A Class of the Generalized Ramanujan Tau Numbers and Their Associated Partition Functions

Abstract

In this paper, the authors derive some believed-to-be new recursion and explicit formulas for the generalized Ramanujan numbers ${\tau }_s\left(n\right)$$(s\in \mathbb{N}\cup \{-1\left\}\right)$, where, as usual, $\mathbb{N}$ is the set of positive integers. The authors consider the associated partition functions and derive connections of the Eisenstein series with the numbers ${\tau }_s\left(n\right)$. Several related corollaries and consequences of each of the presented results are also given. The paper concludes by presenting an open problem that is related to one of these results.

1. Introduction and the Main Recursion Formulas

We define the generalized Ramanujan tau sequence ${\left\{{\tau }_s\left(n\right)\right\}}_{n=1}^{\infty }$ as follows:

$$\prod _{n=1}^{\infty }{(1-z^n)}^s:=\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}(z\in \mathbb{C};|z|<1;z\ne 0;s\in \mathbb{N}\cup \{-1\}),$$

where, as usual, $\mathbb{N}$ denotes the set of positive integers. Euler’s statement and Jacobi’s statement [1] (p. 429) imply that

$$\begin{array}{cc}\hfill \prod _{n=1}^{\infty }(1-z^n)& =\sum _{n=-\infty }^{\infty }{(-1)}^n{z}^{(3n^2+n)/2}=\sum _{n=0}^{\infty }{(-1)}^{{\displaystyle \frac{n(n+1)}{2}}}{z}^{a_n}\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\tau }_1\left(n\right)z^{n-1}\hfill \end{array}$$

(1)

and

$$\begin{array}{c}\hfill \prod _{n=1}^{\infty }{(1-z^n)}^3=\sum _{n=0}^{\infty }{(-1)}^r(2n+1)z^{n(n+1)/2}=\sum _{n=1}^{\infty }{\tau }_3\left(n\right)z^{n-1},\end{array}$$

(2)

where ${\left\{a_j\right\}}_{j=0}^{\infty }$ is a sequence of generalized pentagonal numbers (see, for details, [2] (A001318); see also [3]). In a special case, ${\tau }_{24}\left(n\right)=\tau \left(n\right)$ is a sequence of the Ramanujan tau numbers [2] (A000594). Ramanujan [4] (p. 14) gave the following recursion formula for $\tau \left(n\right)$:

$$\begin{array}{cc}\hfill (n-1)\tau \left(n\right)& =\sum _{1\leqq m\leqq {\alpha }_n}{(-1)}^{m+1}(2m+1)\left(n-1-{\displaystyle \frac{9}{2}}m(m+1)\right)\hfill \\ \hfill & ·\tau \left(n-{\displaystyle \frac{m}{2}}(m+1)\right),\hfill \end{array}$$

(3)

where

$${\displaystyle {\alpha }_n:=\frac{1}{2}\left(\sqrt{8n+1}-1\right).}$$

The recursion formula contains no other known functions. We define the partition function $p\left(n\right)$ as the following generating function (see [2] (A000041)):

$$\prod _{n=1}^{\infty }{(1-z^n)}^{-1}=\sum _{n=0}^{\infty }p\left(n\right)z^n.$$

The function applies in cases when

$${\tau }_{-1}(n+1)=p\left(n\right)(n\in \mathbb{N}\cup \left\{0\right\}).$$

The familiar recursion formula for the generalized Ramanujan tau numbers ${\tau }_s\left(n\right)(s=1,\dots ,24)$ is based upon the Euler transformation and contains the following known divisor function:

$${\sigma }_m\left(n\right):=\sum _{d|n}{d}^m\left(\sigma \left(n\right):={\sigma }_1\left(n\right)\right).$$

From the work of Petojević [5] (p. 46, Eq. (11)) and the Online Encyclopedia of Integer Sequences [2] (A010815, A002107, A010816, …, A000594), we recall the following limit formula:

$${\tau }_s\left(n\right)=-{\displaystyle \frac{s}{n-1}}\sum _{k=1}^{n-1}\sigma \left(k\right){\tau }_s(n-k).$$

(4)

Remark 1.

The method that is described in the proof of [5] (p. 46, Lemma 3) readily demonstrates that Formula (4) is valid for all natural numbers s.

The authors are not aware of the existence of any correct recursion formula for the numbers ${\tau }_s\left(n\right)$$(s=1,\dots ,23)$, which do not contain any known functions. However, the recursion formula

$$(n-1)\tau \left(n\right)=\sum _{1\leqq \left|m\right|\leqq {\beta }_n}\left(n-1-{\displaystyle \frac{25}{2}}m(3m+1)\right)\tau \left(n-{\displaystyle \frac{m}{2}}(3m+1)\right),$$

where

$${\displaystyle {\beta }_n=\frac{1}{6}\left(\sqrt{24n+1}+1\right),}$$

which was given by Lehmer in [6] (p. 487) and [4] (p. 14), contradicts some well-known facts. For example, from Lehmer’s formula for $n=11$ one obtains ${\beta }_{11}\approx 2.88$, which implies that $m\in \{\pm 1,\pm 2\}$. Finally, we find that $\tau \left(11\right)=\mathrm{722,292}$, while the exact value is $\tau \left(11\right)=\mathrm{534,612}$ (see Example 1; see also [7,8,9,10] for some recent developments pertaining to this connection).

Various interesting number sequences and other entities, which are related to the above-defined generalized Ramanujan sequence ${\left\{{\tau }_s\left(n\right)\right\}}_{n-1}^{\infty }$, have indeed continued to be studied widely and extensively in the recent literature on the subject (see, for example, ref. [11,12,13,14,15,16]). Motivated essentially by the aforementioned recent developments, we first prove recursion formulas for the generalized Ramanujan tau numbers ${\left\{{\tau }_s\left(n\right)\right\}}_{n-1}^{\infty }$ associated with Equation (6), which do not have a known function.

Theorem 1.

Let $s,n\in \mathbb{N}$. Then ${\tau }_s\left(1\right)=1$ and

$$\begin{array}{cc}\hfill (n-1){\tau }_s\left(n\right)& =\sum _{1\leqq m\leqq d_n}{(-1)}^{m+1}\left(n-1-{\displaystyle \frac{(s+1)m(3m+1)}{2}}\right){\tau }_s\left(n-{\displaystyle \frac{m(3m+1)}{2}}\right)\hfill \\ \hfill & +\sum _{1\leqq m\leqq b_n}{(-1)}^{m+1}\left(n-1-{\displaystyle \frac{(s+1)m(3m-1)}{2}}\right){\tau }_s\left(n-{\displaystyle \frac{m(3m-1)}{2}}\right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill d_n={\displaystyle \frac{\sqrt{24n-23}-1}{6}}\mathrm{and}{b}_n={\displaystyle \frac{\sqrt{24n-23}+1}{6}}.\end{array}$$

(5)

Proof. 

First of all, Equation (1) produces the following formula:

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}z}}\left\{ln\left(\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^n\right)\right\}={\displaystyle \frac{\mathrm{d}}{\mathrm{d}z}}\left\{lnz+ln{\left(\sum _{n=0}^{\infty }{(-1)}^{{\displaystyle \frac{n(n+1)}{2}}}{z}^{a_n}\right)}^s\right\}.$$

Applying the Cauchy rule for multiplying power series, we get

$$\sum _{n=1}^{\infty }(n-1){\tau }_s\left(n\right)z^n=\sum _{m=1}^{\infty }\sum _{k=1}^m{(-1)}^{{\displaystyle \frac{(m-k+1)(m-k+2)}{2}}}(s{a}_{m-k+1}-k+1){\tau }_s\left(k\right)z^{a_{m-k+1}+k}.$$

Hence, for $m=n+j-a_j-1$ and $k=n-a_j,$ we have $z^n=z^{a_{m-k+1}+k}$, so that

$$(n-1){\tau }_s\left(n\right)=\sum _{1\leqq a_j\leqq n-1}{(-1)}^{{\displaystyle \frac{j(j+1)}{2}}}\left[(s+1)a_j-n+1\right]{\tau }_s(n-a_j).$$

(6)

Therefore, the recursion formula asserted by Theorem 1 follows the following equality:

$$\begin{array}{c}\hfill a_n=\left\{\begin{array}{cc}{\displaystyle \frac{1}{2}}m(3m+1)\hfill & \left(n=2m;m\in \mathbb{N}\right)\hfill \\ \\ {\displaystyle \frac{1}{2}}m(3m-1)\hfill & \left(n=2m-1;m\in \mathbb{N}\right).\hfill \end{array}\right.\end{array}$$

□

Remark 2.

The values of $d_n$ and $b_n,$ given by Equation (5), do not depend on the choice of the number s. Petojević et al. [17] proved Theorem 1 for $s=24$. Also, for $s=-1,$ Theorem 1 holds true (see [18]).

Example 1.

We now calculate $\tau \left(11\right)$ by using Ramanujan’s Formula (3) and Theorem 1:

$$\begin{array}{c}\hfill \tau \left(11\right)={\displaystyle \frac{3\tau \left(10\right)+85\tau \left(8\right)-308\tau \left(5\right)+720\tau \left(1\right)}{10}}=534612\end{array}$$

and

$$\begin{array}{c}\hfill \tau \left(11\right)={\displaystyle \frac{-15\tau \left(10\right)-40\tau \left(9\right)+115\tau \left(6\right)+165\tau \left(4\right)}{10}}=534612.\end{array}$$

2. An Explicit Formula for the Numbers ${\mathit{\tau }}_{\mathit{s}}\left(\mathit{n}\right)$

In this section, we first state and prove the following result.

Theorem 2.

Let $s,n\in \mathbb{N},$$k_i\in \{0,1,2,\cdots ,s\}$ for $i=1,\cdots ,n-1$. Then

$${\tau }_s\left(n\right)=\sum _{{\sum }_{i=1}^{n-1}i·k_i=n-1}{(-1)}^{{\sum }_{i=1}^{n-1}{k}_i}\prod _{i=1}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_i}}\right).$$

Proof. 

Let $m=1,2,\dots ,n-1$. If, in the sum

$$\left({\displaystyle \genfrac{}{}{0pt}{}{s}{0}}\right)z^{m·0}-\left({\displaystyle \genfrac{}{}{0pt}{}{s}{1}}\right)z^{m·1}+\left({\displaystyle \genfrac{}{}{0pt}{}{s}{2}}\right)z^{m·2}-\cdots +{(-1)}^{s-1}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{s-1}}\right)z^{m·(s-1)}+{(-1)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{s}{s}}\right)z^{m·s},$$

where we denote the coefficient $z^{m·k_i}$ by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_i}}\right):=\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_{i,m}}}\right),$$

then the product ${\displaystyle \prod _{m=1}^{n-1}}{(1-z^m)}^s$ is equal to the sum of terms of the following form:

$${(-1)}^{k_{i,1}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_{i,1}}}\right){(-1)}^{k_{i,2}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_{i,2}}}\right)\cdots {(-1)}^{k_{i,n-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_{i,n-1}}}\right)z^{1·k_{i,1}+2·k_{i,2}+\cdots +(n-1)·k_{i,n-1}}.$$

Since we calculate the coefficient with $z^{n-1}$, the following condition must be fulfilled:

$$1·k_{i,1}+2·k_{i,2}+\cdots +(n-1)·k_{i,n-1}=n-1,$$

so that

$${\tau }_s\left(n\right)=\sum _{1·k_1+2·k_2+\cdots +(n-1)·k_{n-1}=n-1}{(-1)}^{k_1+k_2+\cdots +k_{n-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_2}}\right)\cdots \left({\displaystyle \genfrac{}{}{0pt}{}{s}{k_{n-1}}}\right),$$

which evidently proves the formula as claimed. □

Remark 3.

Goran-Dumitru and Merca [13] (Th.5.1, p. 77) proved Theorem 2 for $s=24$.

Example 2.

Making use of the formula proven in Theorem 2, and for $s=24$, we find that

$$\begin{array}{cc}\hfill \tau \left(1\right)& ={(-1)}^0\left({\displaystyle \genfrac{}{}{0pt}{}{24}{0}}\right)=1,\hfill \\ \hfill \tau \left(2\right)& ={(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)=-24,\hfill \\ \hfill \tau \left(3\right)& ={(-1)}^2\left({\displaystyle \genfrac{}{}{0pt}{}{24}{2}}\right)+{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)=252\hfill \\ \hfill \tau \left(4\right)& ={(-1)}^3\left({\displaystyle \genfrac{}{}{0pt}{}{24}{3}}\right)+{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right){(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)+{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)=-1472\hfill \\ \hfill \tau \left(5\right)& ={(-1)}^4\left({\displaystyle \genfrac{}{}{0pt}{}{24}{4}}\right)+{(-1)}^2\left({\displaystyle \genfrac{}{}{0pt}{}{24}{2}}\right){(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)+{(-1)}^2\left({\displaystyle \genfrac{}{}{0pt}{}{24}{2}}\right)\hfill \\ \hfill & +{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right){(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)+{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)=4830.\hfill \\ \hfill \tau \left(6\right)& ={(-1)}^5\left({\displaystyle \genfrac{}{}{0pt}{}{24}{5}}\right)+{(-1)}^3\left({\displaystyle \genfrac{}{}{0pt}{}{24}{3}}\right){(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)+{(-1)}^2\left({\displaystyle \genfrac{}{}{0pt}{}{24}{2}}\right){(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)\hfill \\ \hfill & +{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right){(-1)}^2\left({\displaystyle \genfrac{}{}{0pt}{}{24}{2}}\right)+{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right){(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)\hfill \\ \hfill & +{(-1)}^1\left({\displaystyle \genfrac{}{}{0pt}{}{24}{1}}\right)=-6048.\hfill \end{array}$$

3. The $\mathit{s}$-Partition Function ${\mathit{p}}_{\mathit{s}}\left(\mathit{n}\right)$

We begin this section by introducing the following notations:

$$1·k_1+2·k_2+\cdots +n·k_n=n,s,n\in \mathbb{N},k_i\in \{0,1,2,\cdots ,s\}(i=1,\cdots ,n).$$

(7)

In what follows, we use the following notations and conventions:

• $p_s\left(n\right)$: the number of partitions of n that satisfy Equation (7) (

  Table 1. The s-partition function $p_s\left(n\right)$ for $s=1,2,3,4$ and $n=1,2,\dots ,8$.

  n	1	2	3	4	5	6	7	8
  $p_1\left(n\right)$	1	1	1	1	1	1	1	1
  $p_2\left(n\right)$	1	2	2	4	5	7	9	13
  $p_3\left(n\right)$	1	2	3	4	6	9	12	16
  $p_4\left(n\right)$	1	2	3	5	6	10	13	18

  ).
• $i{p}_s\left(n\right)$: the number of partitions of n that do not satisfy Equation (7) (impossible partitions).
• $c\left(n\right)$: the number of partitions of n that do not contain 1 as a part [2] (A002865).
• $w\left(n\right)$: the nth number [2] (A092885).
• $\underset{mtimes}{\underbrace{x+x+\cdots +x}}:={\sum }_{m}x$.
• $\lfloor ·\rfloor$: the floor function of a real number, which is the largest integer not exceeding that real number.

Theorem 3.

Let $n\in \mathbb{N}$. Then

$$p_{24}(n-1)\geqq p(n-1)-\sum _{i=1}^{\lfloor {\displaystyle \frac{n-1}{25}}\rfloor }p(n-1-i·25).$$

Proof. 

In our proof of Theorem 3, we consider the following cases:

(1)

$0\leqq n-1\leqq 24$: $p_{24}(n-1)=p(n-1)=w(n-1)$.

(2)

$25\leqq n-1\leqq 49$: Based on the results presented in

Table 2. The impossible partition $i{p}_{24}(n-1)$ for $n-1=25,26,27,28,29$.

$\mathit{n}-1$	Impossible Partitions	${\mathit{ip}}_{24}(\mathit{n}-1)$
25	${\sum }_{25}1$	1
26	${\sum }_{26}1$	1
27	${\sum }_{27}1$, ${\sum }_{25}1+2$	2
28	${\sum }_{28}1$, ${\sum }_{26}1+2$, ${\sum }_{25}1+3$	3
29	${\sum }_{29}1$, ${\sum }_{27}1+2$, ${\sum }_{26}1+3$, ${\sum }_{25}1+2+2$, ${\sum }_{25}1+4$	5

, we observe that the impossible partitions for $n-1=29$ are obtained by adding the number 1 to the impossible partitions for $n-1=28$. The total number of such partitions is given by $i{p}_{24}\left(28\right)$. In addition to these, we also consider the impossible partitions ${\sum }_{25}1+2+2$ and ${\sum }_{25}1+4$, whose number is given by $c\left(4\right)$. Therefore, we find that

$$i{p}_{24}\left(29\right)=i{p}_{24}\left(28\right)+c\left(4\right)=3+2=5.$$

We now formulate the following generalization:

$$\begin{array}{cc}\hfill i{p}_{24}(n-1)& =i{p}_{24}(n-2)+c(n-1-25)\hfill \\ \hfill & =i{p}_{24}(n-3)+c(n-2-25)+c(n-1-25)\hfill \\ \hfill & =i{p}_{24}(n-4)+c(n-3-25)+c(n-2-25)+c(n-1-25)\hfill \\ \hfill & ⋮\hfill \\ \hfill & \sum _{k=1}^{n-25}c(n-k-25)=p(n-26),\hfill \end{array}$$

so that

$$p_{24}(n-1)=p(n-1)-p(n-26)=w(n-1).$$

(3)

$50\leqq n-1\leqq 74$: Analogously, we have $i{p}_{24}\left(50\right)=i{p}_{24}\left(49\right)+$ number of partitions ${\sum }_{25}1+25$, ${\sum }_{25}1+23+2$, ${\sum }_{25}1+22+3,\cdots$ + number of partitions ${\sum }_{25}2$. We then have

$$i{p}_{24}\left(50\right)=i{p}_{24}\left(49\right)+c_1\left(25\right)+1=p\left(24\right)+c\left(25\right)+c\left(0\right)=p\left(25\right)+c\left(0\right).$$

For $n-1=51$, the impossible partition ${\sum }_{25}2+1$ is already included in the count of $i{p}_{24}\left(50\right)$. Therefore, we can express the total number of impossible partitions for $n-1=51$ as follows:

$$i{p}_{24}\left(51\right)=i{p}_{24}\left(50\right)+c\left(26\right)+0=p\left(25\right)+c\left(26\right)+c\left(0\right)+c\left(1\right)=p\left(26\right)+p\left(1\right).$$

This leads to the generalization given by

$$i{p}_{24}(n-1)=p(n-26)+p(n-51),$$

so that

$$p_{24}(n-1)=p(n-1)-p(n-26)-p(n-51)=w(n-1).$$

(4)

$75\leqq n-1\leqq 99$: In this case, in addition to the partitions ${\sum }_{25}1$ and ${\sum }_{25}2$, we also consider the partition ${\sum }_{25}3$. Thus, clearly, we can write the total number of impossible partitions as $i{p}_{24}(n-1)=p(n-26)+p(n-51)+p(n-76)$, so that

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)\hfill \\ \hfill & =w(n-1)-p(n-76).\hfill \end{array}$$

(5)

$100\leqq n-1\leqq 124$: For $n-1=100$, the number of impossible partitions of the form

$$\sum _{25}2+50,\sum _{25}2+48+2,\sum _{25}2+47+3,\cdots$$

is given by $c\left(50\right)$. Additionally, the number of impossible partitions of the following form

$$\sum _{25}3+25,\sum _{25}3+23+2,\sum _{25}3+22+3,\cdots$$

is $c\left(25\right)$. We have the impossible partition ${\sum }_{25}4$. This occurs when

$$i{p}_{24}(n-1)=p(n-26)+p(n-51)+p(n-76)+1,$$

which leads to the following expression for $p_{24}(n-1)$:

$$\begin{array}{c}\hfill p_{24}(n-1)=p(n-1)-p(n-26)-p(n-51)-p(n-76)-1.\end{array}$$

For $n-1=101$, the impossible partition ${\sum }_{25}4+1$ is already included in the count of $i{p}_{24}\left(100\right)$. Therefore, we can express the total number of impossible partitions for $n-1=101$ as follows:

$$p_{24}(n-1)=p(n-1)-p(n-26)-p(n-51)-p(n-76)-1-0.$$

For $n-1=102$, the expression for $p_{24}(n-1)$ is given by

$$p_{24}(n-1)=p(n-1)-p(n-26)-p(n-51)-p(n-76)-1-0-1.$$

For $n-1=103$, we adjust the formula as follows:

$$p_{24}(n-1)=p(n-1)-p(n-26)-p(n-51)-p(n-76)-1-0-1-1.$$

In general, we can express $p_{24}(n-1)$ as follows:

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)-p(n-101)\hfill \\ \hfill & =w(n-1)-p(n-76)-p(n-101).\hfill \end{array}$$

(6)

$125\leqq n-1\leqq 149$: For $n-1=125$, the number of impossible partitions of the form

$$\begin{array}{cc}\hfill & \sum _{25}2+75,\sum _{25}2+73+2,\sum _{25}2+72+3,\cdots ,\sum _{25}2+\sum _{25}3,\cdots ,\hfill \\ \hfill & \sum _{25}3+50,\sum _{25}3+48+2,\sum _{25}7+47+3,\cdots ,\sum _{25}3+\sum _{25}2,\cdots ,\hfill \\ \hfill & \sum _{25}4+25,\sum _{25}4+23+2,\sum _{25}4+22+3,\cdots ,\hfill \\ \hfill & \sum _{25}5\hfill \end{array}$$

are given by $c\left(75\right)$, $c\left(50\right)$, $c\left(25\right)$, and 1, respectively. However, we note that the partition ${\sum }_{25}2+{\sum }_{25}3$ has been counted twice. This occurs when

$$i{p}_{24}(n-1)=p(n-26)+p(n-51)+p(n-76)+p(n-101)+1-1,$$

which leads to the following expression for $p_{24}(n-1)$:

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)\hfill \\ \hfill & -p(n-101)-1+1.\hfill \end{array}$$

For $n-1=126$, we obtain the following expression for $p_{24}(n-1)$:

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)\hfill \\ \hfill & -p(n-101)-1-0+1+0.\hfill \end{array}$$

For $n-1=127$, the expression for $p_{24}(n-1)$ is given by

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)\hfill \\ \hfill & -p(n-101)-1-0-1+1+0+1.\hfill \end{array}$$

For $n-1=128$, we have

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)\hfill \\ \hfill & -p(n-101)-1-0-1-1+1+0+1+1.\hfill \end{array}$$

For $n-1=129$, we observe that certain partitions are double-counted. Specifically, the partitions of the following form:

$$\sum _{25}2+\sum _{25}3+2+2\mathrm{and}\sum _{25}2+\sum _{25}3+4$$

are counted more than once. In order to correct this, we adjust the formula as follows:

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)\hfill \\ \hfill & -p(n-101)-p(n-126)+1+0+1+1+2.\hfill \end{array}$$

In this procedure, we mark the number of impossible partitions that are repeated for the number $n-1$ with $r_{24}(n-1)$—that is,

$$r_{24}\left(125\right)=1,r_{24}\left(126\right)=1,r_{24}\left(127\right)=2,r_{24}\left(128\right)=3,\mathrm{and}{r}_{24}\left(129\right)=5.$$

Thus, in general, we can express $p_{24}(n-1)$ as follows:

$$\begin{array}{cc}\hfill p_{24}(n-1)& =p(n-1)-p(n-26)-p(n-51)-p(n-76)-p(n-101)-\hfill \\ \hfill & -p(n-126)+r_{24}(n-1)\hfill \\ \hfill & =p(n-1)-p(n-26)-p(n-51)-p(n-76)-p(n-101)-\hfill \\ \hfill & -p(n-126)+p(n-126)\hfill \\ \hfill & =p(n-1)-p(n-26)-p(n-51)-p(n-76)-p(n-101)\hfill \\ \hfill & =w(n-1)-p(n-76)-p(n-101).\hfill \end{array}$$

If we continue this procedure, we get

$$p_{24}(n-1)=p(n-1)-\sum _{i=1}^{\lfloor {\displaystyle \frac{n-1}{25}}\rfloor }p(n-1-i·25)+r_{24}(n-1).$$

□

Remark 4.

For $s,n\in \mathbb{N},$ what is the value of $p_s(n-1)$? For $s=24,150\leqq n-1$, and $24\ne s\in \mathbb{N}(n\in \mathbb{N}),$ the problem remains open.

4. Connection of the Eisenstein Series with the Numbers ${\mathit{\tau }}_{\mathit{s}}\left(\mathit{n}\right)$

For each positive even integer m, the Eisenstein series $E_m\left(q\right)$ is defined as follows (see [19]; see also the several interesting recent developments reported in [20,21,22,23]):

$$E_m\left(q\right):=1-{\displaystyle \frac{2m}{B_m}}\sum _{n=1}^{\infty }{\sigma }_{m-1}\left(n\right)q^n(\Im \left(z\right)>0),$$

(8)

where $q={\mathrm{e}}^{2\pi \mathrm{i}z}$ is the complex power-series variable and $B_m$ represents the Bernoulli numbers defined by the generating function

$$\sum _{n=0}^{\infty }{B}_n{\displaystyle \frac{x^n}{n!}}={\displaystyle \frac{x}{{\mathrm{e}}^x-1}}.$$

The Eisenstein series are widely applied in the theory of modular forms. In this connection, it seems to be necessary to mention some more facts about modular forms.

Theorem 4.

Let $s,n\in \mathbb{N}$. Then

$$\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}=exp\left(-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n\right)$$

or, equivalently,

$$\prod _{n=1}^{\infty }{(1-z^n)}^s=exp\left(-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n\right).$$

Proof. 

Based on Equation (4), we have the following equalities:

$$\sum _{n=1}^{\infty }n{\tau }_s\left(n\right)z^n=\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^n-s\sum _{n=1}^{\infty }{z}^n\sum _{k=1}^{n-1}\sigma \left(k\right){\tau }_s(n-k),$$

so that

$${\displaystyle \frac{{\displaystyle \sum _{n=1}^{\infty }}n{\tau }_s\left(n\right)z^n}{{\displaystyle \sum _{n=1}^{\infty }}{\tau }_s\left(n\right)z^n}}=1-{\displaystyle \frac{s{\displaystyle \sum _{n=1}^{\infty }}{z}^n{\displaystyle \sum _{k=1}^{n-1}}\sigma \left(k\right){\tau }_s(n-k)}{{\displaystyle \sum _{n=1}^{\infty }}{\tau }_s\left(n\right)z^n}}$$

and that

$${\displaystyle \frac{{\displaystyle \sum _{n=1}^{\infty }}n{\tau }_s\left(n\right)z^n}{{\displaystyle \sum _{n=1}^{\infty }}{\tau }_s\left(n\right)z^n}}=1-s{\displaystyle \sum _{n=1}^{\infty }}\sigma \left(n\right)z^n,$$

that is,

$${\displaystyle \frac{{\displaystyle \sum _{n=1}^{\infty }}n{\tau }_s\left(n\right)z^{n-1}}{{\displaystyle \sum _{n=1}^{\infty }}{\tau }_s\left(n\right)z^n}}={\displaystyle \frac{1}{z}}-s{\displaystyle \sum _{n=1}^{\infty }}\sigma \left(n\right)z^{n-1},$$

which readily yields

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}z}}\left\{ln\left({\displaystyle \sum _{n=1}^{\infty }}{\tau }_s\left(n\right)z^n\right)\right\}={\displaystyle \frac{1}{z}}-s{\displaystyle \sum _{n=1}^{\infty }}\sigma \left(n\right)z^{n-1}.$$

Upon integrating both sides of this last equation, we find that

$$ln\left(\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^n\right)=lnz-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n,$$

so that

$$ln\left(\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}\right)=-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n.$$

Hence, clearly, we conclude that

$$\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}=exp\left(-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n\right).$$

□

Remark 5.

It is also possible to prove Theorem 4 based on the following identity:

$$ln(1-z^n)=-{\int }_0^z{\displaystyle \frac{n{t}^{n-1}}{1-t^n}}\mathrm{d}t\left(\right|z|<1).$$

If we choose $s=24$, that is, ${\tau }_{24}=\tau \left(n\right),$ and set $m=2$ in Definition (8), we obtain the generating function for Ramanujan’s tau numbers using the Eisenstein series $E_2\left(z\right).$ From Theorem 4, we then have

$$\sum _{n=1}^{\infty }\tau \left(n\right)z^n=zexp\left(-24\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n\right),$$

so that

$$\sum _{n=1}^{\infty }\tau \left(n\right)z^n=exp\left({\int }_0^z{\displaystyle \frac{E_2\left(t\right)}{t}}\mathrm{d}t\right).$$

(9)

5. Decomposition of the Numbers ${\mathit{\tau }}_{\mathit{s}}\left(\mathit{n}\right)$

Brent [24] (Theorem 3.1) proved the following formula for computing Ramanujan’s $\tau \left(n\right)$-function values:

$$\tau (n+1)=\sum _{1·k_1+2·k_2+\cdots +(n-1)·k_n=n}{(-24)}^{k_1+k_2+\cdots +k_n}{\displaystyle \frac{\sigma {\left(1\right)}^{k_1}\sigma {\left(2\right)}^{k_2}\cdots \sigma {\left(n\right)}^{k_n}}{1^{k_1}{k}_1!2^{k_2}{k}_2!\cdots n^{k_n}{k}_n!}}.$$

In our next result (see Theorem 5 below), we will prove an analogous decomposition for ${\tau }_s\left(n\right)$ where $s\in \mathbb{N}$.

Theorem 5.

Let $1<n\in \mathbb{N}$ and $s\in \mathbb{N}$. Then

$${\tau }_s\left(n\right)=\sum _{{\sum }_{i=1}^{n-1}i·k_i=n-1}{(-s)}^{{\sum }_{i=1}^{n-1}{k}_i}\prod _{i=1}^{n-1}{\displaystyle \frac{\sigma {\left(i\right)}^{k_i}}{i^{k_i}{k}_i!}}\left(k_i\in \mathbb{N}\cup \left\{0\right\}\right).$$

Proof. 

We present an elementary proof based on Theorem 4 and the expansion of function ${\mathrm{e}}^z$. According to Theorem 4, we have the following equations:

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}& =exp\left(-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n\right)\hfill \\ \hfill & =exp\left(-s{\displaystyle \frac{\sigma \left(1\right)}{1}}{z}^1--s{\displaystyle \frac{\sigma \left(2\right)}{2}}{z}^2-s{\displaystyle \frac{\sigma \left(3\right)}{3}}{z}^3-s{\displaystyle \frac{\sigma \left(4\right)}{4}}{z}^4-\cdots \right).\hfill \end{array}$$

Hence, clearly, the expansion of the function ${\mathrm{e}}^z$ yields

$$\begin{array}{cc}\hfill \sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}& =\left({\displaystyle \frac{{(-s\sigma \left(1\right)z^1)}^0}{1^{0}0!}}+{\displaystyle \frac{{(-s\sigma \left(1\right)z^1)}^1}{1^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(1\right)z^1)}^2}{1^{2}2!}}+\cdots \right)\hfill \\ \hfill & ·\left({\displaystyle \frac{{(-s\sigma \left(2\right)z^2)}^0}{2^{0}0!}}+{\displaystyle \frac{{(-s\sigma \left(2\right)z^2)}^1}{2^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(2\right)z^2)}^2}{2^{2}2!}}+\cdots \right)\hfill \\ \hfill & ·\left({\displaystyle \frac{{(-s\sigma \left(3\right)z^3)}^0}{3^{0}0!}}+{\displaystyle \frac{{(-s\sigma \left(3\right)z^3)}^1}{3^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(3\right)z^3)}^2}{3^{2}2!}}+\cdots \right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & ·\left({\displaystyle \frac{{(-s\sigma (n-1)z^{n-1})}^0}{{(n-1)}^{0}0!}}+{\displaystyle \frac{{(-s\sigma \left(n-1\right)z^{n-1})}^1}{{(n-1)}^{1}1!}}+\cdots \right)\hfill \\ \hfill & ⋮\hfill \end{array}$$

In order to compute the value for $\tau \left(n\right)$, we need to calculate the coefficient of $z^{n-1}$ on the right-hand side of this last equation. For this reason, we consider the following product:

$$\begin{array}{cc}\hfill & \left(1+{\displaystyle \frac{{(-s\sigma \left(1\right)z^1)}^1}{1^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(1\right)z^1)}^2}{1^{2}2!}}+{\displaystyle \frac{{(-s\sigma \left(1\right)z^1)}^3}{1^{3}3!}}+\cdots \right)\hfill \\ \hfill & ·\left(1+{\displaystyle \frac{{(-s\sigma \left(2\right)z^2)}^1}{2^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(2\right)z^2)}^2}{2^{2}2!}}++{\displaystyle \frac{{(-s\sigma \left(2\right)z^2)}^3}{2^{3}3!}}\cdots \right)\hfill \\ \hfill & ·\left(1+{\displaystyle \frac{{(-s\sigma \left(3\right)z^3)}^1}{3^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(3\right)z^3)}^2}{3^{2}2!}}+{\displaystyle \frac{{(-s\sigma \left(3\right)z^3)}^3}{3^{3}3!}}+\cdots \right)\hfill \\ \hfill & ⋮\hfill \\ \hfill & ·\left(1+{\displaystyle \frac{{(-s\sigma \left(n-1\right)z^{n-1})}^1}{{(n-1)}^{1}1!}}+{\displaystyle \frac{{(-s\sigma \left(n-1\right)z^{n-1})}^2}{{(n-1)}^{2}2!}}+{\displaystyle \frac{{(-s\sigma \left(n-1\right)z^{n-1})}^3}{{(n-1)}^{3}3!}}+\cdots \right).\hfill \end{array}$$

In the formation of the coefficient of $z^{n-1}$ in the specified product, each factor contributes a single term from the sum of the type:

$${(-s)}^{k_i}{\displaystyle \frac{{\left[\sigma \left(i\right)\right]}^{k_i}{z}^{i{k}_i}}{i^{k_i}{k}_i!}}(k_i\in \mathbb{N}\cup \left\{0\right\}).$$

It is clear that the following condition must be satisfied:

$$1·k_1+2·k_2+\cdots +(n-1)·k_{n-1}=n-1,$$

so that

$$\begin{array}{cc}\hfill {\tau }_s\left(n\right)& =\sum _{1·k_1+2·k_2+\cdots +(n-1)·k_{n-1}=n-1}{(-s)}^{k_1+k_2+\cdots +k_{n-1}}\hfill \\ \hfill & ·{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^{k_1}{\left[\sigma \left(2\right)\right]}^{k_2}\cdots {\left[\sigma (n-1)\right]}^{k_{n-1}}}{1^{k_1}·k_1!·2^{k_2}·k_2!\cdots {(n-1)}^{k_{n-1}}·k_{n-1}!}}.\hfill \end{array}$$

(10)

□

Example 3.

We use the formula that was proven in Theorem 5. For $s=24,$ we thus find that

$$\begin{array}{c}\hfill \tau \left(2\right)={(-24)}^1{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^1}{1^1·1!}}=-24,\end{array}$$

$$\tau \left(3\right)={(-24)}^2{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^2]}{1^2·2!}}+{(-24)}^1{\displaystyle \frac{{\left[\sigma \left(2\right)\right]}^1}{2^1·1!}}=252,$$

$$\tau \left(4\right)={(-24)}^3{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^3}{1^3·3!}}+{(-24)}^2{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^1}{1^1·1!}}{\displaystyle \frac{{\left[\sigma \left(2\right)\right]}^1}{2^1·1!}}+{(-24)}^1{\displaystyle \frac{{\left[\sigma \left(3\right)\right]}^1}{3^1·1!}}=-1472,$$

and

$$\begin{array}{cc}\hfill \tau \left(5\right)& ={(-24)}^4{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^4}{1^4·4!}}+{(-24)}^3{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^2}{1^2·2!}}{\displaystyle \frac{{\left[\sigma \left(2\right)\right]}^1}{2^1·1!}}+{(-24)}^2{\displaystyle \frac{{\left[\sigma \left(1\right)\right]}^1}{1^1·1!}}{\displaystyle \frac{{\left[\sigma \left(3\right)\right]}^1}{3^1·1!}}\hfill \\ \hfill & +{(-24)}^2{\displaystyle \frac{{\left[\sigma \left(2\right)\right]}^2}{2^2·2!}}+{(-24)}^1{\displaystyle \frac{{\left[\sigma \left(4\right)\right]}^1}{4^1·1!}}=4830.\hfill \end{array}$$

Remark 6.

Recently, Goran-Dumitru and Merca [13] established multiple decompositions of Ramanujan’s τ-function as sums over all partitions.

6. Concluding Remarks and Observations

In this concluding section, we introduce the following notation:

$${\displaystyle \frac{{\partial }^m}{\partial z^m}}\left\{f\left(z\right)\right\}:={\left(f\left(z\right)\right)}^{\left(m\right)}=:f^{\left(m\right)}(m\in {\mathbb{N}}_0).$$

Based on Theorem 4, here we demonstrate another possible decomposition of the Ramanujan $\tau$-function. Let the function $f:=f\left(z\right)$ be given as follows:

$$f=-s\sum _{n=1}^{\infty }{\displaystyle \frac{\sigma \left(n\right)}{n}}{z}^n.$$

We then obtain

$${\left({\left(\sum _{n=1}^{\infty }{\tau }_s\left(n\right)z^{n-1}\right)}^{(n-1)}\right)}_{z=0}={\left({\left({\mathrm{e}}^f\right)}^{(n-1)}\right)}_{z=0},$$

that is,

$$(n-1)!{\tau }_s\left(n\right)={\left({\left({\mathrm{e}}^f\right)}^{(n-1)}\right)}_{z=0}.$$

Let us examine the derivatives of the function ${\mathrm{e}}^f$ at point $z=0$. We thus find that

$$\begin{array}{c}\hfill {\left({\left({\mathrm{e}}^f\right)}^{\left(1\right)}\right)}_{z=0}={\left({\mathrm{e}}^f{f}^{\left(1\right)}\right)}_{z=0}={\left(f^{\left(1\right)}\right)}_{z=0},\end{array}$$

$${\left({\left({\mathrm{e}}^f\right)}^{\left(2\right)}\right)}_{z=0}={\left({\mathrm{e}}^f{\left(f^{\left(1\right)}\right)}^2+{\mathrm{e}}^f{f}^{\left(2\right)}\right)}_{z=0}={\left({\left(f^{\left(1\right)}\right)}^2+f^{\left(2\right)}\right)}_{z=0},$$

$$\begin{array}{cc}\hfill {\left({\left({\mathrm{e}}^f\right)}^{\left(3\right)}\right)}_{z=0}& ={\left({\mathrm{e}}^f{\left(f^{\left(1\right)}\right)}^3+3{\mathrm{e}}^f{f}^{\left(1\right)}{f}^{\left(2\right)}+{\mathrm{e}}^f{f}^{\left(3\right)}\right)}_{z=0}\hfill \\ \hfill & ={\left({\left(f^{\left(1\right)}\right)}^3+3f^{\left(1\right)}{f}^{\left(2\right)}+f^{\left(3\right)}\right)}_{z=0}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\left({\left({\mathrm{e}}^f\right)}^{\left(4\right)}\right)}_{z=0}& ={\left({\mathrm{e}}^f{\left(f^{\left(1\right)}\right)}^4+6{\mathrm{e}}^f{\left(f^{\left(1\right)}\right)}^2{f}^{\left(2\right)}+3{\mathrm{e}}^f{\left(f^{\left(2\right)}\right)}^2\right)}_{z=0}\hfill \\ \hfill & +{\left(4{\mathrm{e}}^f{f}^{\left(1\right)}{f}^{\left(3\right)}+{\mathrm{e}}^f{f}^{\left(4\right)}\right)}_{z=0}\hfill \\ \hfill & ={\left({\left(f^{\left(1\right)}\right)}^4+6{\left(f^{\left(1\right)}\right)}^2{f}^{\left(2\right)}+3{\left(f^{\left(2\right)}\right)}^2+4f^{\left(1\right)}{f}^{\left(3\right)}+f^{\left(4\right)}\right)}_{z=0}.\hfill \end{array}$$

Hence, finally, we choose to conjecture that

$$\begin{array}{c}\hfill (n-1)!{\tau }_s\left(n\right)={\left(\sum _j{b}_j{\left(f^{\left(1\right)}\right)}^{a_{j,1}}{\left(f^{\left(2\right)}\right)}^{a_{j,2}}\cdots {\left(f^{(n-1)}\right)}^{a_{j,n-1}}\right)}_{z=0},\end{array}$$

(11)

where the sum taken over all admissible j and

$$1·a_{j,1}+2·a_{j,2}+\cdots +(n-1)·a_{j,n-1}=n-1.$$

Remarkably, the computation of each of the coefficients

$$b_j\mathrm{and}{a}_{j,i}(i\in \{1,2,3,\cdots ,n-1\}),$$

which are involved in our conjecture as given in (11), remains an open problem.