Note on square-root partitions into distinct parts

Abstract

Let $r_D(n)$ be the number of square-root partitions of n into distinct parts. We will give the asymptotic formula of $r_D(n)$,

$$\begin{aligned} r_D(n)&\sim 2^{-7/6}3^{-1/3}\pi ^{-1/2}\zeta (3)^{1/6}n^{-2/3}\\&\quad \times \exp \Bigg (\frac{3^{4/3}\zeta (3)^{1/3}}{2}n^{2/3}+\frac{\zeta (2)}{2\cdot 3^{1/3}\zeta (3)^{1/3}}n^{1/3}-\frac{\zeta (2)^2}{72\zeta (3)}\Bigg ), \end{aligned}$$

by adjusting the well-known approach of Meinardus.

Appendix A: Expansion of \(X^{-1}\)

In this appendix, we give the expansion of $X^{-1}$. Recall from (2.5) that

$$\begin{aligned} aX^{-3}+bX^{-2}-n=0, \end{aligned}$$

(A.1)

where

$$\begin{aligned} a=3\zeta (3) \quad \text {and}\quad b=\frac{\zeta (2)}{2}. \end{aligned}$$

Let us write

$$\begin{aligned} \mu =n^{-\frac{1}{3}}\quad \text {and}\quad X^{-1}=a^{-\frac{1}{3}}\mu ^{-1}+\xi . \end{aligned}$$

Then (A.1) becomes

$$\begin{aligned} a\big (a^{-\frac{1}{3}}\mu ^{-1}+\xi \big )^{3}+b\big (a^{-\frac{1}{3}}\mu ^{-1}+\xi \big )^{2}-\mu ^{-3}=0, \end{aligned}$$

so that by multiplying by $a^{\frac{2}{3}}\mu ^2$ on both sides, one has

$$\begin{aligned} a\xi \cdot \Big (\big (1+a^{\frac{1}{3}}\mu \xi \big )^2+\big (1+a^{\frac{1}{3}}\mu \xi \big )+1\Big )+b\big (1+a^{\frac{1}{3}}\mu \xi \big )^{2}=0. \end{aligned}$$

(A.2)

Now we may treat $\xi :=\xi (\mu )$ as an implicit function of $\mu$ defined by (A.2). Note that

$$\begin{aligned} \xi (0)=-\frac{b}{3a}=-\frac{\zeta (2)}{18\zeta (3)}. \end{aligned}$$

The implicit function theorem ensures that we may write $\xi (\mu )$ as a power series in $\mu$ in a neighborhood of $\mu =0$. We compute that

$$\begin{aligned} \xi ^\prime (0)=\frac{b^2}{9 a^{5/3}}=\frac{\zeta (2)^2}{36(3\zeta (3))^{5/3}}. \end{aligned}$$

Hence,

$$\begin{aligned} \xi (\mu )=-\frac{\zeta (2)}{18\zeta (3)}+\frac{\zeta (2)^2}{36(3\zeta (3))^{5/3}}\mu +O(\mu ^2) \end{aligned}$$

so that as $n\rightarrow \infty$,

$$\begin{aligned} X^{-1}=\frac{1}{(3\zeta (3))^{1/3}}n^{\frac{1}{3}}-\frac{\zeta (2)}{18\zeta (3)}+\frac{\zeta (2)^2}{36(3\zeta (3))^{5/3}}n^{-\frac{1}{3}}+O(n^{-\frac{2}{3}}). \end{aligned}$$