Multiplicative

[Mitchell Harris]

On Sun, 22 May 2005, David Wilson wrote:

>A023153 through A023161 are all multiplicative.
>
>Basically, if gcd(x, y) = 1, we can use the Chinese remainder theorem to 
>induce a bijection between pairs of cycles modulo x and y to cycles modulo 
>xy.  This means a(xy) = a(x)a(y), and a is multiplicative.  This argument 
>works for cycles of any injection f: Z(n) -> Z(n), including f(n) = n^k, as 
>in these sequences.
>
>a(p^k) is difficult. 

There's the formula for A023153 given by Pieter Moree, but I can't 
figure out what ord_d(2) is or how to use it for the other sequences.

Mitch