### S.Jindal's blog

By S.Jindal, history, 13 months ago, ,

I was trying to calculate this for some integer n <  = 109 :

In other words, summation of summation of ... (m times summations) of divisors of n.

If anyone can help me out with this!

• +4

 » 13 months ago, # |   +11 Let's denote your summation by f(m,n).First, prove that f(m,*) is multiplicative. This can be easily proved by induction, assuming f(m-1,*) are multiplicative, and obviously f(0,*) is 1, which is multiplicative.Hence, factorise n into prime powers, and multiply the answers for individual prime powers.So, now you only need to solve f(m,n) where n is of the form (p^a) where p is some prime.Another observation tells you that p is irrelevant, only the value of 'a' matters, so let's call g(m,a) = f(m,p^a)Writing the obvious recursion from g, by the given definition of f, we get g(m,a) = g(m-1,0) + g(m-1,1) + ... + g(m-1,a)with base case g(0,x) = 1We can see that this is equivalent to distributing 'a' objects in 'm+1' places, where the recursion for g indicates looping on number of objects in the 1st place.So, g(m,a) = C(a+m,m)I believe this solves your problem. I might have missed something here or there or maybe did some off by 1 error, but the basic idea is this only, you'll get a bunch of C(n,r) terms and the function is multiplicative.
•  » » 13 months ago, # ^ |   +5 Perfect!
 » 4 months ago, # |   0 Auto comment: topic has been updated by S.Jindal (previous revision, new revision, compare).
•  » » 4 months ago, # ^ |   +5 xd