A103314: Conjectures for

[David Wilson]

Let a = A103314.   I conjecture that

    [1] a(n) = a(s(n))^(a(n)/s(n))

where s(n) = A007947(n) is the largest squarefree number dividing n.  If I 
knew that all zero-sum subsets of roots of 1 were gotten by set operations 
on cyclical zero-sum subsets, I could probably cob together a proof of [1].

Since s(n) is squarefree, [1] reduces the problem of computing a(n) to that 
of computing a(n) for the squarefree integers.  The squarefree numbers are 
products of distinct primes, and I suspect that expressions for a(n) for 
squarefree n can be formulated based on the number of primes in the product. 
For 0, 1, and 2 primes, we already have such expressions, namely:

    [2] a(1) = 1
    [3] a(p) = 2
    [4] a(pq) = a(pq) = 2^p+2^q-2

(here and henceforth p and q stand for primes).

[2] is of limited application.  [1] and [3] together give

    [5]  a(p^k) = 2^p^(k-1)  (k >= 1)

while [1] and [4] give

    [6]  a(p^j q^k) = (2^p+2^q-2)^(p^(j-1) q^(k-1))  (j, k >= 1).

[5] and [6] allow us to compute A103314(n) for any n with up to 2 prime 
divisors.

The remaining identies noted on A103314(n) also follow from [1]-[4].  These 
are

    [7] a(2^n) = 2^2^(n-1)

This is a special case of [5] with p = 2.

    [8] a(2n) = a(n)^2 (n even)

More succinctly, a(4n) = a(2n)^2.  To show this, let 2n = (2^k)m (k >= 1, m 
odd), giving 4m = (2^(k+1))m.  Note that s is multiplicative, so s(2n) = 
s((2^k)m) = s(2^k)s(m) = 2s(m), and similarly s(4m) = 2s(m).  Therefore

    a(2n) = a(s(2n))^(2n/s(2n)) = a(2s(m))^(2n/(2s(m))) = a(2s(m))^(n/s(m))
    a(4n) = a(s(2n))^(4n/s(4n)) = a(2s(m))^(4n/(2s(m))) = a(2s(m))^(2n/s(m))

from which a(4n) = a(2n)^2 is evident.

    [9] a(2p) = 2^p+2

follows immediately from [4] as noted.

Also, [6] gives a(2^j 3^k) = 10^(2^(j-1) 3^(k-1)) (j, k >= 1), accounting 
for the powers of 10 we see A103314.  It predicts that a(n) = 10^(n/6) when 
n is of the form 2^j 3^k (j, k >= 1).

- David W. Wilson

"Truth is just truth -- You can't have opinions about the truth."
   - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"