Sems that $a(k)=k$?

For $N=6$, $d_1=1$, $d_2=2$, $d_3=3$, $d_4=6$. But $a_1=1$, $a_2=2$, $a_3=2$, $a_4=4$. So not quite $a_k=k$

Hint

gobathegreat wrote: It is given positive integer $N$. Let $d_1$, $d_2$,...,$d_n$ be its divisors and let $a_i$ be number of divisors of $d_i$, $i=1,2,...n$. Prove that $$(a_1+a_2+...+a_n)^2={a_1}^3+{a_2}^3+...+{a_n}^3$$ It was posted here and here before.