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)
Let the monotonicity condition be Cond. 1 and let the condition f(f(n))=3n be Cond. 2. Applying f to both sides of Cond. 2, we have
f(f(f(n)))=f(3n),
or applying Cond. 2 to that,
3f(n)=f(3n),
which we will call Cond. 3.
Let f(1)=a. Then, applying Cond. 2 and 3, we have
f(a)=3
f(3)=3a
Now, since 3<=3a, we have by Cond. 1 that a<=3. We cannot have a=3, since then we have 3=f(3)=9. Also, we cannot have a=1, since then we have f(1)=3=f(3). Thus, we have a=2, giving us the initial conditions
f(1)=2
f(2)=3
f(3)=6
Repeatedly applying Cond. 3 gives us:
f(3)=6
f(9)=18
f(27)=54
f(81)=162
f(243)=486
f(729)=1458
as well as
f(2)=3
f(6)=9
f(18)=27
f(54)=81
f(162)=243
f(486)=729
f(1458)=2187
Applying Cond. 1 to the equations f(729)=1458 and f(1458)=2187 gives us f(n)=n+729 for 729<=n<=1458, implying that f(955)=1684.
Solution
