Thanks for the comment, I will revise the kata tomorrow.

Answer to the first part:

1. Thanks for pointing to the mistake. The edge case should P and S = 0. There should be no cases where P is some number, while S is 0.

Answer to the second part of the comment:

1. It is quite typical in the electrical industry to have various sub-functions to calculate different parameters seperately. In this kata, the seperate function is not required all calulcations can be done in one function.
2. It should not be matter if the PF is 0.8499 or 0.84999 it is still below 0.85 p.u., therefore, a load should be stored in the list.

If I am not wrong, Unnamed was using a mathematical approach and it levarages the periodicity of the sequence in terms of odd and even terms. While most of us is using iterative approach.The provided solution is more efficient for larger values.

Regarding you question:

1. First part (n mod 6)
   If you know the parity of two numbers, you also know the parity of their sum. So the sequence is defined as: f(n) = f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)
   Initial values: f(0)=0, f(1)=1, f(2)=1, f(3)=2, f(4)=4. If you compute the parity, you will be able to see a pattern (i.e.odd, odd, even, even, even, even, odd, odd, even, even, even, even,...). So you will be able to notice the pattern repeats every 6 terms.

2. The list [0,1,2,2,2,2],
   The above pattern (section 1) gives us the cumulative count of odd numbers.
   For n, there are (n/6) complete segments of 6 terms. Therefore, 2x(n/6) gives the number of odd terms in these complete segments.
   The remainder, n mod 6, tells us where in the 6-term segment we are. The list [0,1,2,2,2,2] gives the cumulative count of odd numbers up to that point in the cycle.

In summary, the sequence has a repeating pattern of parity due its linear definition and the properties of of parity over addition. Therefore, this repeating pattern allowed him, Unnamed, to use a mathematical approach.