Eccoci.

We define a permutation as an operation that rearranges the order of the elements $\{1, 2, 3, ..., n\}$. There are $n!$ such permutations, one of which leaves the elements in their initial order. For $n = 3$ we have $3! = 6$ permutations:

• $P_1 =$ keep the initial order
• $P_2 =$ exchange the 1^st and 2^nd elements
• $P_3 =$ exchange the 1^st and 3^rd elements
• $P_4 =$ exchange the 2^nd and 3^rd elements
• $P_5 =$ rotate the elements to the right
• $P_6 =$ rotate the elements to the left

If we select one of these permutations, and we re-apply the same permutation repeatedly, we eventually restore the initial order.
For a permutation $P_i$, let $f(P_i)$ be the number of steps required to restore the initial order by applying the permutation $P_i$ repeatedly.
For $n = 3$, we obtain:

• $f(P_1) = 1$ : $(1,2,3) \to (1,2,3)$
• $f(P_2) = 2$ : $(1,2,3) \to (2,1,3) \to (1,2,3)$
• $f(P_3) = 2$ : $(1,2,3) \to (3,2,1) \to (1,2,3)$
• $f(P_4) = 2$ : $(1,2,3) \to (1,3,2) \to (1,2,3)$
• $f(P_5) = 3$ : $(1,2,3) \to (3,1,2) \to (2,3,1) \to (1,2,3)$
• $f(P_6) = 3$ : $(1,2,3) \to (2,3,1) \to (3,1,2) \to (1,2,3)$

Let $g(n)$ be the average value of $f^2(P_i)$ over all permutations $P_i$ of length $n$.
$g(3) = (1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2)/3! = 31/6 \approx 5.166666667\mathrm e0$
$g(5) = 2081/120 \approx 1.734166667\mathrm e1$
$g(20) = 12422728886023769167301/2432902008176640000 \approx 5.106136147\mathrm e3$

Find $g(350)$ and write the answer in scientific notation rounded to $10$ significant digits, using a lowercase e to separate mantissa and exponent, as in the examples above.