A103314 conjecture

[Mitchell Harris]

On Thu, 19 May 2005, David Wilson wrote:
>T. D. Noe wrote:
>
>> a(3^n) = 2^(3^(n - 1))
>> a(5^n) = 2^(5^(n - 1))
>> a(6^n) = a(6)^(6^(n - 1)) = 10^(6^(n - 1))
>
>(where a(n) = A103314(n)).
>
>These would all follow from my conjecture that
>
>    [1] a(n) = a(s(n))^(n/s(n))
>
>where s(n) = A007947(n) is the squarefree kernel of n (product of primes 
>dividing n).
>
>Now s(p^k) = p (k >= 1), so applying [1] gives
>
>     a(p^k) = a(p)^(p^(k-1))  (k >= 1)
>
>But a(p) = 2 for prime p (have we proved this?), and so
>
>    a(p^k) = 2^p^(k-1).
>
>Special cases p = 3 and p = 5 give your first two observations.
>
>Similarly, s(6) = 6, and so [1] gives
>
>    a(6^k) = a(6)^6^(k-1)
>
>as you observed.  But a(pq) = 2^p+2^q-2, so a(6) = 10, giving
>
>    a(6^k) = 10^6^(k-1).

What I have so far (corroborating this) is:

if p is prime, 
  f(p) = 2 (obvious, either all or none)
  f(p^n) = 2^(p^(n-1)) (there are n copies of a p-star, one vector of the 
star defines the rest so there are p^(n-1) choices, take any subset of those)

if p and q are both prime,
  then f(pq) = 2^p + 2^q - 2 (all subsets of either p or of q, but 
that over counts the 2 of all or none)
  and  f(p^n q^m) = f(pq)^{p^{n-1} q^{m-1}} (this is conjectured (it fits 
the numbers! ) but looks "right" given the resoning above and 
corresponds to the more general conjecture).

The difficulty is with numbers having more than two prime factors 
because of the possibility of interaction. The smallest such n is 
30=2*3*5. a 5-star and 3-star can be combined to make an "off" 6-star 
(Wouter's example). The "base case" of square free numbers needs 
to be worked out. 

What was the reasoning behind a(30) = 146854?

-- 
Mitch