Given: f is a function with domain and range of the positive integers, and f satisfies these two conditions:
(1) f(n+1) > f(n); that is, f is strictly increasing, and
(2) f(f(n)) = 3n
Find all possible values of f(955)
If f(1)=1, then f(f(1))=3, but then f(1)=3.
If f(1)=3, then f(f(1))=3, but then f(3)=3, and the function wouldn't be strictly increasing.
If f(1)=3+K, then f(f(1))=f(3+K)=3, but f would be decreasing.
So, the only possibility is f(1)=2. Thus, f(f(1))=f(2)=3, f(f(2))=f(3)=6, f(6)=9, f(9)=18, f(18)=27, and so on.
THere are no chances but f(4)=7 and f(5)=8, and then f(7)=12, f(8)=15, f(12)=21, f(15)=24. It seems that f(10)=19 and f(11)=20, to fill out spaces.
After many such experiments, it seems that f(x)=x+1 for x=1 to 2; x+3 for x=3 to 6; x+9 for x=9 to 18; x+27 for x=27 to 54, and so on. Thus f(x)=x+729 for x=729 to 1458, and f(955)=1684.
Steps...
