# Problem 483: Repeated permutation
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 ≈
5.166666667e0g(5) = 2081/120 ≈ 1.734166667e1g(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.