Power partitions and saddle-point method

TL;DR

This paper introduces a simplified saddle-point method-based proof for the asymptotic behavior of the number of partitions of integers into k-th powers, improving upon previous complex proofs.

Contribution

It provides a new, more straightforward saddle-point approach to derive the asymptotics of p_k(n), streamlining earlier methods by Wright, Vaughan, and Gafni.

Findings

Simplified proof of asymptotic expansion for p_k(n)

Enhanced clarity over previous methods

Potential for broader application to partition problems

Abstract

For $k\geqslant 1$, 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).

Equations

$$n=a_1+a_2+\cdots +a_d,$$

$$p(n)\sim \frac{\exp (\pi \sqrt{2n/3})}{4\sqrt{3}n}(n\to \infty ).$$

$$n=a_1^k+\cdots +a_d^k$$

$$p_{k}(n)\sim\frac{{\mathfrak{b}}_{k}\exp\big{\{}{\mathfrak{c}}_{k}n^{1/(k+1)}\big{\}}}{n^{(3k+1)/(2k+2)}}\qquad(n\to\infty),$$

$${\mathfrak{a}}_k$$

$${\mathfrak{b}}_k$$

$${\mathfrak{c}}_k$$

$$p_{k}(n)=\frac{\mathfrak{b}_{k}\exp(\mathfrak{c}_{k}(n+h_{k})^{1/(k+1)})}{(n+h_{k})^{(3k+1)/(2k+2)}}\bigg{\{}1+\sum_{1\leqslant j<J}\frac{(-1)^{j}\alpha_{kj}}{(n+h_{k})^{j/(k+1)}}+O\bigg{(}\frac{1}{n^{J/(k+1)}}\bigg{)}\bigg{\}},$$

$$h_k:=\{\begin{array}{ll}0& \text{if k is even,}\\ (-1{)}^{(k+1)/2}(2\pi {)}^{-(k+1)}k!\zeta (k+1)& \text{if k is odd,}\end{array}$$

$$n=({\mathfrak{a}}_{k}X{)}^{1+1/k}-\frac{1}{2}X-\frac{1}{2}\zeta (-k),$$

$$Y=Y_k(n):=(1+1/k){\mathfrak{a}}_k^{1+1/k}{X}^{1/k}-\frac{1}{4}\cdot$$

$$p_{k}(n)=\frac{\exp\big{\{}(k+1){\mathfrak{a}}_{k}^{1+1/k}X^{1/k}-\tfrac{1}{2}\big{\}}}{(2\pi)^{(k+1)/2}X^{3/2}Y^{1/2}}\bigg{\{}1+\sum_{1\leqslant j<J}\frac{\beta_{kj}}{Y^{j}}+O\bigg{(}\frac{1}{Y^{J}}\bigg{)}\bigg{\}}.$$

$$p_{k}(n)=\frac{{\mathfrak{b}}_{k}\exp(\mathfrak{c}_{k}n^{1/(k+1)})}{n^{(3k+1)/(2k+2)}}\bigg{\{}1+\sum_{1\leqslant j<J}\frac{\gamma_{kj}}{n^{j/(k+1)}}+O\bigg{(}\frac{1}{n^{J/(k+1)}}\bigg{)}\bigg{\}}$$

$$\gamma_{11}=-\tfrac{1}{48}{\mathfrak{c}}_{1}-1/{\mathfrak{c}}_{1}=-\sqrt{\frac{2}{3}}\Big{(}\frac{\pi}{48}+\frac{3}{2\pi}\Big{)}.$$

$$F_k(s):=\sum _{n⩾0}{p}_k(n){\mathrm{e}}^{-ns}(\Re es>0),$$

$$p_k(n)=\frac{1}{2\pi \mathrm{i}}{\int }_{\sigma -\mathrm{i}\pi }^{\sigma +\mathrm{i}\pi }{F}_k(s){\mathrm{e}}^{ns}\mathrm{d}s=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{F}_n(\sigma +\mathrm{i}\tau ){\mathrm{e}}^{n\sigma +\mathrm{i}n\tau }\mathrm{d}\tau .$$

$$F_{k}(s)=\prod_{m\geqslant 1}\big{(}1-{\rm e}^{-m^{k}s}\big{)}^{-1}\qquad({\Re e\,}s>0).$$

$$\Phi_{k}(s):=\sum_{m\geqslant 1}\log\Big{(}\frac{1}{1-{\rm e}^{-m^{k}s}}\Big{)}$$

$${\Phi }_k(s)=\sum _{m⩾1}\sum _{n⩾1}\frac{{\mathrm{e}}^{-m^{k}ns}}{n}=\sum _{r⩾1}\frac{w_k(r)}{r}{\mathrm{e}}^{-rs},-{\Phi }_k^{\prime }(s)=\sum _{r⩾1}{w}_k(r){\mathrm{e}}^{-rs}(\Re es>0),$$

$$w_k(r):=\sum _{m^k\mid r}{m}^k(r⩾1).$$

$$\Phi_{k}(\sigma_{n})=\frac{k{\mathfrak{a}}_{k}^{1+1/k}}{\sigma_{n}^{1/k}}+\tfrac{1}{2}\log\Big{(}\frac{\sigma_{n}}{(2\pi)^{k}}\Big{)}+\tfrac{1}{2}\zeta(-k)\sigma_{n}+O\big{(}\sigma_{n}^{J}\big{)},$$

$$(-1)^{m}\Phi_{k}^{(m)}(\sigma_{n})=\frac{{\mathfrak{a}}_{k}^{1+1/k}}{\sigma_{n}^{m+1/k}}\prod_{1\leqslant\ell<m}\Big{(}\ell+\frac{1}{k}\Big{)}-\frac{(m-1)!}{2\sigma_{n}^{m}}-\tfrac{1}{2}\delta_{1m}\zeta(-k)+O(\sigma_{n}^{J}),$$

$${\mathrm{e}}^{-s}=\frac{1}{2\pi \text{i}}{\int }_{2-\text{i}\infty }^{2+\text{i}\infty }\Gamma (z)s^{-z}\mathrm{d}z(\Re es>0)$$

$$\sum _{r⩾1}\frac{w_k(r)}{r^{1+z}}=\zeta (z+1)\zeta (kz)(\Re ez>1/k),$$

$${\Phi }_k(s)=\frac{1}{2\pi \text{i}}{\int }_{2-\text{i}\infty }^{2+\text{i}\infty }\zeta (z+1)\zeta (kz)\Gamma (z)\frac{\mathrm{d}z}{s^z},$$

$$(-1{)}^m{\Phi }_k^{(m)}({\sigma }_n)=\frac{1}{2\pi \text{i}}{\int }_{2-\text{i}\infty }^{2+\text{i}\infty }\zeta (z+1)\zeta (kz)\Gamma (z+m)\frac{\mathrm{d}z}{{\sigma }_n^{z+m}}(m⩾0).$$

$$\zeta (x+\text{i}y){\ll }_{a,b}1+|y{|}^A(a⩽x⩽b,|y|⩾1),$$

$$|\Gamma(x+\text{i}y)|=\sqrt{2\pi}|y|^{x-1/2}{\rm e}^{-\pi|y|/2}\big{\{}1+O_{a,b}\big{(}1/y\big{)}\big{\}}\qquad(a\leqslant x\leqslant b,\,|y|\geqslant 1)$$

$$k^{-1}\zeta (1+k^{-1})\Gamma (k^{-1}){\sigma }_n^{-1/k}=k{\mathfrak{a}}_k^{1+1/k}{\sigma }_n^{-1/k}.$$

$$z^2\zeta (z+1)\zeta (kz)\Gamma (z){\sigma }_n^{-z}$$

$$=(1-\gamma z)\{\zeta (0)+k{\zeta }^{\prime }(0)z\}(1+\gamma z)(1-z\log {\sigma }_n)+O(z^2)$$

$$=\zeta (0)+\{-\zeta (0)\log {\sigma }_n+k{\zeta }^{\prime }(0)\}z+O(z^2).$$

Full text

• We include here some corrections with respect to the published version.

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

[TABLE]

where $d\geqslant 1$ and the $a_{j}$ are positive integers such that $a_{1}\geqslant a_{2}\geqslant\cdots\geqslant a_{d}\geqslant 1$. In 1918, Hardy and Ramanujan [4] proved the asymptotic formula

[TABLE]

by using modular properties of Jacobi’s $\Delta$-function.

More generally, given an integer $k\geqslant 1$, let $p_{k}(n)$ denote the number of partitions of the integer $n$ into $k$-th powers, i.e. the number of solutions to the equation

[TABLE]

where, as before, $d\geqslant 1$ and $a_{1}\geqslant a_{2}\geqslant\cdots\geqslant a_{d}\geqslant 1$. Thus, $p_{1}(n)=p(n)$.

In 1918 too, Hardy and Ramanujan [4] stated without proof the asymptotic formula

[TABLE]

where the constants $\mathfrak{b}_{k}$ and $\mathfrak{c}_{k}$ are defined by

[TABLE]

and $\zeta$ is the Riemann zeta-function. In 1934, introducing a number of complicated objects including generalised Bessel functions, Wright [9] obtained an asymptotic expansion of $p_{k}(n)$: for any integer $k\geqslant 1$, there is a real sequence $\{\alpha_{kj}\}_{j\geqslant 1}$ such that, for any $J\geqslant 1$, we have

[TABLE]

where

[TABLE]

and the implied constant depends at most on $J$ and $k$. Apart from an explicit formula for $\alpha_{k1}$, no further information was given about the $\alpha_{kj}$ 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 [3] generalised the argument to arbitrary, fixed $k$.

Gafni states her result in the following way. Let $X=X_{k}(n)$ denote the real solution to the equation

[TABLE]

and write

[TABLE]

Then, given any $k\geqslant 1$, there is a real sequence $\{\beta_{kj}\}_{j\geqslant 1}$ such that for any fixed $J\geqslant 1$, we have

[TABLE]

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 [5] 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 neighbourhood 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 [2]. We believe that the saddle-point method could still be used in these contexts, with expected substantial simplifications in the analysis.

The constants $\mathfrak{b}_{k}$ and $\mathfrak{c}_{k}$ being defined as in (1.4) and (1.5), we can state the following.

Theorem 1**.**

Let $k\geqslant 1$ be a fixed integer. There is a real sequence $\{\gamma_{kj}\}_{j\geqslant 1}$ such that, for any given integer $J\geqslant 1$, we have

[TABLE]

uniformly for $n\geqslant 1$. The implied constant depends at most on $J$ and $k$.

The coefficients $\gamma_{kj}$ can be made explicit directly from the computations in our proof. For instance, we find that $\gamma_{k1}=-(11k^{2}+11k+2)/(24k{\mathfrak{c}}_{k})$ when $k\geqslant 2$, 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

[TABLE]

It may be seen that $|\gamma_{kj}|$ grows like $\Gamma(j){\rm e}^{O(j)}$ and thus that the series $\sum_{j\geqslant 1}\gamma_{kj}z^{j}$ has radius of convergence 0.

2. Technical preparation

Define

[TABLE]

so that

[TABLE]

According to the principles of the saddle-point method, we aim at selecting the integration abscissa $\sigma$ as a solution $\sigma_{n}$ of $-F_{k}^{\prime}(\sigma)/F_{k}(\sigma)=n$. We plainly have

[TABLE]

Thus, in the same half-plane, we may define a determination of $\log F_{k}(s)$ by the formula

[TABLE]

where the complex logarithms are taken in principal branch. Expanding throughout and inverting summations, we get

[TABLE]

where

[TABLE]

Hence $-\Phi_{k}^{\prime}(\sigma)$ decreases from $+\infty$ to $0+$ on $(0,\infty)$, and so the equation $-\Phi_{k}^{\prime}(\sigma)=n$ has for each integer $n\geqslant 1$ a unique real solution $\sigma_{n}=\sigma_{n}(k)$. Moreover, the sequence $\{\sigma_{n}\}_{n\geqslant 1}$ is decreasing and the trivial estimates $1\leqslant w_{k}(r)\leqslant r^{2}$ yield $1/n\ll\sigma_{n}\ll 1/\sqrt[3]{n}$.

We start with an asymptotic expansion for the derivatives $\Phi_{k}^{(m)}(\sigma_{n})$ in terms of powers of $\sigma_{n}$. It turns out that all coefficients but a finite number vanish.

Lemma 2.1**.**

Let $J\geqslant 1$, $k\geqslant 1$. As $n\to\infty$, we have

[TABLE]

Moreover, for fixed $m\geqslant 1$,

[TABLE]

where $\delta_{1m}$ is Kronecker’s symbol.

Proof.

Considering Mellin’s inversion formula

[TABLE]

and the convolution identity

[TABLE]

we derive from the series representation (2.4) the integral formula

[TABLE]

and in turn

[TABLE]

Using the classical fact that $\zeta(z)$ has finite order in any vertical strip $a\leqslant{\Re e\,}z\leqslant b$ ($a,b\in{\mathbb{R}}$ with $a<b$), or, in other words, satisfies

[TABLE]

for suitable $A=A(a,b)$, and invoking Stirling’s formula in the form

[TABLE]

we may move the line of integration to ${\Re e\,}z=-J-m-\tfrac{1}{2}$.

The shifted integral is clearly $\ll\sigma_{n}^{J}$.

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, $\zeta(z+1)\zeta(kz)$ vanishes at all negative integers $\leqslant-2$, so the corresponding zeros compensate the poles of $\Gamma(z)$ at negative integers $\leqslant-2$.

The residue at $z=1/k$ is equal to

[TABLE]

The residue at $z=0$ is the coefficient of $z$ in the Taylor expansion of

[TABLE]

Since $\zeta(0)=-\frac{1}{2}$ and $\zeta^{\prime}(0)=-\frac{1}{2}\log(2\pi)$, this residue equals $\tfrac{1}{2}\log\{\sigma_{n}/(2\pi)^{k}\}$.

The residue at $z=-1$ equals $\tfrac{1}{2}\zeta(-k)\sigma_{n}$.

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 $(1/k)\Gamma(1+1/k)\zeta(1+1/k)\sigma_{n}^{-1-1/k}$, $-1/2\sigma_{n}$ and $-\tfrac{1}{2}\zeta(-k)$.

When $m\geqslant 2$, the only crossed singularities are two simple poles, at $z=1/k$ and $z=0$, with respective residues $(1/k)\Gamma(m+1/k)\zeta(1+1/k)\sigma_{n}^{-m-1/k}$ and $-\tfrac{1}{2}(m-1)!\sigma_{n}^{-m}$. This proves (2.6). ∎

Lemma 2.2**.**

Let $J\geqslant 1$, $k\geqslant 1$, $m\geqslant 1$.

(i)* There is a real sequence $\{a_{kj}\}_{j\geqslant 1}$ with $a_{k1}=-k/(2{\mathfrak{c}}_{k})$, $a_{k2}=k/(8{\mathfrak{c}}_{k}^{2})$, such that*

[TABLE]

(ii)* There is a real sequence $\{b_{kj}\}_{j\geqslant 1}$ with $b_{k1}=-a_{k1}/k$ such that, as $n\to\infty$, we have*

[TABLE]

(iii)* There is a real sequence $\{b_{kmj}\}_{j\geqslant 1}$ such that, as $n\to\infty$, we have*

[TABLE]

Proof.

We infer from (2.6) that

[TABLE]

This immediately implies (2.10) by Lagrange’s inversion formula — see, e.g. [8, §7.32]. We may obtain an explicit expression for the $a_{kj}$ from the formula

[TABLE]

where $\varrho$ is a fixed, small positive constant and

[TABLE]

This is classically derived from Rouché’s theorem and we omit the details. The values of $a_{k1}$ and $a_{k2}$ 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). ∎

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 $\sigma_{n}$. The next result meets this requirement. Here and in the sequel, all constants $c_{j}$ $(j\geqslant 0)$ are assumed, unless otherwise stated, to depend at most upon $k$.

Lemma 2.3**.**

We have

[TABLE]

Proof.

Noticing that

[TABLE]

we can write

[TABLE]

Thus, there is an absolute positive constant $c_{3}$ such that

[TABLE]

with

[TABLE]

where we have put I:=\big{]}(4\sigma_{n})^{-1/k},(2\sigma_{n})^{-1/k}\big{]}.

If $|\tau|\leqslant 2\pi\sigma_{n}$ and $m\in I$, we have $|m^{k}\tau/(2\pi)|\leqslant\frac{1}{2}$. Thus

[TABLE]

When $2\pi\sigma_{n}<|\tau|\leqslant 2\pi\sigma_{n}^{1-1/3k}$, we proceed similarly, noting that for any integer $h$ with $|\tau|/4\sigma_{n}<2\pi h\leqslant|\tau|/2\sigma_{n}$ there are $\gg(1/\sigma_{n})^{1/k}/h$ integers $m$ in $I$ such that

[TABLE]

This yields the the required estimate

[TABLE]

When $2\pi(\sigma_{n})^{1-1/3k}<|\tau|\leqslant\pi$, we note that (2.18) follows, via the Cauchy-Schwarz inequality, from the inequalities

[TABLE]

provided we can show the modulus of the last exponential sum is, say, $\leqslant(1-c)|I|$ for some positive constant $c$ depending at most upon $k$. Now Dirichlet’s approximation lemma guarantees that there exist integers $a\in{\mathbb{Z}}^{*}$ and $q$, with $1<q\leqslant(1/\sigma_{n})^{1-1/3k}$, $(a,q)=1$,

[TABLE]

If $q\leqslant(1/\sigma_{n})^{1/3k}$, we readily deduce the required estimate from [6, Lemma 2.7 & Theorem 4.2]. If $(1/\sigma_{n})^{1/3k}<q\leqslant(1/\sigma_{n})^{1-1/3k}$, we may apply Weyl’s inequality, as stated for instance in [6, Lemma 2.4], to get that the exponential sum under consideration is $\ll|I|^{1-\varepsilon_{k}}$ for some positive $\varepsilon_{k}$ depending only on $k$. Hence, (2.18) holds in all circumstances. ∎

3. Completion of the proof

Proposition 3.1**.**

Let $k\geqslant 1$, $J\geqslant 1$. Then there is a real sequence $\{e_{kj}\}_{j\geqslant 1}$ such that for any integer $J\geqslant 1$ we have

[TABLE]

Proof.

By (2.2), we have

[TABLE]

From (2.15), we deduce that

[TABLE]

Since these bounds are exponentially small with respect to the expected main term, it only remains to estimate the contribution of the interval ${\mathfrak{I}}:=]-\sigma_{n}^{1+1/(3k)},\sigma_{n}^{1+1/(3k)}[$, corresponding to a small neighbourhood of the saddle-point.

In this range, we have

[TABLE]

where the estimate for the error term follows from (2.6). The same formula ensures that $|\Phi_{k}^{(m)}(\sigma_{n})\tau^{m}|\ll 1$ for $m\geqslant 3$. Thus for $\tau\in{\mathfrak{I}}$, we can write

[TABLE]

where

[TABLE]

Since the contributions from odd powers of $\tau$ vanish, we get

[TABLE]

with

[TABLE]

Extending the range of integration in $I_{m}$ involves an exponentially small error, so we get from the classical formula for Laplace integrals

[TABLE]

Inserting these estimates back into (3.5) and expanding all arising factors $\Phi_{k}^{(m)}(\sigma_{n})$ by (2.12), we obtain (3.1). ∎

Remark. From (3.3) and (3.5) we see that, when $k\geqslant 2$,

[TABLE]

where, in view of (2.6), the quantity inside curly brackets may be replaced by an asymptotic series in powers of $\sigma_{n}^{1/k}$. 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 $1/\sigma_{n}$ agree to any power of $\sigma_{n}$.

We are now in a position to complete the proof of Theorem 1.

We infer from (2.10) and (2.11) that

[TABLE]

with $a_{kj}^{*}:={\mathfrak{a}}_{k}(ka_{k,j+1}+b_{k,j+1})$. Exponentiating and expanding, we get

[TABLE]

with

[TABLE]

It remains to insert back into (3.1) and expand $\sqrt{\sigma_{n}/\Phi_{k}^{\prime\prime}(\sigma_{n})}$ according to (2.10) and (2.12) with $m=2$ to obtain the required asymptotic formula.