An expansion for the number of partitions of an integer

Abstract

We consider p(n) the number of partitions of a natural number n, starting from an expression derived by Báez-Duarte in (Adv Math 125(1):114–120, 1997) by relating its generating function f(t) with the characteristic functions of a family of sums of independent random variables indexed by t. The asymptotic formula for p(n) follows then from a local central limit theorem as $t\uparrow 1$ suitably with $n\rightarrow \infty$. We take further that analysis and compute formulae for the terms that compose that expression, which accurately approximate them as $t\uparrow 1$. Those include the generating function f and the cumulants of the random variables. After develo** an asymptotic series expansion for the integral term, we obtain an expansion for p(n) that can be simplified as follows: for each $N>0$,

$$\begin{aligned} p(n)=\frac{2\pi ^2}{3\sqrt{3}} \,\frac{\text{ e }^{r_n}}{(1+2\,r_n)^2}\, \left( 1-\sum _{\ell = 1}^N\,\frac{D_\ell }{(1+2\,r_n)^\ell }+\,\mathcal R_{N+1}\,\right) . \end{aligned}$$

The coefficients $D_\ell$ are positive and have simple expressions as finite sums of combinatorial numbers, $r_n=\sqrt{\frac{2\pi ^2}{3}\,(n-\frac{1}{24})+\frac{1}{4}}$ and the remainder satisfies $n^{N/2}\,{\mathcal {R}}_{N+1} \rightarrow 0$ as $n\rightarrow \infty$. The cumulants are given by series of rational functions and the approximate formulae obtained could be also of independent interest in other contexts.

Appendix Formulae for higher-order cumulants

In the next lemma we derive a couple of explicit expressions for each cumulant $\kappa _j(t)$, in terms of the Eulerian polynomials $A_j(t)$, that are defined through the following identity

$$\begin{aligned} \sum _{k\ge 0} k^j\,t^k =\frac{A_j(t)}{(1-t)^{j+1}}. \end{aligned}$$

(A.1)

The first four are

$$\begin{aligned} A_0(t)=1,\quad A_1(t)=t,\quad A_2(t)=t+t^2,\quad A_3(t)=t+4t^2+t^3. \end{aligned}$$

(A.2)

More details can be found in Comtet’s book [5].

Lemma A.1

The cumulants $\kappa _j(t)$ satisfy

$$\begin{aligned} \kappa _j(t) \;&= \; \sum _{\ell \ge 1} \,\frac{\ell ^{j-1} \,A_{j}(t^\ell )}{(1-t^\ell )^{j+1}},\quad j\ge 1, \end{aligned}$$

(A.3)

$$\begin{aligned} \kappa _j(t)\;&= \sum _{\ell \ge 1} \,\frac{\ell ^j\,A_{j-1}(t^\ell )}{(1-t^\ell )^j}, \quad j\ge 2. \end{aligned}$$

(A.4)

The above series are absolutely convergent for $|t|<1$.

Proof

Denote by $S_j(t)$ the right-hand side of (A.3). In (3.12), we have seen that $\kappa _1(t)=\sum _{\ell \ge 1}\frac{t^\ell }{(1-t^\ell )^2}$, which equals $S_1(t)$ by (A.2); that proves (A.3) for $j=1$. To conclude, from (3.11) it suffices to show that

$$\begin{aligned} t\partial _t S_j(t)=S_{j+1}(t) \text{ if } j\ge 1. \end{aligned}$$

Using a well-known recurrence relation for the Eulerian polynomials that can be found for instance in [5, p. 292]

$$\begin{aligned} A_{j+1}(t)=t\,(1-t)\,A'_j(t)+(j+1)\,t\,A_j(t), \end{aligned}$$

(A.5)

we compute

$$\begin{aligned} t\partial _t\,\frac{\ell ^{j-1}\,A_{j}(t^\ell )}{(1-t^\ell )^{j+1}}&=\frac{\ell ^j}{(1-t^\ell )^{j+2}}\, \big (t^\ell (1-t^\ell )\,A_j'(t^\ell )\,+(j+1)\, t^\ell \,A_j(t^\ell )\big ) \\&=\frac{\ell ^j}{(1-t^\ell )^{j+2}}\,A_{j+1}(t^\ell ). \end{aligned}$$

Summing up the last expression over $\ell$, we obtain $S_{j+1}$, and conclude the proof of (A.3). To prove (A.4), observe that, from (A.3) and (A.1), if $j\ge 2$,

$$\begin{aligned} \kappa _j(t)&=\sum _{\ell \ge 1} \,\frac{\ell ^{j-1} \,A_{j}(t^\ell )}{(1-t^\ell )^{j+1}} =\sum _{\ell \ge 1} \ell ^{j-1}\sum _{m\ge 0}m^j\,t^{\ell m}\\&=\sum _{m\ge 0}\sum _{\ell \ge 1}m^jt^{\ell m} \ell ^{j-1}= \sum _{m\ge 0}m^j\frac{A_{j-1}(t^m)}{(1-t^m)^j}. \end{aligned}$$

The convergence of the series in (A.3) and (A.4) is clear from the fact that the polynomials $A_j(0)=0$ for any $j\ge 1$. $\square$

The next result is a consequence of Lemma A.1 and a reasoning similar to that leading to prove Lemma 3.3: from the recurrence (3.11), we derive functional equations for $\kappa _j$ for $j>2$, that yield asymptotic formulae as $t\uparrow 1$ for those. The precise statement is given next.

Proposition A.2

The cumulants $\kappa _j(t)$, $j\ge 2$ satisfy the following functional equations

$$\begin{aligned} \kappa _j(t)=\frac{\pi ^2\,j!}{{6\,|\log t|^{j+1}}}-\frac{(j-1)!}{2\,|\log t|^j} +E_j(t), \end{aligned}$$

(A.6)

where the terms $E_j(t)$ are given by the following expression

$$\begin{aligned} E_j(t) = \frac{(j-1)!}{|\log t|^j} \sum _{r=1}^j \left( {\begin{array}{c}j\\ r\end{array}}\right) \left( \frac{-4 \pi ^2}{|\log t|} \right) ^r \frac{\kappa _r (\mathrm{e}^{-\frac{4\pi ^2}{|\log t|}} )}{(r-1)!}, \quad j\ge 2, \end{aligned}$$

(A.7)

and satisfy

$$\begin{aligned} E_j(t)\asymp \mathrm{e}^{-\frac{4\pi ^2}{|\log t|}}. \end{aligned}$$

(A.8)

Proof

Recall that the $\kappa _j(t)$ are obtained from the recurrence (3.11). We already know from Lemma 3.3 and Corollary 3.5 that the proposition holds for $j=2$. Let us start from $\kappa _2(t)$ as given in (3.7), and denote $(t\partial _t)^{(k)}$ the k-th iteration of the operator $t\partial _t$. Use (3.13) with H the first two terms in the right-hand side of (3.7) to obtain directly by induction that for $j>2$,

$$\begin{aligned} (t\partial _t)^{(j-2)}\left( \frac{\pi ^2}{3|\log t|^3}-\frac{1}{2|\log t|^2}\right) = \frac{\pi ^2\,j!}{{6\,|\log t|^{j+1}}}-\frac{(j-1)!}{2\,|\log t|^j}, \end{aligned}$$

which are the first two terms on the right-hand side of (A.6). Thus, the terms $E_j$ also satisfy the recurrence

$$\begin{aligned} E_{j+1}=t\partial _tE_{j}, \quad j\ge 2. \end{aligned}$$

To prove (A.7) for $j>2$ it is enough to verify that the expression satisfies the above recurrence, which follows by a straightforward induction in j.

Now, from (3.11) and (A.4),

$$\begin{aligned} \partial _\lambda \,\kappa _r(\lambda )=\frac{\kappa _{r+1}(\lambda )}{\lambda }= \sum _{\ell \ge 1} \,\frac{\ell ^{r+1}\,A_{r}(\lambda ^\ell )}{\lambda (1-\lambda ^\ell )^{r+1}}. \end{aligned}$$

Since $A_r(\lambda ^\ell )/\lambda$ is a polynomial in $\lambda$ for each $r\ge 1$, the above series is easily seen to be bounded from above by a constant $C_r$ if $\lambda <\frac{1}{2}$, say, so we conclude from the mean value theorem that, if $t>\text{ e }^{-\frac{\pi }{2}}$ (and then $\text{ e }^{-\frac{4\pi ^2}{|\log t|}}<\text{ e }^{-8\pi }<\frac{1}{2}$),

$$\begin{aligned} \kappa _r(\text{ e }^{-4\pi ^2/|\log t|}) \le C_r\,\text{ e }^{-4\pi ^2/|\log t|}. \end{aligned}$$

(A.9)

In addition, recall that the Eulerian polynomials $A_r$ have non-negative coefficients, and coefficient one in the linear terms for $r \ge 1$. We obtain from (A.3), by just taking the first term in each series, the following lower bounds:

$$\begin{aligned} \kappa _r(\text{ e }^{-4\pi ^2/|\log t|}) \ge \frac{ A_r(\text{ e }^{-4\pi ^2/|\log t|})}{(1- \text{ e }^{-4\pi ^2/|\log t|})^{r+1}} \ge \text{ e }^{-4\pi ^2/|\log t|}. \end{aligned}$$

Thus, (A.8) holds. Indeed, from (A.9) and (A.7), we conclude that there is a positive constant $C_j$ that may depend on j such that for any $t>\text{ e }^{-\frac{\pi }{2}}$,

$$\begin{aligned} |E_j(t)|\le \frac{C_j}{|\log t|^{2j}}\,\text{ e }^{-4\pi ^2/|\log t|}. \end{aligned}$$

(A.10)

$\square$

We computed in Lemma A.1 the derivatives at zero of the cumulant generating function $K_{X(t)}$. To estimate the remainder in the Taylor formula, we need also the derivatives of $K_{Z(t)}(\theta )$ at $\theta \ne 0$. Recall that

$$\begin{aligned} K_{Z(t)}(\theta )=K_{X(t)}\big (\theta /\sigma (t)\big )-i \theta \,\kappa _1(t)/\sigma (t). \end{aligned}$$

Lemma A.3

For each $j\ge 2$, $\theta \in {\mathbb {R}}$,

$$\begin{aligned} \partial ^{(j)}_\theta K_{Z(t)}(\theta ) = \frac{i^j}{(\sigma (t))^j}\,\kappa _j(t\, \mathrm{e}^{i\theta /\sigma (t)}), \end{aligned}$$

where the functions $\kappa _j(z)$ are defined for $z\in {\mathbb {C}}, |z|<1$ by formula (A.3):

$$\begin{aligned} \kappa _j(z)= \sum _{\ell \ge 1} \,\frac{\ell ^{j-1} \,A_{j}(z^\ell )}{(1-z^\ell )^{j+1}}. \end{aligned}$$

Proof

Observe first that, from the same reasoning used to prove (A.3), the functions $\kappa _j(z)$ satisfy the recurrence $z\partial _z \kappa _j(z)=\kappa _{j+1}(z)$. From the expression $K_{Z(t)}(\theta )=L(t \text{ e }^{i\theta /\sigma (t)})-L(t)-i\theta \frac{\kappa _1(t)}{\sigma (t)}$ in terms of $L(z):=\sum _{\ell \ge 1}\frac{1}{\ell }\frac{z^\ell }{1-z^\ell }$, it is straightforward to see by differentiating the series that

$$\begin{aligned} \partial _\theta K_{Z(t)}(\theta ) =&\frac{i}{\sigma (t)}\left( t \text{ e }^{i\theta /\sigma (t)}\, \sum _{\ell \ge 1} \,\frac{(t\, \text{ e }^{i\theta /\sigma (t)})^{\ell -1}}{(1-(t\, \text{ e }^{i\theta /\sigma (t)})^\ell )^{2}} -\kappa _1(t)\right) \\ =\,&\frac{i}{\sigma (t)}\,\big (\kappa _1(t\, \text{ e }^{i\theta /\sigma (t)}) -\kappa _1(t)\big ). \end{aligned}$$

Differentiating the last expression, we obtain, using the recurrence for the $\kappa _j$’s,

$$\begin{aligned} \partial ^{(2)}_\theta K_{Z(t)}(\theta )=\left( \frac{i}{\sigma (t)}\right) ^2\,t \,\text{ e }^{i\theta /\sigma (t)} \kappa _1'(t \text{ e }^{i\theta /\sigma (t)})=\frac{i^2}{(\sigma (t))^2}\,\kappa _2(t\, \text{ e }^{i\theta /\sigma (t)}), \end{aligned}$$

which is the desired expression for $j=2$. Successive differentiation using the recurrence yields the general formula. $\square$

Corollary A.4

The derivatives of $K_{Z(t)}(\theta )$ satisfy

$$\begin{aligned}&\mathrm{{(a)}}\quad |\partial ^{(j)}_\theta K_{Z(t)}(\theta ) |\le \frac{\kappa _j(t)}{(\sigma (t))^j}, \\&\mathrm{{(b)}}\quad \frac{1}{j!}\,|\partial ^{(j)}_\theta K_{Z(t_n)}(\theta ) |\le C_j\, \left( \,\frac{2\pi ^2}{3}\,\left( n-\frac{1}{24}\right) + \frac{1}{4}\,\right) ^{\frac{2-j}{4}},\quad j\ge 3, \end{aligned}$$

for some positive constant $C_j$ that may depend on j.

Proof

From Lemma  A.3,

$$\begin{aligned} |\partial ^{(j)}_\theta K_{Z(t)}(\theta )|=\frac{1}{(\sigma (t))^j} \Big |\sum _{\ell \ge 1} \, \frac{\ell ^{j-1} \,A_{j}((t \text{ e }^{i\theta /\sigma (t)})^\ell )}{\big (1-(t \text{ e }^{i\theta /\sigma (t)})^\ell \big )^{j+1}} \Big | \le \frac{1}{(\sigma (t))^j} \sum _{\ell \ge 1} \, \frac{\ell ^{j-1} \,A_{j}(t^\ell )}{(1-t^\ell )^{j+1}}, \end{aligned}$$

where to estimate the denominators we use that $| 1-(t \text{ e }^{i\theta /\sigma (t)})^\ell |\ge 1-t^\ell$, and to estimate the numerators it is enough to observe that the Eulerian polynomials have real positive coefficients. The right-hand side above is precisely that in item (a) in the statement. To prove the inequality (b), we write, from (A.6) for $\kappa _j(t)$ and $\sigma ^2(t)=\kappa _2(t)$:

$$\begin{aligned} \frac{\kappa _j(t)}{j!\,(\sigma (t))^j}=\frac{1}{2}\, \frac{\frac{\pi ^2}{3|\log t|}-\frac{1}{j} +\frac{2\,|\log t|^j}{j!}\,E_j(t)}{\big (\frac{\pi ^2}{3|\log t|}-\frac{1}{2} +|\log t|^2\,E_2(t)\big )^{j/2}}. \end{aligned}$$

For each $j\ge 3$, we take $t=t_n$ above and use (A.10) and (3.19) to conclude that (b) in the statement holds. $\square$