S.Jindal's blog

By S.Jindal, history, 4 weeks ago, In English,

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!

  • Vote: I like it  
  • +4
  • Vote: I do not like it  

4 weeks ago, # |
  Vote: I like it +11 Vote: I do not like it

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) = 1

We 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.