asymptotic approximation for number of partitions of integer that do contain 1 nor 2 Hardy and Ramanujan provided a famous asymptotic approximation to $P(n)$ the number of partitions of an integer $n$ when $n$ gets large. I wonder if  there is an asymptotic approximation to $P_{\mathbb{N} \backslash \{1,2\}}(n)$ the number of partitions of integer $n$ that do not contain neither 1 nor 2. It is easy to prove that $P_{\mathbb{N} \backslash \{1,2\}}(n)/P(n)$ converges to some constant as $n$ goes to infinity. Simulations shows that this constant lies somewhere near 0.04.
 A: Your sequence is A008483 in the OEIS: https://oeis.org/A008483 . This entry gives the formula
$$ P_{\mathbb{N} \backslash \{1,2\}}(n) \sim {\pi^2 \over 12 \sqrt{3} n^2} \exp(\pi \sqrt{2n/3}) $$
and of course there is the well-known Hardy-Ramanujan formula
$$ P(n) \sim {1 \over 4n \sqrt{3}} \exp(\pi \sqrt{2n/3}). $$
Dividing these gives 
$$ {P_{\mathbb{N} \backslash \{1,2\}}(n) \over P(n)} \sim {\pi^2  \over 3n} $$
and so the probability that a partition of $n$ doesn't contain $1$ or $2$ appears to behave like $1/n$ as $n \to \infty$.
As for why this should hold, there's a model of random partitions of larger integers that's of some interest.  Let a partition be denoted by $(Y_1, Y_2, Y_3, \ldots)$ where $Y_j$ is the multiplicity of $j$.  Set a parameter $q$ and let $P(Y_j = k) = (1-q^j) q^{jk}$ be independent random variables.  (So Y_j is a geometric random variable with mean $q^j/(1-q^j)$). Then a partition generated by this procedure, conditional on it being a partition of $n$ (i. e. $\sum_j jY_j = n$) is actually a partition of $n$ chosen uniformly at random.  I learned this model from Corteel, Pittel, Savage, and Wilf (1998) although they credit it to a 1993 paper of Fristedt; this is also what's known as a "Boltzmann sampler" for partitions.  Now it turns out that setting $q = e^{-\pi/{\sqrt{6n}}}$ is a nice way of generating partitions at random which have size near $n$.  Then we have
$$ P(Y_1 = 0, Y_2 = 0) = (1-q) (1-q^2) $$
and since $1-q \approx \pi/\sqrt{6n}$ and $1-q^2 \approx 2\pi/\sqrt{6n}$ we get $P(Y_1 = 0, Y_2 = 0) \approx \pi^2/3n$ as desired.
More generally, by a similar argument we should get that the probability of a partition of $n$ having no parts equal to $1, 2, \ldots, m$ is asymptotically $m! \pi^{m/2} / (6n)^{m/2}$ as $n \to \infty$.
