A103314 conjecture

[David Wilson]

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))
>
> Tony

(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).