A204321 -id:A204321 - cp's OEIS frontend

A204687 a(n) = A204321(n)/4.

1, 3, 33, 4323, 74748993, 22349647743307203, 1998027016999666984531472383425633, 15968447842642350164774423714587589071259011612916483703599782377123

Offset: 1

Alexander R. Povolotsky, Jan 19 2012

nonn

Terms of the sequence can be presented as products of the following distinct prime factors: 1, 3, 3*11, 3*11*131, 3*11*131*17291, 3*11*131*17291*298995971. Alexander R. Povolotsky, Jan 19 2012

Those factors are A121897(m) for m >= 5. - Alexander R. Povolotsky, Jun 17 2012

Cf. A204321, A126604, A121897.

a(n) = product(A126604(k), k=2...n) for n >= 2.

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).

Original entry on oeis.org

0, 1, 2, 20, 374, 21313, 5115140

Offset: 0

Martin Janecke, Oct 12 2015

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.
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..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
From Wolfdieter Lang, Dec 01 2015: (Start)
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.
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} arctanh(x(k)) = log(2)/2 = arctanh(1/3), (which  is about .35).
  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.
(End)

			a(1) = 1, because there is exactly one solution: c(1) = 3.
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:
    OOOOOOOOOOOO
    OIIIIIIIIIIO
    OIIIIIIIIIIO
    OIIIIIIIIIIO
    OOOOOOOOOOOO,
  The second solution yields twenty-four parts of each kind:
    OOOOOOOO
    OIIIIIIO
    OIIIIIIO
    OIIIIIIO
    OIIIIIIO
    OOOOOOOO.

M. Janecke, Kuchengerechtigkeit in n Dimensionen, includes a Python program that finds solutions for a given n, July 2015.
M. Janecke, On a conjecture about cutting hypercubes, Nov 2015.
Martin Janecke, All solutions for n=3
Martin Janecke, All solutions for n=4
Martin Janecke, All solutions for n=5

Cf. A204321.

cp's OEIS Frontend

A204687 a(n) = A204321(n)/4.

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