More possible multiplicative functions

[David Wilson]

A066261 is multiplicative.  A066261 is the second iterate of A066260 which 
is fully multiplicative.  The iterates of fully multplicative functions are 
fully multplicative, as shown below.

Let a^k be the kth iterative of a, that is

    a^0 = I; a^(k+1) = a o a^k  (I = identity function, o = functional 
composition).

(pardon my notation if there is a better one).  Thus a^k(n) = 
a(a(a(...a(a(n))...)))  (a applied k times).

Let a be fully multplicative if a(xy) = a(x)a(y)  (whether or not gcd(x, y) 
= 1).  Fully multiplicative functions are multiplicative, with a(p^e) = 
a(p)^e.  Multiplicative functions are determined by their values on the 
prime powers, while fully multplicative functions are determined by their 
values on the primes.

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Theorem: If a is fully multiplicative, then a^k (k >= 0) is fully 
multiplicative.

Proof:

Base case:  a^0 is fully multiplicative.

Proof: a^0 = I, the identity function, which is fully multiplicative with 
I(xy) = I(x)I(y) = xy.

Recursive case: If a^k is fully multplicative, then so is a^(k+1).

Proof.  Let a^k be fully multplicative.  We then have

    a^(k+1)(xy)
   = a(a^k(xy))   (definition of a^(k+1))
   = a(a^k(x) a^k(y))  (a^k is fully multiplicative)
   = a(a^k(x)) a(a^k(y))  (a is fully multiplicative)
   = a^(k+1)(x) a^(k+1)(y)  (definition of a^(k+1))

and a^(k+1) is fully multplicative, QED.

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This theorem does not generalize to multiplicative functions.  It might be 
nice to not fully multplicative sequences in the EIS.