For the divisor function is $d(n^2)$ related to $d(n)$ knowing also n?

Question

The divisor function d(n) is defined as 'the number of positive divisors of n (including 1 and n)' according to Underwood Dudley.

Is the divisor function $d(n^2)$ related to $d(n)$?

for example

d(10)=4 and $d(10^2)$=9
or d(14)=4 and $d(14^2)=9$

So can one find the $d(n^2)$ from knowing only d(n) and n through some relation or function?

Has any work been done on this problem?

Answer 1

There is no direct way to get from the value of $d(n)$ to the value of $d(n^2)$ without involving $n$. For instance, we have $d(6)=d(8)=3$, but $d(36)=8$ while $d(64)=6$.

Your divisor function is closely related to prime factorisations. If $$ n=2^{e_2}\cdot3^{e_3}\cdot5^{e_5}\cdots $$ (where most of the $e_k$ are $0$), then $$ d(n)=(e_2+1)(e_3+1)(e_5+1)\cdots-1 $$ Squaring $n$ doubles all the $e_k$. For the example above, we have $$ d(6)=(1+1)(1+1)-1=3 d(8)=(3+1)-1=3\\ d(36)=(2+1)(2+1)-1\\ d(64)=(6+1)-1=6 $$ So for each way you can write $d(n)+1$ as a product of natural numbers greater than $1$, there is a different value to $d(n^2)$.

Answer 2

Let $n=\prod\limits_{i=1}^{\omega(n)}p_i^{a_i},\,a_i>0$ so $d(n)=\prod\limits_{i=1}^{\omega(n)}(1+a_i)$. Then $n^2=\prod\limits_{i=1}^{\omega(n)}p_i^{2a_i}$ so $d(n^2)=\prod\limits_{i=1}^{\omega(n)}(1+2a_i)$ which cannot be expressed purely in terms of $d(n)$.