An application of Integer Partitions My following question is;
"Let n be a positive integer.
Prove that the number of partitions of n in which no part appears more than once equals to the  number of partitions into parts not congruent to +1 and -1 .(modulo6) .
i think in this question; we should  start from the number of partitions of n in which the partitions no part appears more than once .
So; its a known formula.
if $S=\left\{n_{1}, n_{2}, \ldots, n_{r}\right\},
$ then
$\sum_{n \geq 0} p\left(n \mid \text { parts in } S, \text { none repeated more than } d \text { times) } q^{n}\right.$

\begin{array}{l}
=\prod_{i=1}^{r}\left(1+q^{n_{i}}+q^{n_{i}+n_{i}}+\cdots+q^{\frac{d \text { times }}{n_{i}+n_{i}+\cdots+n_{i}}}\right) \\
=\prod_{i=1}^{r}\left(1+q^{n_{i}}+q^{2 n_{i}}+\cdots+q^{d n_{i}}\right) \\
=\prod_{i=1}^{r} \frac{\left(1-q^{(d+1) n_{i}}\right)}{\left(1-q^{n_{i}}\right)}=\prod_{n \in S} \frac{1-q^{(d+1) n}}{1-q^{n}}
\end{array}
Please notice that the question says the partitions into part no congruent.I am searching this question.So how can i show this equal? Thanks for your answers.
 A: The original statement (number of partitions of $n$ where no part appears more than once equals to the number of partitions into parts not congruent to $\pm 1\pmod 6$) can be verified to be incorrect, such as in the cases $n=5$ and $n=7.$ The other statement in the comments (the number of partitions of $n$ in which no part appears exactly once equals the number of partitions into parts not congruent to $\pm 1\pmod 6$) holds and we will prove it here.
As with pretty much all proofs of such assertions about partitions, we will use generating functions. We wish to prove that $$\prod_{k=1}^{\infty}{\left(-x^k +\frac{1}{1-x^k}\right)}=\frac{\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5})}{\prod_{k=1}^{\infty }(1-x^k)}.$$
Clearing the denominators, it is equivalent to prove that $$\prod_{k=1}^{\infty}{(1-x^k+x^{2k})}=\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5}).$$
The secret ingredient to this proof is the fact that: The number of partitions of $n$ into parts that are all congruent to $\pm 1 \pmod{6}$ is equal to the number of partitions of $n$ into distinct parts that are all congruent to $\pm 1 \pmod{3}$. Here is a quick proof of this fact from p.4 of An Invitation to the Rogers-Ramanujan Identities by Andrew V. Sills: By difference of squares,
\begin{align*}
\prod_{k= 0}^{\infty}{(1+x^{3k+1})(1+x^{3k+2})} &= \frac{\prod_{k=0}^{\infty}{(1-x^{6k+2})(1-x^{6k+4})}}{\prod_{k=0}^{\infty}{(1-x^{3k+1})(1-x^{3k+2})}}\\
&= \frac{\prod_{k=0}^{\infty}{(1-x^{6k+2})(1-x^{6k+4})}}{\prod_{k=0}^{\infty}{(1-x^{6k+1})(1-x^{6k+4})(1-x^{6k+2})(1-x^{6k+5})}}\\
&= \frac {1}{\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5})}.
\end{align*}
As a result, we know that
\begin{align*}
\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5})&=\frac{1}{\prod_{k= 0}^{\infty}{(1+x^{3k+1})(1+x^{3k+2})}}\\
&=\frac{\prod_{k=1}^{\infty}{(1+x^{3k})}}{\prod_{k=1}^{\infty}{(1+x^k)}}.
\end{align*}
Thus, it suffices to prove that $$\prod_{k=1}^{\infty}{(1-x^k+x^{2k})}=\frac{\prod_{k=1}^{\infty}{(1+x^{3k})}}{\prod_{k=1}^{\infty}{(1+x^k)}}$$
or $$\prod_{k=1}^{\infty}{(1+x^k)(1-x^k+x^{2k})}=\prod_{k=1}^{\infty}{(1+x^{3k})}.$$
By sum of cubes, this is true because $$(1+x^k)(1-x^k+x^{2k})=1+x^{3k}.$$
Infinite products can be equivalent in such strange ways!
