Number sequence: 1, 2, 9, 48, 120, 162

Question

First time entering a puzzle so I have no idea if what I've made is too easy. But it's not on OEIS at least.

Find the next term!

1, 2, 9, 48, 120, 162, __

Answer 1

Strongly inspired by Mahdi Mahmoodian's answer:

$a_n = n \times s$ where $s$ is the sum of digits in all previous numbers in the sequence

$a_1 = 1 \rightarrow$ sum of digits $= 1$

$a_2 = 2 = 2 \times 1 \rightarrow$ sum of digits $= 1 + 2 = 3$

$a_3 = 9 = 3 \times 3 \rightarrow$ sum of digits $= 1 + 2 + 9 = 12$

$a_4 = 48 = 4 \times 12 \rightarrow$ sum of digits $= 1 + 2 + 9 + 4 + 8 = 24$

$a_5 = 120 = 5 \times 24 \rightarrow$ sum of digits $= 1 + 2 + 9 + 4 + 8 + 1 + 2 + 0 = 27$

$a_6 = 162 = 6 \times 27 \rightarrow$ sum of digits $= 1 + 2 + 9 + 4 + 8 + 1 + 2 + 0 + 1 + 6 + 2 = 36$

The answer is:
$a_7 = 252 = 7 \times 36$

Answer 2

I got an idea but it's not complete yet. I will update it as soon as I find something else:

The $i$th number has $i$ as the divisor.

$f(1) = 1 * 1 = 1$
$f(2) = 2 * 1 = 2$
$f(3) = 3 * 3 = 9$
$f(4) = 4* 12 = 48$
$f(5) = 5*24 = 120$
$f(6) = 6 * 27 = 162$

Also, I find something that applies to the first 4 number:

The second divisor of $i$th number is $\sum_{n=1}^{i-1}f(n)$.

Example: $f(4) = 4 * \sum_{n=1}^{3}f(n) =4 * (1 + 2 + 9) = 48$

But after $48$ it doesn't work.

So after all I know one thing about the answer:

It's divisible by 7.