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Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. The chiral colorings of its facets come in pairs, each the reflection of the other.

Also the number of chiral pairs of colorings of the vertices of a regular n-dimensional orthoplex using up to k colors.

Let p = ({b_1},{b_2},...,{b_m}) be an ordered set partition of [n] into m blocks for some m, 1<=m<=n. A descent in p is an ordered pair (x,y) in [n]X[n] such that x is in b_i, y is in b_j, iy.

T(n,binomial(n,2)) = 1 (counts the ordered set partition ({n},{n-1},...,{2},{1})).

For n>=1, T(n,0) = 2^(n-1).

Sum_{k>=0} T(n,k)*2^k = A289545(n).

Sum_{k>=0} T(n,k)*3^k = A347841(n).

Sum_{k>=0} T(n,k)*4^k = A347842(n).

Sum_{k>=0} T(n,k)*5^k = A347843(n).

Sum_{k>=0} T(n,k)*6^k = A385408(n).

Sum_{k>=0} T(n,k)*7^k = A347844(n).

Sum_{k>=0} T(n,k)*8^k = A347845(n).

Sum_{k>=0} T(n,k)*9^k = A347846(n).

T(n,k) is the number of preferential arrangements of n labeled elements with exactly k inversions. For example, there 4 preferential rearrangements of length 3 with 1 inversion: 132, 213, 212, 131. - Kyle Celano, Aug 18 2025

Reversing the string does not leave it unchanged. Only one string from each pair is counted.

Equivalently, the number of nonequivalent strings up to reversal that are not palindromes.

Except for the first term, column k is the "BHK" (reversible, identity, unlabeled) transform of k,0,0,0,... [Corrected by Petros Hadjicostas, Jul 01 2018]

From Petros Hadjicostas, Jul 01 2018: (Start)

Consider the input sequence (c_k(n): n >= 1) with g.f. C_k(x) = Sum_{n>=1} c_k(n)*x^n. Let a_k(n) = BHK(c_k(n): n >= 1) be the output sequence under Bower's BHK transform. It can be proved that the g.f. of BHK(c_k(n): n >= 1) is A_k(x) = (C_k(x)^2 - C_k(x^2))/(2*(1-C_k(x))*(1-C_k(x^2))) + C_k(x). (See the comments for sequences A032096, A032097, and A032098.)

For column k of this two-dimensional array, the input sequence is defined by c_k(1) = k and c_k(n) = 0 for n >= 1. Thus, C_k(x) = k*x, and hence the g.f. of column k (with the term C_k(x) = k*x excluded) is (C_k(x)^2 - C_k(x^2))/(2*(1-C_k(x))*(1-C_k(x^2))) = (1/2)*(k - 1)*k*x^2/((k*x^2 - 1)*(k*x - 1)), from which we can easily prove Howroyd's formula.

(End)

Comment from Bahman Ahmadi, Aug 05 2019: (Start)

We give an alternative definition for the square array A(n,k) = T(n,k) with n >= 2 and k >= 0. A(n,k) is the number of inequivalent "distinguishing colorings" of the path on n vertices using at most k colors. The rows are indexed by n, the number of vertices of the path, and the columns are indexed by k, the number of permissible colors.

A vertex-coloring of a graph G is called "distinguishing" if it is only preserved by the identity automorphism of G. This notion is considered in the context of "symmetry breaking" of simple (finite or infinite) graphs. Two vertex-colorings of a graph are called "equivalent" if there is an automorphism of the graph which preserves the colors of the vertices. Given a graph G, we use the notation Phi_k(G) to denote the number of inequivalent distinguishing colorings of G with at most k colors. This sequence gives A(n,k) = Phi_k(P_n), i.e., the number of inequivalent distinguishing colorings of the path P_n on n vertices with at most k colors.

For n=3, we can color the vertices of P_3 with at most 2 colors in 3 ways such that all the colorings distinguish the graph (i.e., no non-identity automorphism of the path P_3 preserves the coloring) and that all the three colorings are inequivalent.

We have Phi_k(P_n) = binomial(k,2)*k^(n-2) + k*Phi_k(P_(n-2)) for n >= 4; Phi_k(P_2) = binomial(k,2); Phi_k(P_3) = k*binomial(k,2).

(End)

Also called cross polytope and hyperoctahedron. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is an octahedron with eight triangular faces. For n=4, the figure is a 16-cell with sixteen tetrahedral facets. The Schläfli symbol, {3,...,3,4}, of the regular n-dimensional orthoplex (n>1) consists of n-2 threes followed by a four. Each of its 2^n facets is an (n-1)-dimensional simplex. The chiral colorings of its facets come in pairs, each the reflection of the other.

Also the number of chiral pairs of colorings of the vertices of a regular n-dimensional orthotope (cube) using up to k colors.

Row n has length 1 + binomial(n,2) and sum A000110(n) (a Bell number).

For n >= 1, a(n) is the number of chains of binary reflexive symmetric matrices of order n.

The number of chains of strictly upper triangular or strictly lower triangular matrices.

Also, number of chains in power set of (n^2-n)/2 elements.

a(n) is the number of distinct reflexive symmetric fuzzy matrices of order n.

For the denominators see A264389.

This gives the numerators of the rational numbers r(n) = s(1,n), where s(h,k) = Sum_{r=1..(k-1)} (r/k)*(h*r/k - floor(h*r/k)- 1/2), k >=1, are the Dedekind sums. See the Apostol reference, pp. 52, 61-69, 72-73, and the Weisstein link, where GCD(h,k) = 1 is assumed.

s(h,k) = Sum_{r = 1..k} ((r/k))*((h*r/k)) with ((x)) = x - floor(x) - 1/2 if x is not an integer, else 0.

s(h,k) = (Sum_{r=1..(k-1)} cot(Pi*h*r/k)*cot(Pi*r/k))/(4*k), k >= 1, r and h integers. Exercise 11, p. 72 of the Apostol reference.

6*n*s(1,n) = binomial(n-1, 2) = A161680(n-1), n >= 1.