Find a sequence f(1), f(2), f(3), ... of non-negative integers such that 0 occurs in the sequence, all positive integers occur in the sequence infinitely often, and f( f(n163) ) = f( f(n) ) + f( f(361) ).
Solution
Take f(1) = 0, f(361) = 1. For all other n, let f(n) = f(m) if n = m163, or the smallest number not in {f(0), f(1), ... , f(n-1)} otherwise.
Now infinitely many numbers are not 163rd powers, so we certainly get every integer at least once. But if f(k) = n, then f(k163) = n also, so we get all positive integers infinitely often.
Finally, f( f(n163) ) = f( f(n) ) = f( f(n) ) + 0 = f( f(n) ) + f(1) = f( f(n) ) + f( f(361) ).
© John Scholes
jscholes@kalva.demon.co.uk
20 Sep 2002