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)
(In reply to Real numbers
by Brian Smith)
Brian said:"I noticed that if the problem had stated a domain and range of positive real numbers instead of integers, then f(n)=n*sqrt(3)."
... and I notice that, if "k" is an integer,
f(3k)= (3k) * 2 if k is odd, and
f(3k)= (3k) * 1.5 if k is even
and of course sqrt(3) is between 1.5 and 2
It's as if the discrete function given in the problem "wants to be close to" Brian's continuous function, but can't because of the integer restraints. And so the value of the function "jumps" to a nearby function. Similar to electrons in an atom jumping to the next energy ring.
As far as f(955), since 955 is not divisible by 3, the above observation doesn't help. But I'm still not sure I understand the method used to determine f(n) where n>6 and NOT divisible by 3.
Posted by Larry
on 2004-09-04 11:19:17