# [seqfan] Re: Range of one-to-one multiplicative functions

victor miller victorsmiller at gmail.com
Thu Jun 11 19:48:15 CEST 2009

```The following paper is relevant:

"Multiplicative Functions with non-decreasing normal order" by Bryan Birch
in J. London Math. Soc. v 42 (1967), pp 149-151.

In it he proves that if g : N ---> R is a non-decreasing function and |f(n)
- g(n)| <= epsilon g(n)
for all but o(X), n <=X for all X -- i.e. the exceptions are a vanishing
fraction -- and f(n) is totally multiplicative
(i.e. that f(m n) = f(m) f(n) for all non-negative integers m,n), then f(n)
= n^c for some real c.  If the range of f is
the natural number, then it's easy to see that c itself must be a positive
integer.

On Tue, Jun 9, 2009 at 6:43 PM, <franktaw at netscape.net> wrote:

> I've been looking a bit recently at what sets can be the set of values
> taken on by one-to-one multiplicative functions.  For example, it is
> easy to find multiplicative functions onto the positive integers; one
> is induced by any permutation of the primes, or by any permutation of
> the positive integers (acting on the exponents), or by any permutation
> of A050376.
>
> Of course, the kth powers for any k are obtainable, as a(n) = n^k is
> multiplicative.  Less obviously, the powers of k can be obtained as
> k^A052331(n).  We can obtain the powerful numbers with A064549(n), and
> the square-free numbers with (my recently submitted) A160102(n).
>
> (Note that without the constraint that the function be one-to-one, we
> can map onto any set containing 1: just take a(2^k) = s(k), and a(p^k)
> = 1 for any odd prime p.)
>
> I have not been able to settle the question of whether there is a
> one-to-one multiplicative function onto the non-primes (i.e., the
> composite numbers union {1}).  I suspect that the answer is no, but I
> haven't been able to prove it.  A proof (or counter-example) would be
> appreciated.
>