Find #non-isomorphic Abelian groups of order $n$. Seems for non-isomorphic abelian groups, need consider base cases first.
But, can there be any shortcut.
Or is it the best way (though totally doubt, given the number of computations/data it needs) to get formula; and pursue it.

Finite abelian groups are direct sums of their Sylow $p$-subgroups, which are direct sums of cyclic $p$-groups $C_{p^k}$, and this direct sum decomposition uniquely identifies the isomorphism class. So if $n$ has prime factorization $\prod_i p_i^{e_i}$ it follows that
$$\alpha(n) = \prod_i p(e_i)$$
where $p$ is the partition function. This sequence is A000688 on the OEIS. It's actually not hard to figure out the optimal value here for $n \le 200$. For starters, we have
$$\alpha(128) = \alpha(2^7) = p(7) = 15.$$
This is clearly best possible with $n$ a prime power. To beat this we need $n$ to have at least two prime factors, so $\alpha(n)$ consists of a product of at least two values of the partition function. The smallest product of two values of the partition function that exceeds $15$ is $p(3) p(5) = 3 \cdot 7 = 21$, but $2^5 \cdot 3^3 = 864 > 200$ is too large. In general, because $p(1) = 1$ there's no point in including a single copy of any prime, so all exponents need to be at least $2$, and $2^2 \cdot 3^2 \cdot 5^2 = 900 > 200$ is also too large, so we can't use $3$ or more prime factors either. So $\boxed{ \alpha(128) = 15 }$ is actually optimal.
In general, to optimize the value of $\alpha(n)$ over all $n \le N$ we want to optimize $\prod_i p(e_i)$ subject to the constraint that $\prod p_i^{e_i} \le N$, or $\sum e_i \log p_i \le \log N$. The partition function has asymptotic $\log p(n) \propto \sqrt{n}$ so roughly speaking we are trying to optimize $\sum \sqrt{e_i}$ subject to the constraint that $\sum e_i \log p_i \le \log N$. If we relax the $e_i$ to have real values then a Lagrange multiplier argument gives that we want $\frac{1}{\sqrt{e_i}} \propto \log p_i$, or $e_i \propto \frac{1}{(\log p_i)^2}$. But I don't think this kicks in until $N$ is fairly large (because the asymptotic for the partition function won't kick in until the exponents are fairly large) and I expect taking $n$ to be a power of $2$ or close to it (e.g. a product of a power of $2$ and a power of $3$) to be optimal for awhile past $200$.
Edit: In fact the sequence of $n$ such that $\alpha(n)$ is maximized up to $n$ is A046055 and the first entry on that list that is not a power of $2$ is $221184 = 2^{13} \cdot 3^3$. No entry on the list appears to be divisible by $5$ so it looks like they're divisible only by $2$ and $3$ up until $4194304 = 2^{22}$ at least.
