<s>[INST] Partitioning a number into set of coprimes such that their product is maximum and sum of partition is the number itself Example 1)
Let the number be 7. Then we have a set {3,4}, so the product of the numbers is 10 and the numbers are mutually coprime.
Example 2)
for n=12, coprimes set = {3,4,5}, where product is 60 which is maximum.
Example 3)
n = 9, coprimes set = {4,5}, product is 20 which is maximum product.
 [/INST] The product is given by the Landau function, shown in OEIS A000793 and beginning
$$ 1, 2, 3, 4, 6, 6, 12, 15, 20, 30, 30, 60, 60, 84, 105, 140, 210,\\ 210, 420, 420, 420, 420, 840, 840, 1260, 1260, 1540, 2310$$
To get the set you can just factor these numbers.  The set always consists of powers of distinct primes.  The powers are chosen to be "roughly equal".  One of the higher values is $$120120=2^3\cdot 3 \cdot 5\cdot 7 \cdot 11 \cdot 13$$
with sum $47$.  $48$ has the same partition with a $1$ added.  Then at $49$ we get 
$$180180=2^2\cdot 3^2\cdot 5\cdot 7 \cdot 11 \cdot 13$$
If there are $p$ factors the highest prime is about $p \log p$, so we can get a guess for the number of factors by solving $p^2 \log p =n$.  At n=$49$ this says we should have about $5.4$ factors, each about $9$.  We can guess we want $2^2, 2^3,$ or $2^4$ and $3^1$ or $3^2$ and first powers of the others and do a little search.
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