Question on partition of positive integer

Question

I know only basic about partition of a natural number like $p(0)=1$, $p(1)=1$,

$p(2)=2 $ that is $$2=1+1$$

$$2=2$$ and similarly $p(3)=3 $ as

$$3=1+1+1$$

$$3=2+1$$

$$3=3$$ and for further reading of partition of natural number is https://en.wikipedia.org/wiki/Partition_(number_theory)#:~:text=In%20number%20theory%20and%20combinatorics,the%20sum%20becomes%20a%20composition.)

I am solving a problem, for that I need all such partition of natural number n such that it does not include $1's.$ For example for $n=3$ there is only one such partition that is $3=3$. similarly for $n=4$ there are two such partitions that is $4=2+2$ and $4=4.$

Question

So I general for a given arbitrary natural number $n$ how many such partitions exists?

Thanks in advance.

Answer 1

There are $p(n) - p(n-1)$ such partitions; equivalently the number of partitions which have at least one $1$ is just $p(n-1)$ because by removing the one you get a partition of $n-1$.

The formula you wrote down in the comments is the number of derangement of $n$ objects which is different so it’s not clear what you really want. To compare these numbers:

The number of derangements is the number of elements in the symmetric group $S_n$ with no fixed point; the number of partitions without one is the number of conjugacy classes of elements in $S_n$ without fixed points. These are very related but clearly not the same; the first grows faster than exponentially and the second grows slower than exponentially.