Application of Rogers-Ramanujan (types) Identities

Czhang271828

小兔子有一堆胡萝卜，她试着把胡萝卜分成几堆。小兔子先尝试把胡萝卜这样几堆，每堆加上或者减去一根胡萝卜后数量是 $5$ 的倍数，发现有好多种方法。小兔子又把萝卜分成这样几堆，只是没有两堆数量相同或相差 $1$ ，发现又有好多种方法。这个时候小乌龟发现两种分法的可行方法数居然是相等的，聪明的小朋友，这是为什么呢？

比如小兔子有 $6$ 个胡萝卜，那么第一种方法有
\begin{align*}
6&=1+1+1+1+1+1,\\
6&=1+1+4,\\
6&=6.
\end{align*}三种分法。

第二种方法有
\begin{align*}
6&=1+5,\\
6&=2+4,\\
6&=6.
\end{align*}
也是三种分法。

青青子衿

罗杰斯-拉马努金的应用
\begin{align*}
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{5j+1})(1-q^{5j+4})}=1+\sum_{k=1}^{\infty}\dfrac{q^{k^2}}{(1-q)(1-q^2)\cdots(1-q^k)}
\end{align*}

Lemma 2 ( A Rogers-Ramanujan Identity ) The number of partitions of n into parts congruent to 1 or 4 mod 5 is equal to the number of partitions into parts that are neither repeated nor consecutive.
分拆为等于 1 或 4 mod 5 的部分的整数分拆数等于分拆为既不重复也不连续的部分的整数分拆数.

The number of partitions of n such that the adjacent parts differ by at least 2 is the same as the number of partitions of n such that each part is congruent to either 1 or 4 modulo 5.
对于n，所分出的部分两两之间相差至少为 2 的整数分拆数等于所分出的部分都模5余1或4的整数分拆数.

萌新提问：所分出的部分两两之间相差至少为 3 的整数分拆数怎么表示呢？有相应的恒等式吗？

Czhang271828

青青子衿 发表于 2022-5-17 19:42
罗杰斯-拉马努金的应用
\begin{align*}
\prod_{j=1}^{\infty}\dfrac{1}{(1-q^{5j+1})(1-q^{5j+4})}=1+\sum_ ...

根据我在 kuing.cjhb.site/forum.php?mod=viewthread&tid=6607&extra= ... 26typeid%3D5&i=1 中的回答, 整数之"指定间隔划分"的生成函数应为
\[
q^{1+(1+d)+(1+2d)+\cdots(1+ (k-1)d)}\cdot\prod_{j=1}^k\dfrac{1}{1-q^j}.
\]
其中 $d$ 为间隔, 对小白兔第一问题而言, $d=2$.

那么萌新提问里的恒等式一端就是
\[
1+\sum_{k=1}^{\infty}\dfrac{q^{(3k-1)k/2}}{(1-q)(1-q^2)\cdots(1-q^k)}.
\]

据我所知, 没看到什么优美的恒等式.

青青子衿

Czhang271828 发表于 2022-5-18 15:09
\begin{align*}
q^{{\color{red}{k}}+1+(1+d)+(1+2d)+\cdots(1+ (k-1)d)}\cdot\prod_{j=1}^k\dfrac{1}{1-q^j}.
\end{align*}

第二行的公式有笔误

Number of partitions of n with distinct parts p(i) such that if i != j, then |p(i) - p(j)| >= 3.
Also number of partitions of n into distinct parts in which the smallest part is greater than or equal to number of parts.
oeis.org/A025157

Czhang271828

青青子衿 发表于 2022-5-18 19:31
第二行的公式有笔误

Number of partitions of n with distinct parts p(i) such that if  ...

对, 前一步先想着提出一个 $k$, 后一步就直接把 $k$ 放进去求和 thx

以及二楼的乘积式应该从 $0$ 开始吧

青青子衿

Czhang271828 发表于 2022-5-18 21:58
对, 前一步先想着提出一个 $k$, 后一步就直接把 $k$ 放进去求和 thx

以及二楼的乘积式应该从 $ ...

是的，二楼也有笔误，现已更正。

另外，舒尔的文章里有类似于“萌新提问”的恒等式

Schur’s theorem itself deals with gaps ≥ 3, with the additional condition that there are no consecutive multiples of 3 connecting to the congruence condition for parts ±1 (mod 6).
Schur's partition theorem: The numberof partitions of n into parts congruent to 1 or 5 nodulo 6 equals the numberof partitions of n in which the difference bet ween all parts is at least 3 and between multiples of 3 is at least 6.
Schur's partition theorem: The nurmber of partitions of n into distinct parts ≡1,2(mod 3) is equal to the number of partitions of n into parts with minimal difference 3 and with no consecutive multiples of 3.

\begin{align*}
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{6j+1})(1-q^{6j+5})}
&=\prod_{j=0}^{\infty}(1+q^{3j+1})(1+q^{3j+2})\\
&=\prod_{j=0}^{\infty}(1+q^{2j+1}+q^{4i+2})\\
&=\prod_{j=0}^{\infty}\dfrac{1}{1-q^{j}+q^{2j}}
\end{align*}

oeis.org/A003105

The first Göllnitz-Gordon identity states that the number of partitions of n in which the minimal difference between parts is at least 2, and at least 4 between even parts, equals the number of partitions of n into parts congruent to 1, 4, or 7 (mod 8).

格尔尼茨-戈登恒等式
Göllnitz Partition Identities
\begin{align*}
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{8j+1})(1-q^{8j+4})(1-q^{8j+7})}&=1+\sum _{k=1}^{\infty} \frac{q^{k^2}\prod _{m=1}^{k} \left(1+q^{2m-1}\right)}{\prod _{m=1}^k \left(1-q^{2 m}\right)}\\
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{8j+3})(1-q^{8j+4})(1-q^{8j+5})}&=1+\sum _{k=1}^{\infty} \frac{q^{k(k+2)}\prod _{m=1}^{k} \left(1+q^{2m-1}\right)}{\prod _{m=1}^k \left(1-q^{2 m}\right)}
\end{align*}
格尔尼茨-戈登迷你恒等式
Göllnitz Partition Little Identities
\begin{align*}
\prod _{j=0}^{\infty}\dfrac{1}{\left(1-q^{4 j+1}\right) \left(1-q^{8 j+6}\right)}&=1+\sum _{k=1}^{\infty} \frac{q^{k(k+1)}\prod _{m=1}^{k} \left(1+q^{2m-3}\right)}{\prod _{m=1}^k \left(1-q^{2 m}\right)}\\
\prod _{j=0}^{\infty}\dfrac{1}{\left(1-q^{4 j+3}\right) \left(1-q^{8 j+2}\right)}&=1+\sum _{k=1}^{\infty} \frac{q^{k(k+1)}\prod _{m=1}^{k-1} \left(1+q^{2m-1}\right)}{\prod _{m=1}^k \left(1-q^{2m}\right)}
\end{align*}

青青子衿

青青子衿 发表于 2022-5-18 23:13
是的，二楼也有笔误，现已更正。

另外，舒尔的文章里有类似于“萌新提问”的恒等式

有空再补充几个
\begin{align*}
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{3j+1})(1-q^{6j+5})}&=1+\sum_{k=1}^{\infty}\dfrac{q^{(3k-1)k/2}\prod _{m=1}^k \left(1-q^{6m-4}\right)}{\prod _{m=1}^{3 k} \left(1-q^m\right)}\\
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{3j+2})(1-q^{6j+7})}&=1+\sum_{k=1}^{\infty}\dfrac{q^{(3k+1)k/2}\prod _{m=1}^k \left(1-q^{6m-2}\right)}{\prod _{m=1}^{3 k} \left(1-q^{m+1}\right)}\\
\end{align*}



\begin{align*}
\prod _{j=0}^{\infty} \left(1+q^{3 j+3}\right) \left(1+q^{6 j+1}\right)&=1+\sum _{k=1}^{\infty} \frac{q^{3(k+1)k/2} \prod _{m=1}^{k}\left(1+q^{3m-5}\right)}{\prod _{m=1}^{k} \left(1-q^{3m}\right)}\\
\prod _{j=0}^{\infty} \left(1+q^{3 j+3}\right) \left(1+q^{6 j+2}\right)&=1+\sum _{k=1}^{\infty} \frac{q^{3(k+1)k/2} \prod _{m=1}^{k}\left(1+q^{3m-4}\right)}{\prod _{m=1}^{k} \left(1-q^{3m}\right)}\\
\end{align*}

Czhang271828

青青子衿 发表于 2022-5-22 00:01
有空再补充几个
\begin{align*}
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{3j+1})(1-q^{6j+5})}&=1+\sum_{k=1 ...

$\mathfrak {sl}_3$ analogue of the first Rogers–Ramanujan identity
\[
\sum_{\begin{aligned}n_1,n_2=0\\n_2\leq 2 n_1\end{aligned}}\dfrac{q^{n_1^2-n_1n_2+n_2^2}\cdot \prod_{l =1}^{2n_1}(1-q^l)}{\prod_{s=1}^{n_1}(1-q^s)\cdot \prod_{t=1}^{n_2}(1-q^t)\cdot \prod_{k=1}^{n_1+n_2}(1-q^k)}=\prod_{n=1}^\infty\dfrac{1}{(1-q^{7n-1})^2(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n-6})^2}
\]

青青子衿

青青子衿 发表于 2022-5-17 19:42
罗杰斯-拉马努金的应用
\begin{align*}
\prod_{j=0}^{\infty}\dfrac{1}{(1-q^{5j+1})(1-q^{5j+4})}=1+\sum_{k=1}^{\infty}\dfrac{q^{k^2}}{(1-q)(1-q^2)\cdots(1-q^k)}
\end{align*}

\begin{align*}
\prod _{n=0}^{+\infty } \frac{1-q^{n+1}}{\left(1-q^{5 n+1}\right) \left(1-q^{5 n+4}\right)}=\sum _{n=-\infty }^{+\infty } (-1)^n q^{\frac{1}{2} n (5 n+1)}
\end{align*}