Can we find an arithmetic progression ($d\neq0$) which the number of its divisors also form an arithmetic progression ($d\neq0$)? I was solving a problem and then I got this question:
for any length $n>1$, can we find an arithmetic progression ($d\neq0$) that the number of its divisors also form an arithmetic progression? (Common difference $\neq 0$)
for example:

$n=2:$ $1  , 2$
$\sigma_0:$ $1 , 2$


$n=3:$ $1234  , 1235 , 1236$
$\sigma_0:$ $4 , 8 , 12$


$n=4:$ $421  , 422 , 423, 424$
$\sigma_0:$ $2 , 4 , 6, 8$

I tried to code this problem for small common differences but I couldn't code in general, so any ideas how to prove or disprove or code this?
 A: This is a modification of Aryaaaaan's answer aimed at making the sequence of numbers of divisors a nonconstant arithmetic progression.  I will use two facts:  first, the Green-Tao theorem (there exist arbitrarily long arithmetic progressions of primes), and second, the elementary fact that the set of unit fractions $\{1, \frac{1}{2}, \frac{1}{3}, \dots\}$ contains arbitrarily long arithmetic progressions.  The simplest example of the latter is the $n$-term arithmetic progression $\frac{1}{n!}, \frac{2}{n!}, \dots, \frac{n}{n!}$.
Fix a positive integer $n$, an arithmetic progression $p_1, \dots, p_n$ of primes, and a sequence $m_1, \dots, m_n$ whose reciprocals form an arithmetic progression.  Set $N = \prod_{i=1}^n p_i^{m_i-1}$, which has exactly $m_1 m_2 \cdots m_n$ divisors.  Now consider the arithmetic progression $(N p_1, \dots, N p_n)$.  We can compute the number of divisors of each term from its prime factorization:  namely, $N p_i$ has $m_1 m_2 \cdots (m_i+1) \cdots m_n$ divisors.  Dividing out by the number of factors of $N$, we get:
$$
\frac{\sigma_0(N p_i)}{\sigma_0(N)} = \frac{m_i + 1}{m_i} = 1 + \frac{1}{m_i}.
$$
By assumption, these numbers form an arithmetic progression, so the sequence $\sigma_0(N p_i)$ does as well.
For example, if $n = 5$, you can choose $(p_i) = (5, 11, 17, 23, 29)$ and $(m_i) = (30, 20, 15, 12, 10)$, which leads to $N = 5^{29} \cdot 11^{19} \cdot 17^{14} \cdot 23^{11} \cdot 29^{9}$.  Then $5N, 11N, 17N, 23N$, and $29N$ respectively have $1116000, 1134000, 1152000, 1170000$, and $1188000$ divisors.
