Given a number $n$, find $a,b$ such that:
- $a,b$ non-negative integers
- $a+b=n$
- $\Omega(a)+\Omega(b)$ is maximized
$\Omega(n)$ counts the number of prime factors of n (with multiplicity).
For example, for $n = 44100 = 2^2 3^2 5^2 7^2$, we have $\Omega(44100) = \Omega(2^2 3^2 5^2 7^2) = 8$, as the eight prime factors (with repetition) of $n$, are 2, 2, 3, 3, 5, 5, 7, and 7. Obviously $\Omega(1) = 0 $, since 1 is the product of no primes, and for $ p$ prime we have $\Omega(p) = 1 $, since a prime has only one prime factor (itself).
With most of the numbers there are more than one pair of $a$ and $b$ with the same maximal result. With many numbers the highest result achieved by setting $a$ to the highest power of $2$ that is smaller than $n$, but there are also many other cases such as:
$ 17=8+9 \rightarrow 5 \\ 33=9+24 \rightarrow 6 \\ 34=16+18 \rightarrow 7 \\ 43=16+27 \rightarrow 7 \\ 51=24+27 \rightarrow 7 \\ 61=16+45 \rightarrow 7 \\ 66=18+48 \rightarrow 8 \\ 68=32+36 \rightarrow 9 \\ 77=32+45 \rightarrow 8 \\ $
Is there a general rule how to find a pair that will achieve the highest result?