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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.

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  • $\begingroup$ May I ask where you found this problem? Adding this sort of context could help Readers respond in a helpful way more quickly. $\endgroup$ – hardmath Jun 24 '20 at 22:27
  • $\begingroup$ I had found this problem in the one of the books I searched before. But i exactly couldn't remember its source now. i have to look my arshive. $\endgroup$ – user1062 Jun 24 '20 at 22:46
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    $\begingroup$ The statement does not seem to be correct for small $n$ at least. $\endgroup$ – Mark Sapir Jun 24 '20 at 23:03
  • $\begingroup$ i want to ask that i noticed that from a book that i that there is a very similar question but not like that at all. so i get suspicious if this question is right or not. .i am writing that question wait please. $\endgroup$ – user1062 Jun 24 '20 at 23:05
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    $\begingroup$ Okay, the questions is slightly stated wrong. It is about OEIS sequence A007690 "Number of partitions of n in which no part occurs just once". The entry mentions Andrews twice. $\endgroup$ – Somos Jun 25 '20 at 1:03
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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!

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  • $\begingroup$ that's a great answer. Thank you very much :) $\endgroup$ – user1062 Jun 25 '20 at 1:44
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    $\begingroup$ @user1062 I'm glad to have helped you. If you feel that it is appropriate, please click the check mark on the left side to accept the answer, so that it is recorded as resolved. $\endgroup$ – Favst Jun 25 '20 at 1:46
  • $\begingroup$ of course ,i do it. you are also very kind.thanks again. $\endgroup$ – user1062 Jun 25 '20 at 1:49

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