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A263207 Number of integer solutions for Product_{k=1..n}(c(k) + 1) = 2 * Product_{k=1..n}(c(k) - 1) with 1 < c(k) <= c(k+1).

%I A263207 #16 Nov 30 2024 21:02:20
%S A263207 0,1,2,20,374,21313,5115140
%N A263207 Number of integer solutions for Product_{k=1..n}(c(k) + 1) = 2 * Product_{k=1..n}(c(k) - 1) with 1 < c(k) <= c(k+1).
%C A263207 Geometrically interpreted, the sequence a(n) provides the number of distinct ways to cut an n-dimensional 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 k-th dimension.
%C A263207 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.
%C A263207 For any n>=1 the solution given by c(k)=A204321(k) for k=1..n-1 and c(n)=A204321(n)-1 always exists. Conjecture: there is no solution with a greater c(n). - _Martin Janecke_, Dec 01 2015
%C A263207 From _Wolfdieter Lang_, Dec 01 2015: (Start)
%C A263207 In terms of the j-th 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)^(n-j))*sigma(n, j) = 0, with sigma(n, 0) = 1.
%C A263207 If one uses the reciprocals x(k) = 1/c(k) then
%C A263207   0 < x(k+1) <= x(k) <  1 (the c's are all >1, and finite), and the original equation becomes (by taking logarithms)
%C A263207   Sum_{k=1..n} arctanh(x(k)) = log(2)/2 = arctanh(1/3), (which  is about .35).
%C A263207   Here only positive terms appear. It is clear from the monotony of arctanh that all the x(k) < 1/3 for n >= 2, hence c(1) >= 4 for n >= 2.
%C A263207 (End)
%H A263207 M. Janecke, <a href="https://prlbr.de/2015/mehrdimensional-kuchen-schneiden/">Kuchengerechtigkeit in n Dimensionen</a>, includes a Python program that finds solutions for a given n, July 2015.
%H A263207 M. Janecke, <a href="https://prlbr.de/2015/hypercube-cutting-supremum-proof-attempt/">On a conjecture about cutting hypercubes</a>, Nov 2015.
%H A263207 Martin Janecke, <a href="/A263207/a263207.txt">All solutions for n=3</a>
%H A263207 Martin Janecke, <a href="/A263207/a263207_1.txt">All solutions for n=4</a>
%H A263207 Martin Janecke, <a href="/A263207/a263207_2.txt">All solutions for n=5</a>
%e A263207 a(1) = 1, because there is exactly one solution: c(1) = 3.
%e A263207 a(2) = 2, because there are two solutions: c(1) = 4, c(2) = 11 and c(1) = 5, c(2) = 7. The first solution can be illustrated geometrically as a square cut into thirty outer and equally many inner parts:
%e A263207     OOOOOOOOOOOO
%e A263207     OIIIIIIIIIIO
%e A263207     OIIIIIIIIIIO
%e A263207     OIIIIIIIIIIO
%e A263207     OOOOOOOOOOOO,
%e A263207   The second solution yields twenty-four parts of each kind:
%e A263207     OOOOOOOO
%e A263207     OIIIIIIO
%e A263207     OIIIIIIO
%e A263207     OIIIIIIO
%e A263207     OIIIIIIO
%e A263207     OOOOOOOO.
%Y A263207 Cf. A204321.
%K A263207 nonn,more
%O A263207 0,3
%A A263207 _Martin Janecke_, Oct 12 2015