How can I find all increasing sequences $\{a_i\}_{i=1}^{\infty}$ such that $d(x_1+x_2+\cdots+x_k)=d(a_{x_{1}}+a_{x_{2}}+\cdots + a_{x_{k}})$?

Question

How can one find all increasing sequences $\{a_i\}_{i=1}^{\infty}$ such that $$d(x_1+x_2+\cdots+x_k)=d(a_{x_{1}}+a_{x_{2}}+\cdots + a_{x_{k}}),$$ holds for all $k$-tuples $(x_1,x_2,\cdots,x_k)$ of positive integers, where $d(n)$ is number of integer divisors of a positive integer $n$, and $k \geq 3$ is a fixed integer?

A special case of this problem which was given in this year's Iran Olympiad:

Find all increasing sequences $a_1,a_2,a_3,...$ of natural numbers such that for each $i,j\in \mathbb N$, number of the divisors of $i+j$ and $a_i+a_j$ is equal.

Answer 1

Well, for $k$ prime you can use the same argument as $k = 2$: You look at the indices $i_p = k^{p-2}$ for $p$ prime. $k * a_{i_p}$ has $p$ factors and must therefore be of the form $q^{p-1}$ for some prime $q$, but since it is divisible by $k$, $q=k$. So we have an infinite sequence of indices for which $a_n = n$, and we can use the fact that the sequence is increasing to prove $\forall n \space a_n = n $.

Non-prime $k$ is a little trickier, but I think you can do something similar.