Geometrically interpreted, the sequence a(n) provides the number of distinct ways to cut an ndimensional cube orthogonally into equally many outer parts, i.e., those that can be seen from the outside, and inner parts. All cuts must go through the whole body. c(k) is the number of cuts for the kth dimension.
See section Example for all solutions for n=1 and n=2 and section Links for all solutions for n=3, n=4, n=5.
For any n>=1 the solution given by c(k)=A204321(k) for k=1..n1 and c(n)=A204321(n)1 always exists. Conjecture: there is no solution with a greater c(n).  Martin Janecke, Dec 01 2015
From Wolfdieter Lang, Dec 01 2015: (Start)
In terms of the jth elementary symmetric functions sigma(n, j) in the indeterminates [c(1), ..., c[n]] the equation with the products can be rewritten as Sum_{j=0..n} ((1  2*(1)^(nj))*sigma(n, j) = 0, with sigma(n, 0) = 1.
If one uses the reciprocals x(k) = 1/c(k) then
0 < x(k+1) <= x(k) < 1 (the c's are all >1, and finite), and the original equation becomes (by taking logarithms)
Sum_{k=1..n} Artanh(x(k)) = log(2)/2 = Artanh(1/3), (which is about .35).
Here only positive terms appear. It is clear from the monotony of Artanh that all the x(k) < 1/3 for n >= 2, hence c(1) >= 4 for n >= 2.
(End)
