How do I prove that there is no other :$k=9,12,18$ for which this fails :$\sigma^k(114) \equiv 0\mod 6 $? let $\sigma(n)$ be the sum of divisors for a positive integer for example :
$$\sigma(6)=1+2+3+6=12$$ .
I have performed some calculations in wolfram alpha about the  sum divisors of this number: 
$q=114$ such that I got this :$\sigma(114)=240\equiv 0\mod 6 $ and $\sigma(\sigma(114))=744\equiv 0\mod 6 $ and $\sigma(\sigma(\sigma(114)))=1920\equiv 0\mod 6 ,\cdots$ .
My question here is :How do I  prove that there is no other :$k=9,12,18$ for which this fails :$$\sigma^k(114) \equiv 0\mod 6 $$ then what is the place of this number in number thoery  ?
Note (01):* I think that is the only integer fails only for $k=9,12,18$
Note (02) :$\sigma^k(114)=\sigma(\sigma(\sigma(\sigma(114\cdots))))))),k-th $
Thank you for any help .

Hint for fast calculation of sigma in a chain...
Let a number be written as
$$
\prod p^{n_p},
$$
where $p$ are prime numbers. Then we have
$$
\sigma\Big(\prod p^{n_p}\Big) = \prod \frac{p^{n_p+1}-1}{p-1}.
$$
Example:
$$
\sigma(114) = \sigma(2 \times 3 \times 19)
= \frac{2^2-1}{2-1} \frac{3^2-1}{3-1} \frac{19^2-1}{19-1}
= 3 \times 4 \times 20
= 2^4 \times 3 \times 5
$$
$$
\sigma^2(114) = \sigma(2^4 \times 3 \times 5)
= 31 \times 4 \times 6
= 2^3 \times 3 \times 31
$$
$$
\sigma^3(114) = \sigma(2^3 \times 3 \times 31)
= 15 \times 4 \times 32
= 2^7 \times 3 \times 5
$$
$$
\sigma^4(114) = 2^3 \times 3^2 \times 5 \times 17\\
\sigma^5(114) = 2^2 \times 3^4 \times 5 \times 13\\
\sigma^6(114) = 2^2 \times 3 \times 7^2 \times 11\\
\sigma^7(114) = 2^2 \times 3 \times 7^2 \times 19^2\\
\sigma^8(114) = 2^2 \times 3^2 \times 7 \times 19 \times 127\\
\sigma^9(114) = 2^{13} \times 5 \times 7 \times 13\\
$$
and $\sigma^9(114)$ is not dividable by $3$...
So-far, I see no pattern...
You could write a computer program to run the sequence and check it...
