = 14 Aa(1 − a)K 2r(Bx−b−1u ).

We now choose the constant K large enough so that Aa(1 − a)K 2/4 ≥ 2. Then by(2.12) and assumption (2.15), we have

xuI 4 ≤ exp

Bx−b

u − 1

4Baµ2x−b

u

≤ exp

Bx−b

u − 2r(Bx−b−1u )

≤ F (xu)exp

−r(Bx−b−1u )

= o(F (xu)).



2588 YIFAN YANG

Similarly, we see that xuI 2 = o(F (xu)). Hence xuI 3 ≥ F (xu)(1 + o(1)). On theother hand, bounding the integral I 3 trivially, we obtain

xuI 3 ≤ 2µuxue−xu(1−µ)u

P ((1 + µ)u).It follows that

P ((1 + µ)u) ≥ F (xu)exp xu(1 − µ)u − log(2µuxu) + o(1) .

By (2.10), (2.18) and conditions (R1) and (R2) we have

log(2µuxu) log u r(u).

Thus, using the bound (2.15) for F (xu), we obtain

P ((1 + µ)u) ≥ exp

Bx−bu − r(Bx−b−1

u ) + xu(1 − µ)u + O(r(u))

.

Substituting (2.18), (2.19) and (2.20) into the last expression, we see that

log P ((1 + µ)u) ≥ Aua − Aaµua + O(r(u)) = A(1 + µ)aua + O(µua) + O(r(u)).

The last error term can be omitted since r(u) = o(

r(u)ua) = o(µua) by condition

(R1) and (2.10). Moreover, by condition (R2), we have µua = K

r(u)ua ≤K

r((1 + µ)u)(1 + µ)aua. It follows that

log P ((1 + µ)u) ≥ A(1 + µ)aua − C

r((1 + µ)u)(1 + µ)aua

for all sufficiently large u. Since (1 + µ(u))u is a continuous function of u andtends to infinity when u → ∞, for all sufficiently large v, there exists a u such thatv = (1 + µ(u))u. Hence we have

log P (v) ≥ Ava − C

var(v)

for all sufficiently large v. This completes the proof of the proposition.

Proposition 2 is not a complete converse to Proposition 1, as the lower boundin (2.16) is weaker than the lower bound for log P (u) in Proposition 1. Our nextresult gives, under a stronger hypothesis, a complete converse.

Proposition 3. Let P (u) and F (x) be defined as in Proposition 1. Suppose that for some constants B, b > 0 and B = Bb the inequality log F (x) − Bx−b

≤ r(Bx−b−1)(2.15)

holds for all sufficiently small x, where r(u) is a positive differentiable function satisfying conditions (R1) and (R2) in Proposition 1 with a = b/(1 + b). Supposethat, in addition, r(u) satisfies

r(u) ua/2(R3)

as u→ ∞

. Suppose further that the function G(x) = log F (x) satisfies

G(x) = −Bx−b−1 + O

x−1r(Bx−b−1)

(2.22)

and

G(x) x−b−2(2.23)

as x → 0. Then

−Cr(u) ≤ log P (u) − Aua ≤ r(u)(2.24)




for all sufficiently large u, where A and a are determined by (2.3), and C is a positive constant depending on B, b and the constants implicit in (2.22), (2.23) and condition (R3).

Proof. The upper bound in (2.24) follows from Proposition 2. Therefore it remainsto prove the lower bound in (2.24). To this end we use a method of Odlyzko [9].

Given a large number u, we let 0 < µ1 < µ2 < 1 be positive numbers to bechosen later as functions of u. Set u0 = u, u1 = (1 − µ1)u, u2 = (1 − µ2)u, and letx0, x1 and x2 be defined by (2.18) with u = u0, u1 and u2 respectively. Thus, wehave x1 = (1 − µ1)−(1−a)x0 and x2 = (1 − µ2)−(1−a)x0. We consider the function

h(t) = expx1u1 − x1t − expx0u0 + x1u1 − x1u0 − x0t− expx2u2 + x1u1 − x1u2 − x2t.

Since x0 < x1 < x2, we have for t ≥ u0

x1u1 − x1t ≤ x1u1 − x1t + (x1 − x0)(t − u0) = x0u0 + x1u1 − x1u0 − x0t,

and for t ≤ u2

x1u1 − x1t ≤ x1u1 − x1t + (x2 − x1)(u2 − t) = x2u2 + x1u1 − x1u2 − x2t.

Thus h(t) ≤ 0 for t ≤ u2 and t ≥ u0. Now we let

H =

∞0

h(t) dP (t)

= ex1u1F (x1) − ex0u0+x1u1−x1u0F (x0) − ex2u2+x1u1−x1u2F (x2).

Then we see that

H ≤ (P (u0) − P (u2)) × maxu2≤t≤u0

h(t)

≤ P (u0) × maxu2≤t≤u0

ex1u1−x1t = P (u0)ex1(u1−u2).

Hence, we haveP (u0) ≥

ex1u1F (x1) − ex0u0+x1u1−x1u0F (x0)(2.25)

− ex2u2+x1u1−x1u2F (x2)

e−x1(u1−u2).

We now show that, with a suitable choice of µ1 and µ2, ex0u0+x1u1−x1u0F (x0) +ex2u2+x1u1−x1u2F (x2) ≤ ex1u1F (x1)/2, and thus the last expression has the sameorder of magnitude as F (x1)ex1u2 .

By assumptions (2.22) and (2.23), there are positive constants C 1 and C 2 suchthat

−Bx−b−1 − C 1x−1r(Bx−b−1) ≤ G(x) ≤ −Bx−b−1 + C 1x−1r(Bx−b−1)

and

G

(x) ≥ C 2x−b−2

when x is sufficiently small. Thus, Taylor’s formula yields that

G(x1) − G(x0) ≥ G(x0)(x1 − x0) +1

2min

x0≤ξ≤x1

G(ξ)(x1 − x0)2(2.26)

≥ (x1 − x0)(−Bx−b−10 − C 1x−1

0 r(Bx−b−10 ))

+1

2C 2x−b−2

1 (x1 − x0)2.



2590 YIFAN YANG

We now choose µ1 to be of the form µ1 = Kxb0r(Bx−b−1

0 ), where K is a largeconstant independent of u0. We note that, by (2.8) and (2.20), µ1 is also equal toKB (Aa)−1u−a

0 r(u0). Thus, by condition (R3), we have

µ1 ≥ KC 3u−a/20(2.27)

for all sufficiently large u0, where C 3 is a positive constant depending on the con-stant implicit in condition (R3). Moreover, by condition (R1), this choice of µ1

satisfies µ1 → 0 as u0 → ∞. It follows that when u0 is sufficiently large,

x1 − x0 = x0((1 − µ1)−(1−a) − 1)

≤ 2(1 − a)µ1x0,

≥ (1 − a)µ1x0.

Hence, recalling that, by (2.20), Bx−b−10 = u0, we have from (2.26)

G(x1) − G(x0) ≥ −(x1 − x0)u0 − 2C 1(1 − a)µ1r(u0)

+ 12

C 2x−b−20 (1 − µ1)(1−a)(b+2)(1 − a)2µ2

1x20

≥ −(x1 − x0)u0 − 2C 1(1 − a)µ1r(u0) +1

4C 2(1 − a)2µ2

1x−b0

for all sufficiently large u0. We now choose K sufficiently large in terms of B, b, C 1and C 2 so that the second term on the right-hand side of the last expression is lessthan one half of the last term uniformly for all sufficiently large u0, and so that thelast term is at least 2 log 4 (which is possible by (2.8) and (2.27)), i.e., so that

2C 1(1 − a)µ1r(Bx−b−10 ) ≤ 1

8C 2(1 − a)2µ2

1x−b0

and

18

C 2(1 − a)2µ21x−b

0 ≥ log4

for all sufficiently large u0. Then it follows that

G(x1) − G(x0) ≥ −(x1 − x0)u0 − 2C 1(1 − a)µ1r(Bx−b−10 ) +

1

4C 2(1 − a)2µ2

1x−b0

≥ −(x1 − x0)u0 +1

8C 2(1 − a)2µ2

1x−b0

≥ −(x1 − x0)u0 + log 4.

Hence

ex1u1F (x1) ≥ 4ex0u0+x1u1−x1u0F (x0)(2.28)

for all sufficiently large u0. Similarly, if we choose µ2 = 2µ1, then

ex1u1F (x1) ≥ 4ex2u2+x1u1−x1u2F (x2)(2.29)

for all sufficiently large u0. Therefore, by (2.25), (2.28) and (2.29), we have for allsufficiently large u0

P (u0) ≥ 1

2ex1u1F (x1)e−(u1−u2)x1 =

1

2F (x1)ex1u2 = exp G(x1) + x1u2 + O(1) .




On the other hand, by (2.15) and the definition of x1 and u2, we have

G(x1) + x1u2 ≥ Bx−b1 − r(Bx−b−1

1 ) + x1u2

= B(1 − µ1)(1−a)bx−b0 − r(Bx−b−11 ) + (1 − µ1)−(1−a)x0(1 − 2µ1)u0

= Bx−b0 + x0u0 + O(µ1x−b

0 ) − r(Bx−b−11 )

= Bx−b0 + x0u0 + O(r(Bx−b−1

0 )) − r(Bx−b−11 )

for all sufficiently large u0. By condition (R2), we have r(Bx−b−11 ) ≤ r(Bx−b−1

0 ).It follows that, by (2.19) and (2.20),

log P (u0) ≥ G(x1) + x1u2 + O(1) ≥ Bx−b0 + x0u0 − Cr(Bx−b−1

0 ) = Aua0 − Cr(u0)

for all sufficiently large u0, where C is a constant depending on C 1, C 2, B and b.This completes the proof of the proposition.

3. Some Lemmas on Dirichlet Series and Mellin Transforms

Let w(n) be a non-negative function defined on the set of positive integers. Let

f w(s) =

∞n=1

w(n)

ns(3.1)

be the Dirichlet series generated by w(n), and suppose that f w(s) has finite abscissaof absolute convergence σa. We define F w(x) for x > 0 by

F w(x) =∞

n=1

(1 − e−nx)−w(n).

We note that in the case w(n) = Λ(n), we have f w(s) = −ζ (s)/ζ (s), and F w(x) isthe generating function for pΛ(n).

In this section we will prove some lemmas relating analytic properties of f w(s)to the behavior of F w(x).

Lemma 1. For any κ > max(0, σa) we have

log F w(x) =1

2πi

κ+i∞

κ−i∞

Γ(s)ζ (1 + s)f w(s)x−s ds.(3.2)

Furthermore, for any T > 1 we have

log F w(x) =1

2πi

κ+iT

κ−iT

Γ(s)ζ (1 + s)f w(s)x−s ds(3.3)

+ O

ζ (1 + κ)f w(κ)x−κT κ+1/2e−πT /2

,

where the O-constant is absolute.

Proof. The proof of (3.2) can be found in Meinardus [7, p. 390].To prove (3.3), we observe that, by Stirling’s formula,

1

2πi

κ+i∞

κ+iT

+

κ−iT

κ−i∞

Γ(s)ζ (1 + s)f w(s)x−s ds

ζ (1 + κ)f w(κ)x−κ

∞T

tκ+1/2e−πt/2 dt

ζ (1 + κ)f w(κ)x−κT κ+1/2e−πT /2.

Thus (3.3) follows.



2592 YIFAN YANG

Lemma 2. For 0 < x < 1 and Re s > max(0, σa) we have

1

0

xs−1 log F w

(x) dx = Γ(s)ζ (1 + s)f w

(s)−

hw

(s),

where hw(s) is an entire function.

Proof. Suppose that s satisfies Re s > max(0, σa). Then 1

0

xs−1 log F w(x) dx =

∞0

− ∞

1

xs−1 log F w(x) dx

=

∞0

− ∞

1

xs−1

∞k=1

∞n=1

w(n)

ke−knx dx

=

∞

k=1

∞

n=1

Γ(s)w(n)

nsks+1− w(n)

k

∞1

xs−1e−knxdx

= Γ(s)ζ (1 + s)f w(s) − hw(s),

where interchanging the order of integration and summation is justified by theabsolute convergence of the double series involved. The function hw(s) here isgiven by

hw(s) =

∞k=1

∞n=1

w(n)

k

∞1

xs−1e−knx dx =

∞k=1

∞n=1

w(n)

k1+snsΓ(s,kn),

where Γ(s, u) = ∞

uxs−1e−x dx is the incomplete Gamma function. Using the

estimate

|Γ(s, u)| σ1,σ2 ∞

u

e−x/2 dx σ1,σ2 e−u/2 (σ1 ≤ Re s ≤ σ2, u ≥ 1)

and the bound w(n) nσa+1 (which follows from the convergence of f w(s) ats = σa + 1), we see that the double series defining hw(s) converges uniformly inany strip σ1 ≤ Re s ≤ σ2, and hence represents an entire function of s.

Lemma 3 (Landau). (i) Let g(n) be a function defined on the set of positiveintegers and of constant sign for all sufficiently large n. Suppose that the Dirichlet series

∞n=1

g(n)n−s

has finite abscissa of absolute convergence σa. Then s = σa is a singularity of the

function represented by the Dirichlet series.(ii) Let g(x) be an integrable function on [0, 1] and of constant sign for all suffi-ciently small x. Suppose that the Dirichlet integral 1

0

g(x)xs−1 dx

has finite abscissa of absolute convergence σa. Then s = σa is a singularity of the function represented by the Dirichlet integral.




Proof. Part (i) is the classical version of a well-known theorem of Landau (see, e.g.,Ingham [5, p. 88]). To obtain part (ii), we note that the given integral can bewritten as

∞

1

h(u)u−s du

with h(u) = g(1/u)/u and x = 1/u. By the integral version of Landau’s lemma(Ingham [5, p. 88]) the last integral has a singularity at s = σa.

Lemma 4. Assume that for some constants B > 0 and b > 0

log F w(x) = Bx−b1 + o(1)(3.4)

as x → 0+. Then the Dirichlet series f w(s) has abscissa of convergence b and satisfies

f w(s) =1

s

−b

B

Γ(b)ζ (1 + b)+ o

|s − b|σ

−b

(3.5)

as s → b, while σ = Re s > b.

Proof. By Lemma 2 we have

f w(s) =1

Γ(s)ζ (1 + s)

1

0

xs−1 log F w(x) dx + hw(s)

,(3.6)

where hw(s) is an entire function. On the other hand, the hypothesis (3.4) impliesthat 1

0

xs−1 log F w(x) dx = B

1

0

xs−b−1 dx + o

1

0

xσ−b−1 dx

(3.7)

=B

s − b+ o

1

σ − b

as σ → b+. Combining these two estimates yields (3.5).To show that the Dirichlet series f w(s) has abscissa of absolute convergence b, we

observe that, by (3.6) and (3.7), f w(s) is analytic on the half-plane s : Re s > b,but has a singularity at the real point s = b. Since w(n) is non-negative, Lemma 3implies that f w(s) has abscissa of absolute convergence b.

Lemma 5. Suppose that

gw(s) = f w(s) − B

Γ(b)ζ (1 + b)

1

s − b

can be meromorphically continued to a half-plane s : Re s > σ0, where σ0 < b.Let θ be the least upper bound of the real parts of the singularities of gw(s), and suppose that gw(s) is analytic at the real point s = θ. Then, for any > 0,

log F w(x) − Bx−b = Ω±(x−(θ−)) as x → 0.

Proof. Let be given and consider the function

Φ(x) = log F w(x) − Bx−b − x−(θ−)

with Mellin transform

φ(s) =

1

0

xs−1Φ(x) dx.



2594 YIFAN YANG

Let σa be the abscissa of absolute convergence of φ(s). By Lemma 2, we have forRe s > b

φ(s) = Γ(s)ζ (1 + s)f w(s) − hw(s) −B

s − b −1

s − (θ − ) ,(3.8)

where hw(s) is an entire function. By the definition of θ, for any δ > 0, the right-hand side of (3.8) has singularities on the half-plane s : Re s > θ − δ; thus wehave

σa ≥ θ.(3.9)

Now, if Φ(x) ≤ 0 for all sufficiently small x, then Lemma 3 implies that thereal point s = σa is a singularity of φ(s). But by the assumption of Lemma 5,φ(s) is analytic at the real point s = θ and has no singularities to the right of s = θ. Thus σa < θ, which contradicts (3.9). It follows that Φ(x) changes signinfinitely often, i.e., log F w(x) − Bx−b = Ω+(x−(θ−)). A similar argument giveslog F w(x) − Bx−b = Ω−(x−(θ−)).

4. Proof of Theorems 1-3

We begin by proving two lemmas which show that the function pΛ(n) in ourtheorems can be replaced by its summatory function P Λ(u) =

n≤u pΛ(n).

Lemma 6. For any integer k ≥ 15 and n ≥ 0, we have pΛ(n + k) ≥ pΛ(n).

Proof. We first observe that, for any odd prime power q,

∞n=0

( pΛ(n + q) − pΛ(n))xn +

q−1j=0

pΛ( j)x−q+j

= (x−q − 1)

∞

m=1

(1 − xm)−Λ(m) = x−q(1 − xq)−(Λ(q)−1)

m=q

(1 − xm)−Λ(m).

Since Λ(q)− 1 > 0 when q is an odd prime power, expanding each factor in the lastexpression yields a series with non-negative coefficients. Thus we have pΛ(n + q) ≥

pΛ(n) for all n ≥ 0. Iterating this inequality with q = 3 and q = 5, we see that pΛ(n + 3k + 5l) ≥ pΛ(n) for all non-negative integers n, k and l. Since everyinteger ≥ 15 can be written as a linear combination of 3 and 5 with non-negativecoefficients, the lemma follows.

Lemma 7. Let n ∈ N. Let r(u) be a function satisfying r(u)(log u)−1 → ∞ asu → ∞, and r(u + v) r(u) uniformly for all sufficiently large u and 0 ≤ v ≤ 15.Then

log pΛ(n) = 2

ζ (2)n1/2 + O(r(n))

as n→ ∞

if and only if

log P Λ(u) = 2

ζ (2)u1/2 + O(r(u))

as u → ∞. A similar statement holds if O is replaced by Ω±.

Proof. By Lemma 6, we have

pΛ(n) ≤ P Λ(n) =n

k=0

pΛ(k) ≤ (n + 1) pΛ(n + 15)(4.1)




for all integers n ≥ 0. Suppose that, for some positive constant C ,

2 ζ (2)u1/2 − Cr(u) ≤ log P Λ(u) ≤ 2 ζ (2)u1/2 + Cr(u)

for all sufficiently large u. Then (4.1) implies that

2

ζ (2)(n − 15)1/2 − log(n − 14) − Cr(n − 15) ≤ log pΛ(n) ≤ 2

ζ (2)n1/2 + Cr(n)

for all sufficiently large integers n. Hence, by the assumptions on r(u), we obtain

log pΛ(n) = 2

ζ (2)n1/2 + O(r(n)).

The proof of the converse implication and of the analogous implications betweenΩ±-estimates is similar.

By Lemma 7, it suffices to prove Theorems 1-3 with P Λ(u) in place of pΛ(n).

Proof of Theorem 1. By the Vinogradov-Korobov zero-free region for the Riemannzeta function (Titchmarsh [15, p. 135]), we have ζ (s)

= 0 for s = σ + it satisfying

σ ≥ η(t) = 1 − C

(log t∗)2/3(log2 t∗)1/3,

where logk is the k times iterated logarithm, t∗ = max(10, |t|), and C is a positiveconstant. Furthermore, in the same region we have

ζ (s)

ζ (s) log t∗,(4.2)

provided |s − 1| ≥ δ for some positive constant δ.Applying Lemma 1 with w(n) = Λ(n) and noting that in this case f w(s) =

−ζ (s)/ζ (s), we obtain, for 0 < x < 1,

log F Λ(x) = −1

2πi κ+iT

κ−iT Γ(s)ζ (1 + s)

ζ (s)

ζ (s) x

−s

ds + O

e

−T ζ (κ)

ζ (κ) x

−κ,

where 1 < κ < 2 and T > 10 are positive numbers to be chosen later. We shiftthe path of integration to the path consisting of the segments γ 1 : σ − iT : κ ≥σ > η(T ), γ 2 : η(t) + it : −T ≤ t < −10, γ 3 : η(t) + it : −10 ≤ t < 10,γ 4 : η(t) + it : 10 ≤ t < T and γ 5 : σ + iT : η(t) ≤ σ ≤ κ.

The estimate

|Γ(s)| (1 + |t|)σ+1/2e−π|t|/2 e−|t| (1

2≤ σ ≤ 2, t ∈ R)

(which follows from Stirling’s formula) and (4.2) imply that γ1,5

e−T (log T )x−κ

and γ2,4

T

10

e−t(log t)x−η(t) dt x−η(T ).

We also have γ3

x−η(10).



2596 YIFAN YANG

Thus, we obtain

log F Λ(x) = ζ (2)x−1 + O

e−T ζ

(κ)ζ (κ) x−κ

+ O

e−T (log T )x−κ

+ O

x−η(T )

.

(4.3)

We now assume that x is sufficiently small so that log3 x−1 is defined and ≥ 1. Weset κ = 1 + 1/ log x−1 and

T =C log x−1

(log2 x−1)2/3(log3 x−1)1/3,

where C is the constant appearing in the definition of η(t). Then x−κ = ex−1 andζ (κ)/ζ (κ) log x−1. Hence, noting that

1 − η(T ) ∼ C

(log2 x−1)2/3(log3 x−1)1/3

as x → 0, we obtain from (4.3)

log F Λ(x) = ζ (2)x−1 + O

x−1 exp

− (C/2)log x−1

(log2 x−1)2/3(log3 x−1)1/3

.(4.4)

We now apply Proposition 2 with B = B = ζ (2), b = 1 and the quantities A anda determined by (2.3), i.e.,

a = 1/2 and A = a−1 (Ba/(1 − a))1−a

= 2

ζ (2).(4.5)

Setting y = Bx−b−1 = ζ (2)x−2 and

r(y) = y1/2 exp

− (C/5) log y

(log2 y)2/3(log3 y)1/3

,

it easily follows from (4.4) that

| log F Λ(x) − ζ (2)x−1| ≤ r(ζ (2)x−2)

for all sufficiently small x. Hence, by Proposition 2, we may conclude that, withc = C/10,

log P Λ(u) = Aua + O(

r(u)ua)

= 2

ζ (2)u1/2 + O

u1/2 exp

− c log u

(log2 u)2/3(log3 u)1/3

.

This completes the proof of Theorem 1.

Proof of Theorem 2. We will prove only the Ω−-estimate; the Ω+-estimate can beproved similarly.

Applying the corollary to Proposition 1 with P (u) = P Λ(u), F (x) = F Λ(x),

A = 2

ζ (2), a = 1/2, B = B = ζ (2), b = 1 and r(u) = u1/4, we see that it suffices

to show that

log F Λ(x) − ζ (2)x−1 = Ω−(x−1/2).(4.6)

Let θ be the least upper bound of the real parts of the zeros of the Riemann zetafunction. If the Riemann Hypothesis is false, i.e., if θ > 1/2, we apply Lemma 5with w(n) = Λ(n), f w(s) = −ζ (s)/ζ (s), B = ζ (2) and b = 1. The function gw(s)in Lemma 5 then is given by −ζ (s)/ζ (s) − 1/(s − 1). Thus gw(s) is a meromorphicfunction in Re s > 0, whose singularities are exactly the zeros of ζ (s). In particular,the least upper bound of the real parts of these singularities is equal to the least




upper bound θ of the real parts of the zeros of ζ (s). Moreover, since ζ (s) has nozeros on the real line, gw(s) is analytic at s = θ. Thus, Lemma 5 implies that, forany > 0,

log F Λ(x) − ζ (2)x−1 = Ω±(x−(θ−)).

In particular, we can choose = θ − 1/2 and the assertion of Theorem 2 follows.We therefore assume that the Riemann Hypothesis is true.

Let 1/2 + iγ be one of the non-trivial zeros of the Riemann zeta function and letm be its multiplicity. Let C be a positive constant such that

C < m|Γ(1/2 + iγ )ζ (3/2 + iγ )|(4.7)

and define

Φ(x) = log F Λ(x) − ζ (2)x−1 + Cx−1/2.

If (4.6) is false, then there exists x0 > 0 such that Φ(x) ≥ 0 holds for all x ≤ x0.By Lemma 3, the integral

φ(s) =

1

0

Φ(x)xs−1 dx

has a singularity at the real point s = σa, the abscissa of absolute convergence of φ(s). On the other hand, applying Lemma 2 with f w(s) = −ζ (s)/ζ (s), we obtain

φ(s) = −Γ(s)ζ (1 + s)ζ (s)

ζ (s)− hΛ(s) − ζ (2)

s − 1+

C

s − 1/2,(4.8)

where hΛ(s) is an entire function. Since the last expression has no singularity onthe real axis to the right of 1/2, the abscissa of absolute convergence σa of theintegral φ(s) must be ≤ 1/2. For σ > 1/2 we have

|φ(σ + iγ )

| ≤ 1

x0 |Φ(x)

|xσ−1 dx +

x0

0

Φ(x)xσ−1 dx(4.9)

=

1

x0

|Φ(x)| − Φ(x)xσ−1 dx +

1

0

Φ(x)xσ−1 dx

≤ 1

x0

2|Φ(x)|x−1/2 dx + φ(σ).

We multiply both sides by σ −1/2 and let σ → 1/2+. The right-hand side becomeslimσ→1/2+(σ − 1/2)φ(σ), which, by (4.8), is equal to

limσ→1/2+

(σ − 1/2)

− Γ(σ)ζ (1 + σ)ζ (σ)

ζ (σ)− ζ (2)

σ − 1+

C

σ − 1/2

= C.

On the other hand, applying (4.8) to φ(σ + iγ ), we see that the left-hand side of

(4.9) becomes (after multiplying by σ − 1/2 and letting σ → 1/2+)

limσ→1/2+

(σ − 1/2)

− Γ(σ + iγ )ζ (1 + σ + iγ )ζ (σ + iγ )

ζ (σ + iγ )

− ζ (2)

σ − 1 + iγ +

C

σ − 1/2 + iγ

= m|Γ(1/2 + iγ )ζ (3/2 + iγ )|.

Hence m|Γ(1/2 + iγ )ζ (3/2 + iγ )| ≤ C , which contradicts (4.7). Therefore (4.6)holds, and the proof of Theorem 2 is complete.



2598 YIFAN YANG

Proof of Theorem 3. Let θ be the least upper bound for the real parts of the zerosof the Riemann zeta function. It suffices to show that if α is a real number suchthat the estimate

log P Λ(u) = 2

ζ (2)u1/2 + O(uα/2)(4.10)

holds as u → ∞, then α ≥ θ.By Proposition 1, (4.10) implies that

log F Λ(x) = ζ (2)x−1 + O(x−α).

Thus the integral

φ(s) =

1

0

xs−1

log F Λ(x) − ζ (2)x−1

dx

defines an analytic function on the half-plane s : Re s > α. On the other hand,by Lemma 2 we have, for Re s > 1,

φ(s) = −Γ(s)ζ (1 + s) ζ (s)ζ (s)

− ζ (2)s − 1

− hΛ(s),(4.11)

where hΛ(s) is an entire function. Since φ(s) is analytic in Res > α, the latterrelation remains valid in this larger region, and the right-hand side of (4.11) isanalytic in Res > α. Therefore ζ (s) cannot have zeros on the half-plane s : Re s >α, i.e., we have θ ≤ α, as claimed.

5. A New Proof of Theorem B

Let θ be the least upper bound for the real parts of the zeros of the Riemannzeta function. We may assume 1/2 ≤ θ < 1, for if θ = 1, the asserted estimatefollows from Theorem 1.

By Lemma 1 we have, for 0 < x < 1 and κ > 1,

log F Λ(x) = − 1

2πi

κ+i∞

κ−i∞

Γ(s)ζ (1 + s)ζ (s)

ζ (s)x−s ds.

By standard arguments (as, for example, in the proof of Theorems 28 of Ingham[5]) we now shift the line of integration to the vertical line Re s = −1/2, taking intoaccount the residues of the integrand at s = 1, 0 and at the zeta-zeros. This gives

log F Λ(x) = ζ (2)x−1 −

ρ

Γ(ρ)ζ (1 + ρ)x−ρ − ζ (0)

ζ (0)log x−1

− 1

2πi

−1/2+i∞

−1/2−i∞

Γ(s)ζ (1 + s)ζ (s)

ζ (s)x−s ds,

where ρ runs over non-trivial zeta-zeros, counted with multiplicities.By the symmetry of the zeta-zeros and the assumption that θ < 1, we have

|ζ (1 + ρ)| ≤ ζ (1 + (1 − θ)) 1

for all zeta-zeros ρ. Using Stirling’s formula and the boundT ≤Im ρ≤T +1

1 log T,




we obtain

ρ

Γ(ρ)ζ (1 + ρ)x−ρ ρ=β+iγ

e−|γ|x−β x−θ.

Also, for Re s = −1/2 we have ζ (s)/ζ (s) log(|s| +2) and ζ (1 + s) (|s| + 1)1/2

(Theorems 9 and 27 in [5]). Thus −1/2+i∞

−1/2−i∞

Γ(s)ζ (1 + s)ζ (s)

ζ (s)x−s ds

x1/2

∞−∞

e−|t|(|t| + 1)1/2 log(|t| + 2) dt x1/2.

Hence, we obtain

log F Λ(x) = ζ (2)x−1 + O(x−θ).(5.1)

Similarly, we have, for κ > 1,

ddx

log F Λ(x) = 12πi

κ+i∞

κ−i∞

sΓ(s)ζ (1 + s) ζ

(s)ζ (s)

x−s−1 ds(5.2)

= −ζ (2)x−2 + O(x−(θ+1))

and

d2

dx2log F Λ(x) = − 1

2πi

κ+i∞

κ−i∞

s(1 + s)Γ(s)ζ (1 + s)ζ (s)

ζ (s)x−s−2 ds(5.3)

= 2ζ (2)x−3 + O(x−(θ+2)).

By (5.1)-(5.3), the function F (x) = F Λ(x) satisfies the hypotheses (2.15), (2.22)and (2.23) of Proposition 3 with b = 1, B = B = ζ (2) and r(u) = Kuθ/2, whereK is a sufficiently large constant. Moreover, the function r(u) satisfies conditions(R1), (R2) and (R3) of that proposition. Hence the conclusion (2.24) of Proposition

3 holds with A and a given by (4.5). Therefore

log P Λ(u) = Aua + O(r(u)) = 2

ζ (2)u1/2 + O(uθ/2)

as u → ∞, which is the claimed result.

Acknowledgment

The author wishes to thank Professor A. Hildebrand for helpful advice on theexposition of the paper.

References

1. N. A. Brigham, On a certain weighted partition function , Proc. Amer. Math. Soc. 1 (1950),192–204. MR 11:582c

2. G. A. Freiman, Inverse problems of the additive theory of numbers, Izv. Akad. SSSR. Ser.Mat. 19 (1955), 275-284. (Russian) MR 17:239c

3. G. H. Hardy and S. Ramanujan, Asymptotic formulae in combinatory analysis, Proc. London

Math. Soc. (2) 17 (1918), 75–115.4. A. E. Ingham, A Tauberian theorem for partitions, Ann. of Math. (2) 42 (1941), 1075–1090.

MR 3:166a

5. , The distribution of prime numbers, Cambridge University Press, Cambridge, 1990.MR 91f:11064

6. E. E. Kohlbecker, Weak asymptotic properties of partitions, Trans. Amer. Math. Soc 88

(1958), 346–365. MR 20:2309

http://www.ams.org/mathscinet-getitem?mr=11:582c






http://www.ams.org/mathscinet-getitem?mr=3:166a



http://www.ams.org/mathscinet-getitem?mr=91f:11064



http://www.ams.org/mathscinet-getitem?mr=20:2309










2600 YIFAN YANG

7. G. Meinardus, Asymptotische Aussagen ¨ uber Partitionen , Math. Z. 59 (1954), 388–398.

MR 16:17e8. , ¨ Uber Partitionen mit Differenzenbedingungen , Math. Z. 61 (1954), 289–302.

MR 16:905a9. A. M. Odlyzko, Explicit Tauberian estimates for functions with positive coefficients, J. Com-

put. Appl. Math. 41 (1992), 187–197. MR 94f:11086

10. B. Richmond, Asymptotic relations for partitions, J. Number Theory 7 (1975), 389–405.MR 52:3095

11. , A general asymptotic result for partitions, Canad. J. Math. 27 (1975), 1083–1091.

MR 52:560412. K. F. Roth and G. Szekeres, Some asymptotic formulae in the theory of partitions, Quart. J.

Math., Oxford Ser. (2) 5 (1954), 241–259. MR 16:797b13. W. Schwarz, Schwache asymptotische Eigenschaften von Partitionen , J. Reine Angew. Math.

232 (1968), 1–16. MR 38:4433

14. , Asymptotische Formeln f¨ ur Partitionen , J. Reine Angew. Math. 234 (1969), 174–178.MR 40:7217

15. E. C. Titchmarsh, The theory of the Riemann zeta-function , 2nd ed., The Clarendon Press,Oxford University Press, New York, 1986. MR 88c:11049

Department of Mathematics, University of Illinois, Urbana, Illinois 61801

E-mail address: [email protected]

http://www.ams.org/mathscinet-getitem?mr=16:17e















http://www.ams.org/mathscinet-getitem?mr=16:797b









http://www.ams.org/mathscinet-getitem?mr=88c:11049












Yang Y - Partitions Into Primes - Trans. Amer. Math. Soc. 352 (2000) No.6 2581-2600

Documents

Transcript of Yang Y - Partitions Into Primes - Trans. Amer. Math. Soc. 352 (2000) No.6 2581-2600