4 ms·
Not a coincidence. sigma(n) is largest when n has lots of large factors. Since d is a factor if n/d is, it's easiest to guarantee this by making sure that n ha
by impendia 7y ago
Not a coincidence.
sigma(n) is largest when n has lots of large factors. Since d is a factor if n/d is, it's easiest to guarantee this by making sure that n has lots of small factors.
Factorials, by construction, are good examples of numbers with many small factors.
- karma_fountain 7y agoIs that fact that 7 is prime relevant?
- lvh 7y agoSomewhat: you want the number to be as unsmooth (having few small factors) as possible. Since the factorial is the product of all numbers up to a given number, that happens when the top number is a prime, since the next number is necessarily divisible by 2 (and therefore more likely to be smooth).
- pfortuny 7y agoPossibly that all the odd factors of 5040 are prime and they are all the primes less than pr equal to the largest prime divisor of 5040. Only happens for 7!
- impendia 7y agoYes. For example, 7! is already divisible by 8. The smallest integer that 8! is divisible by, and that 7! isn't, is 32 -- much less small.
- mywittyname 7y agoInteresting. So would one way to approach a proof be to define the upper bounds of sigma(n)? If so, have people attempted this? It seems like something that could be done, since you know sigma(n) cannot contain factors larger than n/2 and the number of factors is finite.