Would I be justified in saying that a number $N$, that is the product of the first $k$ odd primes, would have the largest sum of factors than all odd numbers less than $N$?
ex. if $k = 4; N = 3 \cdot 5 \cdot 7 \cdot 11$
It seems pretty obvious and I have tested for $k < 8$.
Would I need to somehow prove this statement if I needed to use it somewhere and how would I go on to prove it?
Best Answer
No, as $N$ gets larger you will want to have more factors of the small primes. The first example I found is that $$3\cdot 5 \cdot 7 \cdot 11 \cdot 13\cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47\approx 3.074e17$$ while $$3^3\cdot 5^2 \cdot 7 \cdot 11 \cdot 13\cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43\approx 2.943e17$$ The first has a sum of factors of about $9.018e17$ while the second has a sum of factors of about $9.707e17$ There may be smaller ones, but I doubt it. Note that I replaced $47$ with $3^2\cdot 5=45$
Deleting $47$ divided the sum of factors by $48$. Multiplying by $3^2 \cdot 5$ multiplied the number of factors by $\frac {40 \cdot 31}{4 \cdot 6}=51\frac 23$.
I would expect most examples would come by replacing the prime just above a combination of small factors with the small factors, so other places to look would be replacing $107$ with $3 \cdot 5 \cdot 7=105$ and replacing $137$ with $3^3\cdot 5=135$