m|n => a(m)|a(n)

[Gerald McGarvey]

Suppose a(k*l) = a(k)*a(l) for all k and l,
then m|n => n = k*m => a(n) = a(k*m) = a(k)*a(m) and so a(m)|a(n),
so any totally multiplicative arithmetic function has this property.

-- Gerald

At 08:26 PM 4/11/2005, Christopher Hanusa wrote:

>Hello Seqfans,
>     I've just come upon a sequence that satisfies the condition "if m 
> divides n then a(m) divides a(n)".  (It's related to the number of domino 
> tilings of some sequence of regions.)
>
>Two other sequences that satisfy this property are
>0)  a(n)=n (trivially)
>1)  Fibonacci indexed where f(1)=f(2)=1
>
>Is there anything known about such sequences in general?  or maybe 
>not-so-much-in-general?  I realize that there are infinitely many 
>sequences that satisfy this --- values for a(p) for p prime are free, and 
>a(p^i q^j)=k*lcm(a(p^i q^{j-1}),a(p^{i-1} q^j)), where k is free, etc. But 
>maybe there is some theory around that can help me learn more.  How might 
>one go about finding a closed form?  Can you think of any other examples? 
>What do you think of when you think of this property?
>
>Thanks for any insights,
>--Chris