Let N be the set of natural numbers {1, 2, 3, ... }. Let Z be the integers. Define d **:** N → Z by d(1) = 0, d(p) = 1 for p prime, and d(mn) = m d(n) + n d(m) for any integers m, n. Determine d(n) in terms of the prime factors of n. Find all n such that d(n) = n. Define d_{1}(m) = d(m) and d_{n+1}(m) = d(d_{n}(m)). Find lim_{n→∞} d_{n}(63).

**Solution**

d(p^{a}) = a p^{a-1} by a trivial induction on a. Hence for n = p^{a}q^{b} ... we have d(n) = n ( a/p + b/q + ... ) by a trivial induction on the number of primes.

Hence d(n) = n iff a/p + b/q + ... = 1. Multiplying through by all the prime denominators gives integral terms. All but the first are clearly divisible by p, so the first must be also. Hence a/p = 1 and b/q etc are zero. In other words, n = p^{p} for some prime p.

We find d_{1}(63) = 51, d_{2}(63) = 20, d_{3}(63) = 24, d_{4}(63) = 44, d_{5}(63) = 48. Now suppose n is divisible by 16. Then n = 2^{k}p^{a}q^{b} ... , where p, q, ... are all odd and k >= 4. Hence d(n) = n( 4/2 + a/p + b/q + ... ). Now all of 2n, na/p, nb/q, ... are integral and multiples of 16. So d(n) is at least twice n and a multiple of 16. So if we have d_{k}(m) divisible by 16, then d_{h+k}(m) ≥ 2^{h} which diverges. Hence d_{k}(63) tends to infinity.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002