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?

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
$$ f(x)=\prod_{k\geq 1}\frac{1-x^{3k}}{1-x^k} = \prod_{k\geq 1}\frac{1}{1-x^k}\prod_{k\geq 1}(1-x^{3k}) = \prod_{n\geq 1}\frac{1}{1-x^n}\prod_{\substack{n\geq 1\\ 3\mid n}}(1-x^n) = g(x).$$
