Asymptotics of the Number of Restricted Partitions

Abstract

Let $0<t_1\le t_2\le t_3\le\dotsm$ be a given sequence of real numbers, and let $\rho(t):=\sum_{j\colon t_j\le t}1$ be its counting function. We assume that $\rho(t)=\rho_0t^\gamma(1+o(1))$ as $t\to\infty$ for some $\rho_0,\gamma>0$. For any $T>0$ and $k\in\overline{\mathbb{N}}=\mathbb{N}\cup\{\infty\}$, let $N(k,T)$ be the number of solutions $\{n_j\}_{j=1}^k$, $n_j\in \mathbb{N}_0=\mathbb{N}\cup\{0\}$, of the inequality $\sum_{j=1}^{k}n_jt_j\le T$. We are interested in the asymptotics of $\log N(k,T)$ as $T\to\infty$.

Appendix A. Proof of Lemma 2

Let us introduce the unified notation

$$\Phi_m(k,\xi)=(-1)^m \frac{\partial^m\Phi}{\partial\xi^m}(k,\xi), \qquad m=0,1,2,$$

for the three functions (3.2) and (3.8). They can be represented as

$$\Phi_m(k,\xi) =\xi^{-m}F_m(k,\xi),\quad \text{where}\quad F_m(k,\xi)=\sum_{j=1}^{k}G_m(\xi t_j),$$

(A.1)

$$G_0(x) =\log\frac{1}{1-e^{-x}},\quad G_1(x)=\frac{x}{e^x-1},\quad G_2(x)=\frac{x^2 e^x}{(e^x-1)^2}.$$

(A.2)

The functions $G_m(x)$ are smooth and positive on $(0,\infty)$ and, together with all derivatives, exponentially decay as $x\to\infty$. Further, $G_1(x)$ and $G_2(x)$ are smooth on $[0,\infty)$ and satisfy $G_1(0)=G_2(0)=1$, while $G_0(x)$ has a logarithmic singularity at $x=0$. Let us compute $F=F_m$, where $m=0$, $1$, or $G_2$. We write $\rho_{\text{as}}(t)=t^\gamma$ and $\tau_k=k^{1/\gamma}$. Then

$$\begin{aligned} \, F(k,\xi) =&\int_0^{t_k}G(\xi t)d\rho(t) =\int_0^{\tau_k}G(\xi t)d\rho_{\text{as}}(t) +\int_{\tau_k}^{t_k}G(\xi t)d\rho_{\text{as}}(t) {}+\int_0^{t_k}G(\xi t)d(\rho(t)-\rho_{\text{as}}(t)) \\ =&\int_0^{\tau_k}G(\xi t)d\rho_{\text{as}}(t) -\xi\int_{\tau_k}^{t_k}G'(\xi t)\rho_{\text{as}}(t)\,dt +G(\xi t)\rho_{\text{as}}(t)\big|_{\tau_k}^{t_k} +G(\xi t)(\rho(t)-\rho_{\text{as}}(t))\big|_{0}^{t_k} \\&-\xi\int_0^{t_k}G'(\xi t)(\rho(t)-\rho_{\text{as}}(t))\,dt. \end{aligned}$$

The integrated terms give

$$\begin{aligned} \, G(\xi t)\rho_{\text{as}}(t)\big|_{\tau_k}^{t_k} +G(\xi t)(\rho(t)-\rho_{\text{as}}(t))\big|_{0}^{t_k} &=G(\xi t_k)\rho(t_k)-G_m(\xi\tau_k)\rho_{\text{as}}(\tau_k) =k(G(\xi t_k)-G(\xi\tau_k)) \\ &=\xi\tau_k^\gamma\int_{\tau_k}^{t_k}G'(\xi t)\,dt. \end{aligned}$$

Here we have used the fact that $\rho_{\text{as}}(\tau_k)=\rho(t_k)=k$; further, $\rho(t)-\rho_{\text{as}}(t)=-\rho_{\text{as}}(t)=-t^\gamma$ for $t<t_1$, and hence $\lim_{t\to0}G(\xi t)(\rho(t)-\rho_{\text{as}}(t))=0$ even for the case of $m=0$, where $G$ has a logarithmic singularity at zero. We make the change of variables $x=\xi t$ and finally write

$$F(k,\xi)=\xi^{-\gamma}(\gamma I(k,\xi)-R_1(k,\xi)-R_2(k,\xi)),$$

where (we omit the arguments $(k,\xi)$ for brevity):

$$ \begin{aligned} \, I &= \int_0^{\xi\tau_k}G(x)x^{\gamma-1}\,dx, \quad R_1 =\int_{\xi\tau_k}^{\xi t_k}G'(x) (x^\gamma-(\xi\tau_k)^\gamma)\, dx. \\ R_2 &=\int_0^{\xi t_k}G'(x)\theta\biggl(\frac x\xi\biggr)x^\gamma\,dx, \quad \theta(t)=\frac{\rho(t)-\rho_{\text{as}}(t)}{\rho_{\text{as}}(t)} \xrightarrow{\,t\to\infty\,}0. \end{aligned}$$

(A.3)

We will prove that $R_j/I\to0$ uniformly with respect to $k\in\mathbb{N}$ as $\xi\to0$ and hence $F\sim\gamma\xi^{-\gamma} I$. As a by-product, we will also obtain asymptotic estimates for $F$.

First, let us make some preliminary remarks. Note that $\theta(t)=-1$ for $t\in(0,t_1)$, and so we readily see that $\theta(t)$ is uniformly bounded on $(0,\infty)$. Further, $\tau_k^\gamma-t_k^\gamma=\rho(t_k)-\rho_{\text{as}}(t_k) =\theta(t_k)t_k^\gamma$, and hence $\tau_k/t_k=(1+\theta(t_k))^{1/\gamma}$. Consequently, there exists a $C_0>1$ such that

$$C_0^{-1}\le \frac{t_k}{\tau_k}\le C_0\quad \text{for all $k\in\mathbb{N}$ and } \lambda_k:=\frac{t_k}{\tau_k}-1\xrightarrow{\,k\to\infty\,}0.$$

As a first consequence, we note that $x^\gamma\le C_0^\gamma(\xi\tau_k)^\gamma$ on the integration interval in the expression for $R_1$ in (A.3), and so

$$ \lvert{R_1}\rvert\le\widetilde R_1=\operatorname{const} \lambda_k(\xi\tau_k)^{\gamma+1} \max_{\alpha\in[C_0^{-1},C_0]} \lvert{G'(\alpha\xi\tau_k)}\rvert.$$

(A.4)

Now let us carry out the desired estimates. We do this separately for $m=1,2$ and for $m=0$.

Appendix A.1. 1. Case of $m\in\{1,2\}$.

Since $G(0)=1$, $G(x)>0$ for all $x\in[0,\infty)$, and is a convergent integral $\int_{0}^{\infty} x^{\gamma-1}G(x)\,dx$ converges, the formula for $I$ in (A.3) implies

$$ I\asymp (\xi\tau_k)^\gamma\text{ for $\xi\tau_k\le1$ and } I\asymp 1\text{ for $\xi\tau_k\ge1$.}$$

(A.5)

Since $G'(x)$ is continuous on $[0,\infty)$ and decays exponentially as $x\to\infty$, it follows from (A.4) and (A.5) that

$$ \frac{R_1}{I}=O(\lambda_k\xi\tau_k)\text{ for $\xi\tau_k\le1$ and } \frac{R_1}{I}=O(\lambda_k)\text{ for $\xi\tau_k\ge1$.}$$

(A.6)

Further, for $R_2$ we obtain

$$\begin{aligned} \, &\lvert{R_2}\rvert\le\operatorname{const}\int_0^{\xi t_k}\biggl\vert\theta\biggl(\frac x\xi\biggr)\biggr\vert x^\gamma\,dx \le \operatorname{const}(\xi\tau_k)^{\gamma+1}\widetilde\lambda_k \text{ for $\xi\tau_k\le1$,} \\ &\text{where }\widetilde\lambda_k=\int_{0}^{1} \lvert{\theta(yt_k)}\rvert y^\gamma\,dy, \quad\text{and $\widetilde\lambda_k\xrightarrow{\,k\to\infty\,}0$} \end{aligned}$$

by Lebesgue’s dominated convergence theorem. For $\xi\tau_k\ge1$, we have

$$\lvert{R_2}\rvert\le\int_0^{\infty}\biggl\vert G'(x)\theta\biggl(\frac x\xi\biggr)\biggr\vert x^\gamma\,dx=:R(\xi)\xrightarrow{\,\varepsilon\to0\,}0,$$

again by Lebesgue’s dominated convergence theorem. Thus,

$$ \frac{R_2}{I}=O(\widetilde \lambda_k\xi\tau_k)\text{ for $\xi\tau_k\le1$ and } \frac{R_2}{I}\stackrel{\xi\to0}\rightrightarrows0\text{ for $\xi\tau_k\ge1$.}$$

(A.7)

Now we make use of the following lemma:

Lemma A.1.

Let $f(k,\xi)=g(k)h(\xi k^{1/\gamma})$ , $(k,\xi)\in\mathbb{N}\times (0,1)$ , where $g(k)\le G$ and $h(x)\le H$ are nonnegative bounded functions, $g(k)\xrightarrow{\,k\to\infty\,}0$ and $h(x)\xrightarrow{\,x\to0\,}0$ . Then for each $\varepsilon>0$ there exists a $\delta>0$ such that $f(k,\xi)<\varepsilon$ whenever $\xi<\delta$ .

Proof.

Without loss of generality, we can assume that $g(k)$ and $h(x)$ are monotone functions. Now, for an arbitrary $k_0\in\mathbb{N}$, $f(k,\xi)\le g(k_0)H$ for $k\ge k_0$ and $f(k,\xi)\le Gh(\xi k_0^{1/\gamma})$ for $k\le k_0$, and we conclude that

$$f(k,\xi)\le \max\{g(k_0)H,Gh(\xi k_0^{1/\gamma})\}$$

for all $k$ and $\xi$. Given an $\varepsilon>0$, pick a $k_0$ such that $g(k_0)H<\varepsilon$ and then take a $\delta>0$ such that $Gh(\xi k_0^{1/\gamma})<\varepsilon$ for $\xi<\delta$.

It follows from (A.6) by virtue of this lemma with $g(k)=\lambda_k$ and $h(x)=\min\{1,x\}$ that we have $R_1/I\xrightarrow{\,\varepsilon\to0\,}0$ uniformly with respect to $k$. To prove that $R_2/I\xrightarrow{\,\varepsilon\to0\,}0$ uniformly with respect to $k$, we use the first relation in (A.7) and apply Lemma A.1 with $g(k)=\widetilde\lambda_k$ and with the same $h(x)$; this proves the desired uniform convergence in the domain $\xi\tau_k\le1$, while for $\xi\tau_k\ge1$ we already have what we need by the second relation in (A.7).

Appendix A.2. 2. Case of $m=0$.

We integrate by parts in the expression (A.3) for $I$ with

$$G(x)=G_0(x)=\log(1-e^{-x})^{-1},$$

obtaining

$$ \gamma I=I_1+I_2,\quad I_1=(\xi\tau_k)^\gamma\log\frac{1}{1-e^{-\xi\tau_k}},\quad I_2=\int_0^{\xi\tau_k}\frac{x^\gamma\,dx}{e^x-1}.$$

(A.8)

By the same reasoning as for (A.5), we have

$$ I_2\asymp (\xi\tau_k)^\gamma\text{ for $\xi\tau_k\le1$ and } I_2\asymp 1\text{ for $\xi\tau_k\ge1$.}$$

(A.9)

The function $\log(1-e^{-x})^{-1}$ tends to $\infty$ as $x\to0$ and decays exponentially as $x\to\infty$. Hence, in view of (A.9), the term $I_1$ dominates in the sum $I_1+I_2$ for small $\xi\tau_k$, and the term $I_2$ dominates for large $\xi\tau_k$, and so we can write

$$ I\asymp (\xi\tau_k)^\gamma\log\frac{1}{1-e^{-\xi\tau_k}}\text{ for $\xi\tau_k\le1$ and } I\asymp 1\text{ for $\xi\tau_k\ge1$.}$$

(A.10)

Now note that $G'(x)=-(e^x-1)^{-1}$, and so we can conclude from (A.4) and that

$$ \begin{aligned} \, \frac{R_1}{I}&=O\biggl(\lambda_k \biggl(\log\frac{1}{1-e^{-\xi\tau_k}}\biggr)^{-1}\biggr)\text{ for $\xi\tau_k\le1$ and }\\ \frac{R_1}{I}&=O(\lambda_k)\text{ for $\xi\tau_k\ge1$.} \end{aligned}$$

(A.11)

Now we apply Lemma A.1 with $g(k)=\lambda_k$ and

$$h(x)=\biggl(\log\frac{1}{1-e^{-x}}\biggr)^{-1},\quad x\le1;\qquad h(x)=1,\quad x>1$$

and find that $R_1/I\xrightarrow{\,\varepsilon\to0\,}0$ uniformly with respect to $k$. As to the convergence of $R_2/I$, the reasoning remains the same as in the case of $m=1,2$ with the only difference that

$$\widetilde\lambda_k=\int_{0}^{1} \lvert{\theta(yt_k)}\rvert y^{\gamma-1}\,dy.$$

One factor $y$ has been used to compensate for the zero of the denominator $e^y-1$.

Formulas (A.10) and (A.5) give the desired asymptotic estimates (3.13), while formulas (A.3) and (A.8) for $I$ and (A.2) for $G_j$, together with the definition of the incomplete zeta function, result in the asymptotics (3.14) and (3.15). The proof of Lemma 3.1 is complete. □