@Goblin8093 ooo very good question. Ends up you can reverse this formula using the equivalence of the two.
If the number of divisors =2 then you can pick the smallest prime.
Ends up the other cases are very similar to a 'reverse engineering' of the mechanism driving the d-function. d(n)= (e_0 + 1)(e_1 +1)...(e_k + 1) where e_i is the exponent of a prime number (in order of the primes). Grost did a paper on these before. Look up "Divisor" on Wolfram Mathworld and I think it's there.
@Goblin8093 Neverless, take a few examples of this and break it apart. You know based on d(p^n) = n+1. Then strip this property apart and just get back the original result by applying the same sort of procedure used to break the number into it's pieces originally except raise each multiple by a prime value once you group them (and remove the value effectively). I believe that should give you what you need but, I warn you prime power decomposition is a costly algorithm. I'll send you a couple egs
I am working on perfect numbers, and this is really helpful.
Thanks a lot.
Is there an algorythm to figure out not just the number of factors of a given number, but all the factors themselves?
Thanks.
Kurtlane 1 year ago
Great simple explanation, it is really helpful, but i have a question:
"How can i get the smallest number which makes a given number of divisors ?"
Ex: What is the minimum number which makes 20 divisors ?
because i want to apply it on greater number of divisors like 500 divisors !
Goblin8093 1 year ago
@Goblin8093 ooo very good question. Ends up you can reverse this formula using the equivalence of the two.
If the number of divisors =2 then you can pick the smallest prime.
Ends up the other cases are very similar to a 'reverse engineering' of the mechanism driving the d-function. d(n)= (e_0 + 1)(e_1 +1)...(e_k + 1) where e_i is the exponent of a prime number (in order of the primes). Grost did a paper on these before. Look up "Divisor" on Wolfram Mathworld and I think it's there.
Entertainmentwf 1 year ago
@Goblin8093 Neverless, take a few examples of this and break it apart. You know based on d(p^n) = n+1. Then strip this property apart and just get back the original result by applying the same sort of procedure used to break the number into it's pieces originally except raise each multiple by a prime value once you group them (and remove the value effectively). I believe that should give you what you need but, I warn you prime power decomposition is a costly algorithm. I'll send you a couple egs
Entertainmentwf 1 year ago
man, I wish I knew what you were saying but I'm going back to school to start out at the bottom. we're talking about Elementary arithmatic. =[
runtubefantasy 1 year ago
@runtubefantasy Awe if you need any help with those topics feel free to send me a message or post any questions you have here :). I'll happily help!
Entertainmentwf 1 year ago