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For a positive integer k, let d(k) be the number of divisors of k. So d(1)=1, d(p) =2 if p is a prime, d(6)=4, and d(12)=6.

What is the precise asymptotics of SUM_{k=1}^n 1/(kd(k))

## Background:

1) This came up on the side in the polymath5 project.

2) There, Tim Gowers wrote: If nobody knows the answer, maybe that’s one for Mathoverflow, where I imagine a few minutes would be enough.

3) Asked: 14:17 Jerusalem time. (The first accurate answer: 17:44 Jerusalem time.)

4) Looking only at primes or only at integers with a typical number of divisors suggested a loglogn behavior, but looking at semiprimes indicates the sum is larger. I dont know how much larger.

5) I couldn't find an asnwer on the web. If there is an easy way searching for an answer that I missed this will be interesting too.

$\sum_{k=1}^n 1/(kd^2(k))$ ?

For a positive integer k, let d(k) be the number of divisors of k. So d(1)=1, d(p) =2 if p is a prime, d(6)=4, and d(12)=6.

What is the precise asymptotics of SUM_{k=1}^n 1/(kd(k))

## Background:

1) This came up on the side in the polymath5 project.

2) There, Tim Gowers wrote: If nobody knows the answer, maybe that’s one for Mathoverflow, where I imagine a few minutes would be enough.

4) Looking only at primes or only at integers with a typical number of divisors suggested a loglogn behavior, but looking at semiprimes indicates the sum is larger. I dont know how much larger.

5) I couldn't find an asnwer on the web. If there is an easy way searching for an answer that I missed this will be interesting too.

# Followup:

$\sum_{k=1}^n 1/(kd^2(k))$ ?

3 added 3 characters in body

For a positive integer k, let d(k) be the number of divisors of k. So d(1)=1, d(p) =2 if p is a prime, d(6)=4, and d(12)=6.

What is the precise asymptotics of SUM_{k=1}^n 1/(kd(k))

## Background:

1) This came up on the side in the polymath5 project.

2) There, Tim Gowers wrote: If nobody knows the answer, maybe that’s one for Mathoverflow, where I imagine a few minutes would be enough.