 ## Repeated permutation Published on Sunday, 5th October 2014, 10:00 am; Solved by 164;
Difficulty rating: 100%

### Problem 483

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:
- P1 = keep the initial order
- P2 = exchange the 1st and 2nd elements
- P3 = exchange the 1st and 3rd elements
- P4 = exchange the 2nd and 3rd elements
- P5 = rotate the elements to the right
- P6 = 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 Pi, let f(Pi) be the number of steps required to restore the initial order by applying the permutation Pi repeatedly.
For n = 3, we obtain:
- f(P1) = 1 : (1,2,3) → (1,2,3)
- f(P2) = 2 : (1,2,3) → (2,1,3) → (1,2,3)
- f(P3) = 2 : (1,2,3) → (3,2,1) → (1,2,3)
- f(P4) = 2 : (1,2,3) → (1,3,2) → (1,2,3)
- f(P5) = 3 : (1,2,3) → (3,1,2) → (2,3,1) → (1,2,3)
- f(P6) = 3 : (1,2,3) → (2,3,1) → (3,1,2) → (1,2,3)

Let g(n) be the average value of f2(Pi) over all permutations Pi of length n.
g(3) = (12 + 22 + 22 + 22 + 32 + 32)/3! = 31/6 ≈ 5.166666667e0
g(5) = 2081/120 ≈ 1.734166667e1
g(20) = 12422728886023769167301/2432902008176640000 ≈ 5.106136147e3

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.