Find the smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n 
What is the  smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n?

This  is a question in Joseph A.Gallian's book, and the  answer is $n=36$ and the  groups are :
$\mathbb Z_9× \mathbb Z_4$;
$\mathbb Z_3×\mathbb Z_3×\mathbb Z_4$;
$\mathbb Z_9×\mathbb Z_2×\mathbb Z_2$;
$\mathbb Z_3×\mathbb Z_3×\mathbb Z_2×\mathbb Z_2$;
Now my question is when $n=16$ then the group is isomorphic to Z(16) and hence abelian.
And $\mathbb Z_{16}$ is of order $16$ so there are $4$ non isomorphic abelian groups:
$\mathbb Z_{16}$;
$\mathbb Z_8×\mathbb Z_2$;
$\mathbb Z_4×\mathbb Z_2×\mathbb Z_2$;
$\mathbb Z_2×\mathbb Z_2×\mathbb Z_2 ×\mathbb Z_2$;
What is wrong in my answer?
 A: The number of non-isomorphic Abelian groups depends on the partition number of the exponents in the prime factorization. Let $P(k)$ be the number of ways to partition $k$ (for example, $P(4) = 5$, since we have $4, 3+1, 2+2, 2+1+1$, and $1+1+1+1$). Then for $n=p_1^{e_1} \ldots p_r^{e_r}$, there are
$$\prod_i P(e_i)$$
non-isomorphic Abelian groups of order $n$. You missed $2+2$ among the ways to partition $4$, corresponding to $\mathbb Z/(2^2\mathbb Z) \times \mathbb Z/(2^2\mathbb Z) = \mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$.
You can use this formula to answer the original question. The first few partition numbers are:
P(1) = 1    1
P(2) = 2    2, 1+1
P(3) = 3    3, 2+1, 1+1+1
P(4) = 5    4, 3+1, 2+2, 2+1+1, 1+1+1+1
P(5) = 7    5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1

and they obviously increase from here. Now, if there are $4$ Abelian groups of order $n$, then we must have $e_1=e_2=2$, and we choose as bases the two smallest primes. Thus the number we are looking for is $2^23^2=36$.
