But where does come from? Do you assume that divisors of n are of magnitude ? That isn't true.

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

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,

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