>>11229146We can see that K's elements are 2, 3, 5, 6, 7...
Let's start with the sequence of the exponents of 2: 2, 4, 8, ...
You can then say 1=(2^1)-1, 3=(2^2)-1, 7=(2^3)-1
In other words, 1=((k_1)^1-1), 3=((k_1)^2-1), 7=((k_1)^3-1)
Thus, f(1)=(a_1, b_1), f(3)=(a_1,b_2), f(7)=(a_1, b_3)...
Then, repeating for the sequence of 3's exponents
f(2)=(a_2, b_1), f(8)=(a_2,b_2), f(26)=(a_2, b_3)...
And so forth.