### tahsin's blog

By tahsin, history, 12 months ago, ,

What is the upper bound of total number of divisors of divisors of a number ?

• +41

 » 12 months ago, # |   +28 If a is divisor of x and y and both x and y are divisors of z, do you count a twice?
•  » » 12 months ago, # ^ |   +6 yes
•  » » » 12 months ago, # ^ | ← Rev. 3 →   -10 then you may consider the upper bound to be , as the upper bound on number of divisors is (verified upto n = 1018)
•  » » » » 12 months ago, # ^ |   +8 But where does come from? Do you assume that divisors of n are of magnitude ? That isn't true.
•  » » » » » 12 months ago, # ^ |   +37 I don't know if I was high writing that... My bad :(
•  » » » » 12 months ago, # ^ | ← Rev. 3 →   +28 Let f(n) be the number of divisors of divisors of n. If we are going to use a bound for d(n), we may use the identity: where rad(n) and ω(n) are the product and the number of distinct prime divisors of n, respectively. That formula can be obtained by noting that f is multiplicative (being the Dirichlet convolution of d and 1, or the triple Dirichlet convolution of 1), and multiplying everything after getting .Now, using the simple estimate (which comes from the fact that d(n) = (α1 + 1)·...·(αω(n) + 1)and increasing each term by 1 multiplies each bracket by at most 3/2)we get According to the last column of this table, talking only "competitive programming numbers" into account: this bound is better than the trivial bound by ~ an order of magnitude, but should also not be very far from the truth - the worst cases have several prime factors, with only the exponent on the prime number $2$ being significant.Of course, the asymptotic behaviour of d(n) has already been well-explained here, and even better on What's New. I couldn't obtain a real-world bound using this kind of approach, though.Also, Unable to parse markup [type=CF_TEX] is buggy here. I think there is a problem with parsing the comments.
 » 12 months ago, # | ← Rev. 3 →   +28 if your purpose is not mathematical , you can approximately find by using brute force (I mean , you don't have a time limit in your computer)