A325005 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the facets of a regular n-dimensional orthotope using up to k colors.
1, 3, 1, 6, 6, 1, 10, 21, 10, 1, 15, 55, 56, 15, 1, 21, 120, 220, 126, 21, 1, 28, 231, 680, 715, 252, 28, 1, 36, 406, 1771, 3060, 2002, 462, 36, 1, 45, 666, 4060, 10626, 11628, 5005, 792, 45, 1, 55, 1035, 8436, 31465, 53130, 38760, 11440, 1287, 55, 1
Offset: 1
Examples
Array begins with A(1,1): 1 3 6 10 15 21 28 36 45 55 ... 1 6 21 55 120 231 406 666 1035 1540 ... 1 10 56 220 680 1771 4060 8436 16215 29260 ... 1 15 126 715 3060 10626 31465 82251 194580 424270 ... 1 21 252 2002 11628 53130 201376 658008 1906884 5006386 ... 1 28 462 5005 38760 230230 1107568 4496388 15890700 50063860 ... 1 36 792 11440 116280 888030 5379616 26978328 115775100 436270780 ... 1 45 1287 24310 319770 3108105 23535820 145008513 752538150 3381098545 ... For A(1,2) = 3, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses one color for each vertex.
Links
- Robert A. Russell, Table of n, a(n) for n = 1..325
- Robin Chapman, answer to Coloring the faces of a hypercube, Math StackExchange, September 30, 2010.
Crossrefs
Programs
-
Mathematica
Table[Binomial[Binomial[d-n+2,2]+n-1,n],{d,1,11},{n,1,d}] // Flatten
Formula
A(n,k) = binomial(n + binomial(k+1,2) - 1, n).
A(n,k) = Sum_{j=1..2n} A325009(n,j) * binomial(k,j).
A(n,k) = A325004(n,k) - A325006(n,k) = (A325004(n,k) + A325007(n,k)) / 2 = A325006(n,k) + A325007(n,k).
G.f. for row n: Sum_{j=1..2n} A325009(n,j) * x^j / (1-x)^(j+1).
Linear recurrence for row n: T(n,k) = Sum_{j=0..2n} binomial(-2-j,2n-j) * T(n,k-1-j).
G.f. for column k: 1/(1-x)^binomial(k+1,2) - 1.
Comments