For k 1, denote by p k (n) the number of partitions of an integer n into k-th powers. In this note, we apply the saddle-point method to provide a new proof for the well-known asymptotic expansion of p k (n). This approach turns out to significantly simplify those of Wright (1934), Vaughan (2015) and Gafni (2016)

Li, Yali

Tenenbaum, Gérald

Wu, Jie

English

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POWER PARTITIONS AND SADDLE-POINT
METHOD
Gérald Tenenbaum, Jie Wu, Yali Li
To cite this version:
Gérald Tenenbaum, Jie Wu, Yali Li. POWER PARTITIONS AND SADDLE-POINT METHOD.
2019. ￿hal-02323099￿
POWER PARTITIONS AND SADDLE-POINT METHOD
GE´RALD TENENBAUM, JIE WU AND YALI LI
Abstract. For k > 1, denote by pk(n) the number of partitions of an integer n into k-th
powers. In this note, we apply the saddle-point method to provide a new proof for the well-
known asymptotic expansion of pk(n). This approach turns out to significantly simplify those
of Wright (1934), Vaughan (2015) and Gafni (2016).
1. Introduction
Let p(n) denote, as usual, the number of unrestricted partitions of an integer n, i.e. the
number of solutions to the equation
n = a1 + a2 + · · ·+ ad,
where d > 1 and the aj are positive integers such that a1 > a2 > · · · > ad > 1. In 1918, Hardy
and Ramanujan [5] proved the asymptotic formula
(1.1) p(n) ∼ exp(pi
√
2n/3)
4
√
3n
(n→∞),
by using modular properties of Jacobi’s ∆-function.
More generally, given an integer k > 1, let pk(n) denote the number of partitions of the
integer n into k-th powers, i.e. the number of solutions to the equation
n = ak1 + · · ·+ akd
where, as before, d > 1 and a1 > a2 > · · · > ad > 1. Thus, p1(n) = p(n).
In 1918, Hardy and Ramanujan [5] also stated without proof the asymptotic formula
(1.2) pk(n) ∼ bk exp(ckn
1/(k+1))
n(3k+1)/(2k+2)
(n→∞),
where the constants bk and ck are defined by
ak := {k−1ζ(1 + k−1)Γ(1 + k−1)}k/(k+1),(1.3)
bk :=
ak
(2pi)(k+1)/2
√
(1 + 1/k)
,(1.4)
ck := (k + 1)ak,(1.5)
and ζ is the Riemann zeta-function. In 1934, introducing a number of complicated objects
including generalised Bessel functions, Wright [9] obtained an asymptotic expansion of pk(n):
for any integer k > 1, there is a real sequence {αkj}j>1 such that, for any J > 1, we have
(1.6) pk(n) =
bk exp(ck(n+ hk)
1/(k+1))
(n+ hk)(3k+1)/(2k+2)
{
1 +
∑
16j<J
(−1)jαkj
(n+ hk)j/(k+1)
+O
(
1
nJ/(k+1)
)}
,
Date: May 25, 2019.
2010 Mathematics Subject Classification. 05A17, 11N37, 11P82.
Key words and phrases. Asymptotic estimates, partitions, partitions into powers, saddle-point method.
1
2 GE´RALD TENENBAUM, JIE WU AND YALI LI
where
hk :=
{
0 if k is even,
(−1)(k+1)/2(2pi)−(k+1)k!ζ(k + 1) if k is odd,
and the implied constant depends at most on J and k. Apart from an explicit formula for
αk1, no further information was given about the αkj beyond the statement that they depend
only on k and j and that they “may be calculated with sufficient labour for any given values
of k, j”.
Of course, taking J = 1 in (1.6) yields an effective form of (1.2).
More recently, appealing to a relatively simple implementation of the Hardy-Littlewood
circle method, Vaughan [7] obtained an explicit version of (1.6) in the case k = 2 and Gafni [4]
generalised the argument to arbitrary, fixed k.
Gafni states her result in the following way. Let X = Xk(n) denote the real solution to the
equation
(1.7) n = (akX)
1+1/k − 12X − 12ζ(−k),
and write
(1.8) Y = Yk(n) := (1 + 1/k)a
1+1/k
k X
1/k − 14 ·
Then, given any k > 1, there is a real sequence {βkj}j>1 such that for any fixed J > 1, we
have
(1.9) pk(n) =
exp
(
(k + 1)a
1+1/k
k X
1/k − 12
)
(2pi)(k+1)/2X3/2Y 1/2
{
1 +
∑
16j<J
βkj
Y j
+O
(
1
Y J
)}
.
It may be checked that the asymptotic formulae (1.6) and (1.9) match each other.
In this note, our aim is to provide a new proof of (1.2), and indeed also of (1.6) and (1.9),
by applying the saddle-point method along lines very similar to those employed in [6] in the
case of p(n).
Our approach appears to be significantly simpler than those of the quoted previous works.
Indeed, as mentioned above, Wright’s method, which provides a sharper error term, rests on
the introduction of generalised Bessel functions and so may be regarded as far more difficult.
Note that Vaughan [7] does motivate his study by the search of a ‘relatively simple argument’.
Our method simplifies the matter further. Actually, the saddle-point method may be seen
as a crude version of the circle method in which the major arcs are reduced to a single neigh-
bourhood of one point. In the present context, it actually provides the same accuracy.
We thank the referee for pointing out to us that the Vaughan-Gafni method has been
very recently generalised to enumerate other classes of partitions—see [1] and [3]. We believe
that the saddle-point method could still be used in these contexts, with expected substantial
simplifications in the analysis.
The constants bk and ck being defined as in (1.4) and (1.5), we can state the following.
Theorem 1. Let k > 1 be a fixed integer. There is a real sequence {γkj}j>1 such that, for any
given integer J > 1, we have
(1.10) pk(n) =
bk exp(ckn
1/(k+1))
n(3k+1)/(2k+2)
{
1 +
∑
16j<J
γkj
nj/(k+1)
+O
(
1
nJ/(k+1)
)}
uniformly for n > 1. The implied constant depends at most on J and k.
POWER PARTITIONS AND SADDLE-POINT METHOD 3
The coefficients γkj can be made explicit directly from the computations in our proof. For
instance, we find that γk1 = −(11k2 + 11k + 2)/(24kck) when k > 2, in accordance with the
expression given by Wright. (It can be checked, after some computations, that it matches
Gafni’s formula too.) We also have
γ11 = − 148c1 − 1/c1 = −
√
2
3
( pi
48
+
3
2pi
)
.
It may be seen that |γkj | grows like Γ(j)eO(j) and thus that the series
∑
j>1 γkjz
j has radius
of convergence 0.
2. Technical preparation
Define
(2.1) Fk(s) :=
∑
n>0
pk(n)e
−ns (<e s > 0),
so that
(2.2) pk(n) =
1
2pii
∫ σ+ipi
σ−ipi
Fk(s)e
ns ds =
1
2pi
∫ pi
−pi
Fk(σ + iτ)e
nσ+inτ dτ.
According to the principles of the saddle-point method, we aim at selecting the integration
abscissa σ as a solution σn of −F ′k(σ)/Fk(σ) = n. We plainly have
(2.3) Fk(s) =
∏
m>1
(
1− e−mks)−1 (<e s > 0).
Thus, in the same half-plane, we may define a determination of logFk(s) by the formula
Φk(s) :=
∑
m>1
log
( 1
1− e−mks
)
where the complex logarithms are taken in principal branch. Expanding throughout and
inverting summations, we get
(2.4) Φk(s) =
∑
m>1
∑
n>1
e−mkns
n
=
∑
r>1
wk(r)
r
e−rs, −Φ′k(s) =
∑
r>1
wk(r)e
−rs (<e s > 0),
where
wk(r) :=
∑
mk | r
mk (r > 1).
Hence −Φ′k(σ) decreases from +∞ to 0+ on (0,∞), and so the equation −Φ′k(σ) = n has for
each integer n > 1 a unique real solution σn = σn(k). Moreover, the sequence {σn}n>1 is
decreasing and the trivial estimates 1 6 wk(r) 6 r2 yield 1/n σn  1/ 3
√
n.
We start with an asymptotic expansion for the derivatives Φ
(m)
k (σn) in terms of powers of σn.
It turns out that all coefficients but a finite number vanish.
Lemma 2.1. Let J > 1, k > 1. As n→∞, we have
(2.5) Φk(σn) =
ka
1+1/k
k
σ
1/k
n
+ 12 log
( σn
(2pi)k
)
+ 12ζ(−k)σn +O
(
σJn
)
.
Moreover, for fixed m > 1,
(2.6) (−1)mΦ(m)k (σn) =
a
1+1/k
k
σ
m+1/k
n
∏
16`<m
(
`+
1
k
)
− (m− 1)!
2σmn
− 12δ1mζ(−k) +O(σJn),
4 GE´RALD TENENBAUM, JIE WU AND YALI LI
where δ1m is Kronecker’s symbol.
Proof. Considering Mellin’s inversion formula
e−s =
1
2pii
∫ 2+i∞
2−i∞
Γ(z)s−z dz (<e s > 0)
and the convolution identity
(2.7)
∑
r>1
wk(r)
r1+z
= ζ(z + 1)ζ(kz) (<e z > 1/k),
we derive from the series representation (2.4) the integral formula
(2.8) Φk(s) =
1
2pii
∫ 2+i∞
2−i∞
ζ(z + 1)ζ(kz)Γ(z)
dz
sz
,
and in turn
(2.9) (−1)mΦ(m)k (σn) =
1
2pii
∫ 2+i∞
2−i∞
ζ(z + 1)ζ(kz)Γ(z +m)
dz
σz+mn
(m > 0).
Using the classical fact that ζ(z) has finite order in any vertical strip a 6 <e z 6 b (a, b ∈ R
with a < b), or, in other words, satisfies
ζ(x+ iy)a,b 1 + |y|A (a 6 x 6 b, |y| > 1),
for suitable A = A(a, b), and invoking Stirling’s formula in the form
|Γ(x+ iy)| =
√
2pi|y|x−1/2e−pi|y|/2{1 +Oa,b(1/y)} (a 6 x 6 b, |y| > 1)
we may move the line of integration to <e z = −J −m− 12 .
The shifted integral is clearly  σJn .
Let us first consider the case m = 0 in (2.9). Then the crossed singularities are a pole of
order 2 at z = 0, and two simple poles at z = 1/k and z = −1. Indeed, ζ(z + 1)ζ(kz) vanishes
at all negative integers 6 −2, so the corresponding zeros compensate the poles of Γ(z) at
negative integers 6 −2.
The residue at z = 1/k is equal to
k−1ζ(1 + k−1)Γ(k−1)σ−1/kn = ka
1+1/k
k σ
−1/k
n .
The residue at z = 0 is the coefficient of z in the Taylor expansion of
z2ζ(z + 1)ζ(kz)Γ(z)σ−zn = zζ(z + 1)ζ(kz)Γ(z + 1)σ
−z
n
= (1− γz){ζ(0) + kζ ′(0)z}(1 + γz)(1− z log σn) +O(z2)
= ζ(0) + {−ζ(0) log σn + kζ ′(0)}z +O(z2).
Since ζ(0) = −12 and ζ ′(0) = −12 log(2pi), this residue equals 12 log{σn/(2pi)k}.
The residue at z = −1 equals 12ζ(−k)σn.
This completes the proof of (2.5).
When m = 1, the three crossed singularities are simple poles. The residues at z = 1/k,
z = 0 and z = −1 are respectively (1/k)Γ(1 + 1/k)ζ(1 + 1/k)σ−1−1/kn , −1/2σn and −12ζ(−k).
When m > 2, the only crossed singularities are two simple poles, at z = 1/k and z = 0,
with respective residues (1/k)Γ(m+ 1/k)ζ(1 + 1/k)σ
−m−1/k
n and −12(m− 1)!σ−mn . This proves
(2.6). 
POWER PARTITIONS AND SADDLE-POINT METHOD 5
Lemma 2.2. Let J > 1, k > 1, m > 1.
(i) There is a real sequence {akj}j>1 with ak1 = −k/(2ck), ak2 = k/(8c2k), such that
(2.10) σn =
ak
nk/(k+1)
{
1 +
∑
16j<J
akj
nj/(k+1)
+O
( 1
nJ/(k+1)
)}
(n→∞).
(ii) There is a real sequence {bkj}j>1 with bk1 = −ak1/k such that, as n→∞, we have
(2.11) Φk(σn) = kakn
1/(k+1)
{
1 +
∑
16j<J
bkj
nj/(k+1)
+O
( 1
nJ/(k+1)
)}
+ 12 log
( σn
(2pi)k
)
.
(iii) There is a real sequence {bkmj}j>1 such that, as n→∞, we have
(2.12)
(−1)mΦ(m)k (σn) + 12δ1mζ(−k)
=
n(mk+1)/(k+1)
am−1k
∏
16`<m
(
`+
1
k
){
1 +
∑
16j<J
bkmj
nj/(k+1)
+O
( 1
nJ/(k+1)
)}
.
Proof. We infer from (2.6) that
(2.13) n =
a
1+1/k
k
σ
1+1/k
n
− 1
2σn
− 12ζ(−k) +O(σJn).
This immediately implies (2.10) by Lagrange’s inversion formula — see, e.g. [8, §7.32]. We
may obtain an explicit expression for the akj from the formula
(2.14) σn =
ak
2piink/(k+1)
∮
|z−1|=%
zG′(z)
G(z)
dz +O
( 1
n(k+J)/(k+1)
)
where % is a fixed, small positive constant and
G(z) := z−1−1/k − 1− 1
2akzn1/(k+1)
− ζ(−k)
2n
·
This is classically derived from Rouche´’s theorem and we omit the details. The values of ak1
and ak2 may be retrieved from the above or by formally inserting (2.10) into (2.13).
Inserting (2.14) back into (2.5) and (2.6) immediately yields (2.11) and (2.12). 
The following statement appears in [2, Lemma 6.3]. Here and in the sequel, we employ the
standard notation ‖t‖ := minn∈Z |n− t| (t ∈ R).
Lemma 2.3. Let ϑ ∈ R, a ∈ Z, q ∈ N with (a, q) = 1, ϑ = a/q + β, |β| 6 1/q2, t ∈ N,
0 6 v < q. Then there exist at most six integers r with 0 6 r < q such that
(2.15) ‖ϑ(tq + r)‖ ∈ (v/q, (v + 1)/q].
With the aim of applying Laplace’s method to evaluate the integral on the right-hand side of
(2.2), we need to show that it is dominated by a small neighbourhood of the saddle-point σn.
The next result meets this requirement. Here and in the sequel, all constants cj (j > 0) are
assumed, unless otherwise stated, to depend at most upon k.
Lemma 2.4. We have
(2.16)
|Fk(σn + iτ)|
|Fk(σn)| 6
e−c1τ
2σ
−(2+1/k)
n if |τ | 6 2piσn,
e−c2σ
−1/k
n if 2piσn < |τ | 6 pi.
6 GE´RALD TENENBAUM, JIE WU AND YALI LI
Proof. Noticing that∣∣1− e−mk(σn+iτ)∣∣2 = ∣∣1− e−mkσn∣∣2 + 4e−mkσn sin2(12mkτ)
>
∣∣1− e−mkσn∣∣2 + 16e−mkσn‖mkτ/(2pi)‖2,
we can write
|Fk(σn + iτ)|2
|Fk(σn)|2 6
∏
m>1
(
1 +
16‖mkτ/(2pi)‖2
emkσn(1− e−mkσn)2
)−1
6
∏
(4σn)−1/k<m6(2σn)−1/k
(
1 +
16‖mkτ/(2pi)‖2
emkσn(1− e−mkσn)2
)−1
.
Thus, there is an absolute positive constant c3 such that
(2.17)
|Fk(σn + iτ)|
|Fk(σn)| 6 e
−c3S(τ ;σn)
with
S(τ ;σn) :=
∑
(4σn)−1/k<m6(2σn)−1/k
‖mkτ/(2pi)‖2.
If |τ | 6 2piσn and m 6 (2σn)−1/k, we have |mkτ/(2pi)| 6 12 . Thus
(2.18) S(τ ;σn) =
∑
(4σn)−1/k<m6(2σn)−1/k
m2kτ2/4pi2  τ2σ−(2+1/k)n .
When 2piσn < |τ | 6 pi, Dirichlet’s approximation lemma guarantees that there exist integers
a ∈ Z∗ and q ∈ [1, σ−1/kn ] such that
|τ/(2pi)− a/q| 6 σ1/kn /q 6 1/q2.
According to Lemma 2.3, there are  q integers v ∈ [1, q) such that the inequality
‖mkτ/(2pi)‖  v/q
holds for  σ−1/kn /q integers m from the interval [(4σn)−1/k, (2σn)−1/k]. Hence S(τ ;σn) 
σ
−1/k
n . 
3. Completion of the proof
Proposition 3.1. Let k > 1, J > 1. Then there is a real sequence {ekj}j>1 such that for any
integer J > 1 we have
(3.1) pk(n) =
exp(nσn + Φk(σn))√
2piΦ′′k(σn)
{
1 +
∑
26j<J
ekj
nj/(k+1)
+O
(
1
nJ/(k+1)
)}
(n→∞).
Proof. By (2.2), we have
(3.2) pk(n) =
enσn
2pi
∫ pi
−pi
eΦk(σn+iτ)+inτ dτ.
From (2.16), we deduce that
(3.3)
∫
2piσn<|τ |6pi
eΦk(σn+iτ)+inτ dτ  eΦk(σn)−c4σ−1/kn ,∫
σ
1+1/(3k)
n <|τ |62piσn
eΦk(σn+iτ)+inτ dτ  eΦk(σn)−c4σ−1/(3k)n .
POWER PARTITIONS AND SADDLE-POINT METHOD 7
Since these bounds are exponentially small with respect to the expected main term, it only
remains to estimate the contribution of the interval I := ]−σ1+1/(3k)n , σ1+1/(3k)n [, corresponding
to a small neighbourhood of the saddle-point.
In this range, we have
Φk(σn + iτ) =
∑
06m62J+1
Φ
(m)
k (σn)
m!
(iτ)m +O
(
τ2J+2
σ
1/k+2J+2
n
)
,
where the estimate for the error term follows from (2.6). The same formula ensures that
|Φ(m)k (σn)τm|  1 for m > 3. Thus for τ ∈ I, we can write
eΦk(σn+iτ)+inτ
= eΦk(σn)−
1
2
Φ′′k(σn)τ
2
{
1 +
∑
16`62J
1
`!
( ∑
36m62J+1
Φ
(m)
k (σn)
m!
(iτ)m
)`
+O
(
τ2J+2
σ
1/k+2J+2
n
)}
= eΦk(σn)−
1
2
Φ′′k(σn)τ
2
{
1 +
∑
16`62J
1
`!
∑
3`6m6(2J+1)`
λk,`,m(n)τ
m +O
(
τ2J+2
σ
1/k+2J+2
n
)}
,
where
(3.4) λk,`,m(n) := i
m
∑
36m1,...,m`62J+1
m1+···+m`=m
∏
16r6`
Φ
(mr)
k (σn)
mr!
·
Since the contributions from odd powers of τ vanish, we get
(3.5)
∫
I
eΦk(σn+iτ)+inτ dτ = eΦk(σn)
{
I0 +
∑
16`62J
1
`!
∑
3`62m6(2J+1)`
λk,`,2m(n)Im +O(R)
}
,
with
Im :=
∫
I
e−
1
2
Φ′′k(σn)τ
2
τ2m dτ, R := σ−1/k−2J−2n
∫
I
e−
1
2
Φ′′k(σn)τ
2
τ2J+2 dτ.
Extending the range of integration in Im involves an exponentially small error, so we get from
the classical formula for Laplace integrals
Im =
√
2pi(2m)!
m!2mΦ′′k(σn)m+1/2
+O
(
e−c5n
1/(3k+3)
)
, R  σ1+(J+1/2)/kn 
1√
Φ′′k(σn)nJ/(k+1)
·
Inserting these estimates back into (3.5) and expanding all arising factors Φ
(m)
k (σn) by (2.12),
we obtain (3.1). 
Remark. From (3.3) and (3.5) we see that, when k > 2,
(3.6) pk(n) =
enσn+Φk(σn)√
2piΦ′′k(σn)
{
1− 2k
2 + 5k + 2
24kck
(σn
ak
)1/k
+O
(
σ2/kn
)}
where, in view of (2.6), the quantity inside curly brackets may be replaced by an asymptotic
series in powers of σ
1/k
n . Inserting (2.5) and (2.10) in the main term, we thus get a formula
which is very close to, but simpler than (1.9), since it follows from (2.13) that X and 1/σn
agree to any power of σn.
We are now in a position to complete the proof of Theorem 1.
8 GE´RALD TENENBAUM, JIE WU AND YALI LI
We infer from (2.10) and (2.11) that
nσn + Φk(σn) = ckn
1/(k+1) +
∑
16j<J
a∗kj
nj/(k+1)
+O
( 1
nJ/(k+1)
)
+ 12 log
(
σn
(2pi)k
)
with a∗kj := ak(kak,j+1 + bk,j+1). Exponentiating and expanding, we get
(3.7)
exp(nσn + Φ(σn))
=
√
σn
(2pi)k/2
exp
(
ckn
1/(k+1)
){
1 +
∑
16`<J
1
`!
( ∑
16j<J
a∗kj
nj/(k+1)
)`
+O
( 1
nJ/(k+1)
)}
=
√
σn
(2pi)k/2
exp
(
ckn
1/(k+1)
){
1 +
∑
16j<J
fkj
nj/(k+1)
+O
( 1
nJ/(k+1)
)}
with
fkj :=
∑
16`<J
1
`!
∑
16j1,...,j`<J
j1+···+j`=j
a∗kj1 · · · a∗kj` .
It remains to insert back into (3.1) and expand
√
σn/Φ′′k(σn) according to (2.10) and (2.12)
with m = 2 to obtain the required asymptotic formula.
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Institut E´lie Cartan de Lorraine, Universite´ de Lorraine,
BP 70239, 54506 Vandœuvre-le`s-Nancy Cedex, France
E-mail address: gerald.tenenbaum@univ-lorraine.fr
CNRS LAMA 8050, Laboratoire d’Analyse et de Mathe´matiques Applique´es,
Universite´ Paris-Est Cre´teil, 94010 Cre´teil Cedex, France
E-mail address: jie.wu@math.cnrs.fr
Yali Li, School of Mathematics and Statistics, Henan University, Kaifeng, Henan 475004,
P. R. China


POWER PARTITIONS AND SADDLE-POINT METHOD

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POWER PARTITIONS AND SADDLE-POINT METHOD

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