<s>[INST] Generating functions (with symbol method) of special partitions of natural numbers I want to show that the generating function of the number of partitions where every summand appears at most twice of every natural number n equals the number of paritions of n into summands which are not divisible by 3. 
I know that for the partitions $\mathcal{\tilde{P}}$ where every summand appears at most once it holds that: 
$\tilde{\mathcal{P}} = (\epsilon + 1)\times (\epsilon + 2) \times (\epsilon + 3) \times \dotsc$ and therefore $P(z) = (1+z)(1+z^2)(1+z^3) + \dotsc$. 
So now I'm wondering if the partitions $\mathcal{P}$ where every summand appears at motst twice is described as 
1) $\mathcal{P} = (\{\epsilon\} + \{1\} + \{1,1\})\times (\{\epsilon\} + \{2\} + \{2,2\}) \times (\{\epsilon\} + \{3\} + \{3,3\}) \times \dotsc$ and therefore corresponds with 
1a)
$P(z) = (1+z+2z)(1+z^2+2z^2)+(1+z^3+2z^3) \dotsc$ or with 
1b) 
$P(z) = (1+ z + z)(1+ z^2 + z^4)(1+z^3+z^6)\dotsc$ 
or if the description is 
2) $\mathcal{P} = (\epsilon + 1)\times(\epsilon + 1) \times (\epsilon + 2) \times (\epsilon + 2) \times \dotsc$ and therefore corresponds with 
$P(z) = (1+z)^2(1+z^2)^2(1+z^3)^2\dotsc$? 
The second partition, the one into parts that are not divisible by 3, I think the description is: 
$\mathcal{P}^{\ast} = SEQ(1) \times SEQ(2) \times SEQ(4) \times \dotsc$ 
and therefore $P^{\ast}(z) = \frac{1}{1-z}\frac{1}{1-z^2}\frac{1}{1-z^4}\frac{1}{1-z^5}\dotsc$. 
So is any of the options described in 1) correct, and if so, which one or why not, and is 2) correct?
 [/INST] The GF of the partitions where every summand appears at most twice is 
$$f(x)=(1+x+x^2)(1+x^2+x^4)\cdot\ldots = \prod_{k\geq 1}(1+x^k+x^{2k})$$
while the GF of the partitions into summands which are not divisible by 3 is 
$$ g(x) = \frac{1}{1-x}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x^4}\cdot\frac{1}{1-x^5}\cdot\ldots = \prod_{\substack{k\geq 1\\3\nmid k}}\frac{1}{1-x^k}. $$
It is not difficult to check that $f(x)=g(x)$. Indeed
$$ f(x)=\prod_{k\geq 1}\frac{1-x^{3k}}{1-x^k} = \prod_{k\geq 1}\frac{1}{1-x^k}\prod_{k\geq 1}(1-x^{3k}) = \prod_{n\geq 1}\frac{1}{1-x^n}\prod_{\substack{n\geq 1\\ 3\mid n}}(1-x^n) = g(x).$$
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