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There is no known closed formula. Nicely however, the coefficients can be computed recursively. This answer in a related problem shows that for $n\ge 3$ the coefficients $a_n(k)$ (which are the finite q-Pochhammer symbols $(q;q)_n$ ) in $$\prod_{j=1}^{n}(1-q^j)=\sum_{k=0}^{n(n+1)/2}a_n(k)q^k$$ satisfy $$a_n(k)=\begin{cases} a_{n-1}(k) & 0\le k\le n-1 \\ a_{n-1}(k)-a_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ -a_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases}$$ with the base case $a_2 = (1 , -1, -1, 1)$

Let's use this recursion for the first few cases of $n$:

\begin{align} n=2: &1 , -1, -1, 1\\ n=3: &1 , -1, -1, 0, 1, 1 ,-1 \\ n=4: &1 , -1, -1, 0, 0, \color{red}{2} ,0, 0, -1,-1,1 \end{align}

So for $n=4$ we have $\prod_{j=1}^{4}(1-q^j) = 1 - q - q^2 + \color{red}{2} q^5 - q^8 - q^9 + q^{10}$. The value marked in red shows that there are coefficients which differ from those of the pentagonal number theorem which are $\in \{-1,0,1\}$.

Will there be $n$ for which the coefficients are in the range $[−h \; h]$? Yes. The reference arxiv.org/pdf/1705.07504.pdf from 2017 deals with that and gives limits, see table 1 on page 8 there. The range increases. For example, with $n=30$, you have $h=34$.

Using the recursive procedure outlined above, further values $h$ of the range can be computed easily and with very short processing time by a simple program script. Clearly, it would be practically impossible to obtain these results for mediumsize or large $n$ by expanding $\prod_{j=1}^{n}(1-q^j)$ which entails $2^n$ many product terms which then need to be distributed amongst $n(n+1)/2$ summands with their correct signs. Due to alternating signs, far less than (as a rule of thumb) the average number of assignments, $m = \frac{2^n}{n(n+1)/2}$, result as the coefficients in the sum. Note the alternating sign annihilation effect in the pentagonal number theorem which is so drastic that $ h \in \{-1,0,1\}$. Here, with finite $n$, the $h$-values yet rise drastically as $n$ is increased, we have: \begin{align} n=31: & \; h = 41 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 41 \, q^{240}\\ n=32: & \; h = 50 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 50 \, q^{264}\\ n=33: & \; h = 56 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 56 \, q^{270}\\ & \cdots\\ n=40: & \; h = 196 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 196 \, q^{410}\\ n=50: & \; h = 1111 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 1111 \, q^{637}\\ & \cdots\\ n=100: & \; h = 11,493,312 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 11493312 \, q^{2525}\\ n=200: & \; h = 2,436,994,475,366,700 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 2436994475366700 \, q^{10050}\\ \end{align}

There is no known closed formula. Nicely however, the coefficients can be computed recursively. This answer in a related problem shows that for $n\ge 3$ the coefficients $a_n(k)$ (which are the finite q-Pochhammer symbols $(q;q)_n$ ) in $$\prod_{j=1}^{n}(1-q^j)=\sum_{k=0}^{n(n+1)/2}a_n(k)q^k$$ satisfy $$a_n(k)=\begin{cases} a_{n-1}(k) & 0\le k\le n-1 \\ a_{n-1}(k)-a_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ -a_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases}$$ with the base case $a_2 = (1 , -1, -1, 1)$

Let's use this recursion for the first few cases of $n$:

\begin{align*} n=2: &1 , -1, -1, 1\\ n=3: &1 , -1, -1, 0, 1, 1 ,-1 \\ n=4: &1 , -1, -1, 0, 0, \color{red}{2} ,0, 0, -1,-1,1 \end{align*}

So for $n=4$ we have $\prod_{j=1}^{4}(1-q^j) = 1 - q - q^2 + \color{red}{2} q^5 - q^8 - q^9 + q^{10}$. The value marked in red shows that there are coefficients which differ from those of the pentagonal number theorem which are $\in \{-1,0,1\}$.

Will there be $n$ for which the coefficients are in the range $[−h \; h]$? Yes. The reference arxiv.org/pdf/1705.07504.pdf from 2017 deals with that and gives limits, see table 1 on page 8 there. The range increases. For example, with $n=30$, you have $h=34$.

Using the recursive procedure outlined above, further values $h$ of the range can be computed easily and with very short processing time by a simple program script. Clearly, it would be practically impossible to obtain these results for mediumsize or large $n$ by expanding $\prod_{j=1}^{n}(1-q^j)$ which entails $2^n$ many product terms which then need to be distributed amongst $n(n+1)/2$ summands with their correct signs. Due to alternating signs, far less than (as a rule of thumb) the average number of assignments, $m = \frac{2^n}{n(n+1)/2}$, result as the coefficients in the sum. Note the alternating sign annihilation effect in the pentagonal number theorem which is so drastic that $ h \in \{-1,0,1\}$. Here, with finite $n$, the $h$-values yet rise drastically as $n$ is increased, we have: \begin{align*} n=31: & \; h = 41 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 41 \, q^{240}\\ n=32: & \; h = 50 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 50 \, q^{264}\\ n=33: & \; h = 56 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 56 \, q^{270}\\ & \cdots\\ n=40: & \; h = 196 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 196 \, q^{410}\\ n=50: & \; h = 1111 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 1111 \, q^{637}\\ & \cdots\\ n=100: & \; h = 11,493,312 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 11493312 \, q^{2525}\\ n=200: & \; h = 2,436,994,475,366,700 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 2436994475366700 \, q^{10050}\\ \end{align*}

There is no known closed formula. Nicely however, the coefficients can be computed recursively. This answer in a related problem shows that for $n\ge 3$ the coefficients $a_n(k)$ (which are the finite q-Pochhammer symbols $(q;q)_n$ ) in $$\prod_{j=1}^{n}(1-q^j)=\sum_{k=0}^{n(n+1)/2}a_n(k)q^k$$ satisfy $$a_n(k)=\begin{cases} a_{n-1}(k) & 0\le k\le n-1 \\ a_{n-1}(k)-a_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ -a_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases}$$ with the base case $a_2 = (1 , -1, -1, 1)$

Let's use this recursion for the first few cases of $n$:

$$ \begin{align} n=2: &1 , -1, -1, 1\\ n=3: &1 , -1, -1, 0, 1, 1 ,-1 \\ n=4: &1 , -1, -1, 0, 0, \color{red}{2} ,0, 0, -1,-1,1 \end{align} $$

So for $n=4$ we have $\prod_{j=1}^{4}(1-q^j) = 1 - q - q^2 + \color{red}{2} q^5 - q^8 - q^9 + q^{10}$. The value marked in red shows that there are coefficients which differ from those of the pentagonal number theorem which are $\in \{-1,0,1\}$.

Will there be $n$ for which the coefficients are in the range $[−h \; h]$? Yes. The reference arxiv.org/pdf/1705.07504.pdf from 2017 deals with that and gives limits, see table 1 on page 8 there. The range increases. For example, with $n=30$, you have $h=34$.

Using the recursive procedure outlined above, further values $h$ of the range can be computed easily and with very short processing time by a simple program script. Clearly, it would be practically impossible to obtain these results for mediumsize or large $n$ by expanding $\prod_{j=1}^{n}(1-q^j)$ which entails $2^n$ many product terms which then need to be distributed amongst $n(n+1)/2$ summands with their correct signs. Due to alternating signs, far less than (as a rule of thumb) the average number of assignments, $m = \frac{2^n}{n(n+1)/2}$, result as the coefficients in the sum. Note the alternating sign annihilation effect in the pentagonal number theorem which is so drastic that $ h \in \{-1,0,1\}$. Here, with finite $n$, the $h$-values yet rise drastically as $n$ is increased, we have: $$ \begin{align} n=31: & \; h = 41 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 41 \, q^{240}\\ n=32: & \; h = 50 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 50 \, q^{264}\\ n=33: & \; h = 56 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 56 \, q^{270}\\ & \cdots\\ n=40: & \; h = 196 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 196 \, q^{410}\\ n=50: & \; h = 1111 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 1111 \, q^{637}\\ & \cdots\\ n=100: & \; h = 11,493,312 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 11493312 \, q^{2525}\\ n=200: & \; h = 2,436,994,475,366,700 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 2436994475366700 \, q^{10050}\\ \end{align} $$

There is no known closed formula. Nicely however, the coefficients can be computed recursively. This answer in a related problem shows that for $n\ge 3$ the coefficients $a_n(k)$ (which are the finite q-Pochhammer symbols $(q;q)_n$ ) in $$\prod_{j=1}^{n}(1-q^j)=\sum_{k=0}^{n(n+1)/2}a_n(k)q^k$$ satisfy $$a_n(k)=\begin{cases} a_{n-1}(k) & 0\le k\le n-1 \\ a_{n-1}(k)-a_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ -a_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases}$$ with the base case $a_2 = (1 , -1, -1, 1)$

Let's use this recursion for the first few cases of $n$:

\begin{align} n=2: &1 , -1, -1, 1\\ n=3: &1 , -1, -1, 0, 1, 1 ,-1 \\ n=4: &1 , -1, -1, 0, 0, \color{red}{2} ,0, 0, -1,-1,1 \end{align}

So for $n=4$ we have $\prod_{j=1}^{4}(1-q^j) = 1 - q - q^2 + \color{red}{2} q^5 - q^8 - q^9 + q^{10}$. The value marked in red shows that there are coefficients which differ from those of the pentagonal number theorem which are $\in \{-1,0,1\}$.

Will there be $n$ for which the coefficients are in the range $[−h \; h]$? Yes. The reference arxiv.org/pdf/1705.07504.pdf from 2017 deals with that and gives limits, see table 1 on page 8 there. The range increases. For example, with $n=30$, you have $h=34$.

Using the recursive procedure outlined above, further values $h$ of the range can be computed easily and with very short processing time by a simple program script. Clearly, it would be practically impossible to obtain these results for mediumsize or large $n$ by expanding $\prod_{j=1}^{n}(1-q^j)$ which entails $2^n$ many product terms which then need to be distributed amongst $n(n+1)/2$ summands with their correct signs. Due to alternating signs, far less than (as a rule of thumb) the average number of assignments, $m = \frac{2^n}{n(n+1)/2}$, result as the coefficients in the sum. Note the alternating sign annihilation effect in the pentagonal number theorem which is so drastic that $ h \in \{-1,0,1\}$. Here, with finite $n$, the $h$-values yet rise drastically as $n$ is increased, we have: \begin{align} n=31: & \; h = 41 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 41 \, q^{240}\\ n=32: & \; h = 50 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 50 \, q^{264}\\ n=33: & \; h = 56 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 56 \, q^{270}\\ & \cdots\\ n=40: & \; h = 196 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 196 \, q^{410}\\ n=50: & \; h = 1111 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 1111 \, q^{637}\\ & \cdots\\ n=100: & \; h = 11,493,312 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 11493312 \, q^{2525}\\ n=200: & \; h = 2,436,994,475,366,700 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 2436994475366700 \, q^{10050}\\ \end{align}

There is no known closed formula. Nicely however, the coefficients can be computed recursively. This answer in a related problem shows that for $n\ge 3$ the coefficients $a_n(k)$ (which are the finite q-Pochhammer symbols $(q;q)_n$ ) in $$\prod_{j=1}^{n}(1-q^j)=\sum_{k=0}^{n(n+1)/2}a_n(k)q^k$$ satisfy $$a_n(k)=\begin{cases} a_{n-1}(k) & 0\le k\le n-1 \\ a_{n-1}(k)-a_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ -a_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases}$$ with the base case $a_2 = (1 , -1, -1, 1)$

Let's use this recursion for the first few cases of $n$:

$$ \begin{align} n=2: &1 , -1, -1, 1\\ n=3: &1 , -1, -1, 0, 1, 1 ,-1 \\ n=4: &1 , -1, -1, 0, 0, \color{red}{2} ,0, 0, -1,-1,1 \end{align} $$

So for $n=4$ we have $\prod_{j=1}^{4}(1-q^j) = 1 - q - q^2 + \color{red}{2} q^5 - q^8 - q^9 + q^{10}$. The value marked in red shows that there are coefficients which differ from those of the pentagonal number theorem which are $\in \{-1,0,1\}$.

Will there be $n$ for which the coefficients are in the range $[−h \; h]$? Yes. The reference arxiv.org/pdf/1705.07504.pdf from 2017 deals with that and gives limits, see table 1 on page 8 there. The range increases. For example, with $n=30$, you have $h=34$.

Using the recursive procedure outlined above, further values $h$ of the range can be computed easily and with very short processing time by a simple program script. These values rise drastically as $n$ is increased, we have: $$ \begin{align} n=31: & \; h = 41 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 41 \, q^{240}\\ n=32: & \; h = 50 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 50 \, q^{264}\\ n=33: & \; h = 56 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 56 \, q^{270}\\ & \cdots\\ n=40: & \; h = 196 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 196 \, q^{410}\\ n=50: & \; h = 1111 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 1111 \, q^{637}\\ & \cdots\\ n=100: & \; h = 11,493,312 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 11493312 \, q^{2525}\\ n=200: & \; h = 2,436,994,475,366,700 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 2436994475366700 \, q^{10050}\\ \end{align} $$

Clearly, it would be practically impossible to obtain these results by expanding $\prod_{j=1}^{n}(1-q^j)$ which entails $2^n$ many terms which then need to be accumulated.

There is no known closed formula. Nicely however, the coefficients can be computed recursively. This answer in a related problem shows that for $n\ge 3$ the coefficients $a_n(k)$ (which are the finite q-Pochhammer symbols $(q;q)_n$ ) in $$\prod_{j=1}^{n}(1-q^j)=\sum_{k=0}^{n(n+1)/2}a_n(k)q^k$$ satisfy $$a_n(k)=\begin{cases} a_{n-1}(k) & 0\le k\le n-1 \\ a_{n-1}(k)-a_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ -a_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases}$$ with the base case $a_2 = (1 , -1, -1, 1)$

Let's use this recursion for the first few cases of $n$:

$$ \begin{align} n=2: &1 , -1, -1, 1\\ n=3: &1 , -1, -1, 0, 1, 1 ,-1 \\ n=4: &1 , -1, -1, 0, 0, \color{red}{2} ,0, 0, -1,-1,1 \end{align} $$

So for $n=4$ we have $\prod_{j=1}^{4}(1-q^j) = 1 - q - q^2 + \color{red}{2} q^5 - q^8 - q^9 + q^{10}$. The value marked in red shows that there are coefficients which differ from those of the pentagonal number theorem which are $\in \{-1,0,1\}$.

Will there be $n$ for which the coefficients are in the range $[−h \; h]$? Yes. The reference arxiv.org/pdf/1705.07504.pdf from 2017 deals with that and gives limits, see table 1 on page 8 there. The range increases. For example, with $n=30$, you have $h=34$.

Using the recursive procedure outlined above, further values $h$ of the range can be computed easily and with very short processing time by a simple program script. Clearly, it would be practically impossible to obtain these results for mediumsize or large $n$ by expanding $\prod_{j=1}^{n}(1-q^j)$ which entails $2^n$ many product terms which then need to be distributed amongst $n(n+1)/2$ summands with their correct signs. Due to alternating signs, far less than (as a rule of thumb) the average number of assignments, $m = \frac{2^n}{n(n+1)/2}$, result as the coefficients in the sum. Note the alternating sign annihilation effect in the pentagonal number theorem which is so drastic that $ h \in \{-1,0,1\}$. Here, with finite $n$, the $h$-values yet rise drastically as $n$ is increased, we have: $$ \begin{align} n=31: & \; h = 41 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 41 \, q^{240}\\ n=32: & \; h = 50 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 50 \, q^{264}\\ n=33: & \; h = 56 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 56 \, q^{270}\\ & \cdots\\ n=40: & \; h = 196 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 196 \, q^{410}\\ n=50: & \; h = 1111 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 1111 \, q^{637}\\ & \cdots\\ n=100: & \; h = 11,493,312 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 11493312 \, q^{2525}\\ n=200: & \; h = 2,436,994,475,366,700 \; {\mathrm{with}} \; {\mathrm{the}} \; {\mathrm{term:}} \; 2436994475366700 \, q^{10050}\\ \end{align} $$