Recent content by soandos

Thank you, any ideas where there are implementations of it, or should I just write my own?

I don't want to know if its prime or not, I want the factorization. If i want to know if its prime, I just have to check my list of primes.

It also seems that since PrimePi(x) grows faster than Sqrt(x) doing it the "smart" way will continually get worse thant the brute force way.

I realize that this might seems to be a strange question, but after doing some coding i realized the following. to brute force the factorization of all numbers less than one million takes around 665 million tests (i.e. does this number divide the original). to do it "smarter" (least i...

thank you.

to restate the question differently (I hope these statements are equivalent) given the prime factors of a number and their powers, how many proper divisors does it have?

I want ALL possible combinations, of every length, using only the elements in the list. 2,2,3 is indistinct from 2,3,2.

Combinations? a,b,c a,b b,c a,c a b c 0 so 8 = 3!/1! + 2(the null and the full) (1 being number of repeated elements) doing that for my original set gives 158 if i don't differentiate between what gets repeated (i.e. 13!/12! + 2 = 158) but it seems that there are more.

is the fact that its 2^13 coincidence?

sum of binomials = 8192 dividing by 7! then 4! does not work... Am I correct in assuming that 160 is the correct answer?

How can one find the number of combinations of a list such as {2,2,2,2,2,2,2,3,3,3,3,5,7} (7 2s, 4 3s, 1 5, 1 7). I am pretty sure that the answer is 160 (the number of proper divisors of 9!). How can I figure it out using combinations/permutations or some other method?

I got 824. Did i implement it correctly?

why is P = 1 + x^{33} + x^{67} as opposed to P = 1 + x^{33} + x^{66} ?

(x + x^5 + x^10 + x^25 + x^50)^100. the x^100 co-efficient is one... unless i messed up.

It seems related to the knapsack problem, there there is no issue of weights, and the value is not to be maximized, but equal to a given number.