Abelian groups of order n.

Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?

Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n.

Can anyone please help me understand. Thank you.

By the fundamental theorem for finite abelian groups the number of abelian groups of order $n=p_1^{n_1}\dots p_k^{n_k}$ is the product of the partition numbers of $n_i$.

Note that the partition number of $2$ is $2$ and the partition number of $4$ is $5$. Since $10^6=2^6\cdot 5^6$ such an $n$ therefore exists.

Given any number $n$, the number of different (up to isomorphism) abelian groups of order $n$ is dependant on the prime factorization of $n$. If $n$ is a prime power $p^k, k\in \Bbb N$, then the number of abelian groups of order $n$ is exactly the number of distinct partitions of $k$. Example: For $p = 3, k = 4$ and thus $n = 81$, there are $5$ different abelian groups: $$ \Bbb Z_{81} \quad \Bbb Z_{27}\times \Bbb Z_3 \quad \Bbb Z_{9} \times \Bbb Z_9 \quad \Bbb Z_9 \times \Bbb Z_3 \times \Bbb Z_3 \quad (\Bbb Z_3)^4 $$ Now, if there are multiple primes in the factorization of $n$, the number of partitions allowed by each one are just multiplied together. Example: If $n = 2^53^7 = 69,984$, then there are $7\cdot 15 = 105$ different abelian groups, since $5$ has $7$ partitions and $7$ has $15$.

Now, to construct a million, we need $2^65^6$ different partitions total. The easiest way (maybe the only way) is to let $n$ have be $6$ primes to the power $2$ ($2$ partitions) and $6$ primes to the power $4$ (five partitions). Thus, I believe the smallest such $n$ is given by $$ n = 2^43^45^47^411^413^417^219^223^229^231^237^2 \\= 49,659,789,817,537,838,957,341,175,342,490,000 $$

If $n=\prod_{i=1}^rp_i^{k_i}$, then the number of distinct abelian groups of order $n$ is given by $$ \prod_{i=1}^rp(k_i), $$ where $p(k)$ denotes the number of partitions of $k$. Now $10^6=p(k_1)\cdots p(k_r)$ has to be solved. But $p(2)=2$ and $p(4)=5$, so we can solve it.