A293500 Number of orientable strings of length n using a maximum of k colors, array read by descending antidiagonals, T(n,k) for n >= 1 and k >= 1.

0, 0, 0, 0, 1, 0, 0, 3, 2, 0, 0, 6, 9, 6, 0, 0, 10, 24, 36, 12, 0, 0, 15, 50, 120, 108, 28, 0, 0, 21, 90, 300, 480, 351, 56, 0, 0, 28, 147, 630, 1500, 2016, 1053, 120, 0, 0, 36, 224, 1176, 3780, 7750, 8064, 3240, 240, 0, 0, 45, 324, 2016, 8232, 23220, 38750, 32640, 9720, 496, 0

Offset: 1

Author

Andrew Howroyd, Oct 10 2017

Keywords

• nonn
• tabl
• easy

Comments

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)

Examples

			Array begins:
======================================================
n\k| 1   2    3     4      5      6       7       8
---|--------------------------------------------------
1  | 0   0    0     0      0      0       0       0...
2  | 0   1    3     6     10     15      21      28...
3  | 0   2    9    24     50     90     147     224...
4  | 0   6   36   120    300    630    1176    2016...
5  | 0  12  108   480   1500   3780    8232   16128...
6  | 0  28  351  2016   7750  23220   58653  130816...
7  | 0  56 1053  8064  38750 139320  410571 1046528...
8  | 0 120 3240 32640 195000 839160 2881200 8386560...
...
For T(4,2)=6, the chiral pairs are AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, and BABB-BBAB.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1275
• B. Ahmadi, F. Alinaghipour and M. H. Shekarriz, Number of Distinguishing Colorings and Partitions, arXiv:1910.12102 [math.CO], 2019.
• C. G. Bower, Transforms (2).

Crossrefs

Columns 2-5 for n > 1 are A032085, A032086, A032087, A032088.

Column 6 is A320524.

Rows 2-6 are A161680, A006002(n-1), A083374, A321672, A085744.

Cf. A277504, A293496.

Cf. A003992 (oriented), A277504 (unoriented), A321391 (achiral).

Programs

Mathematica

Table[Function[n, (k^n - k^(Ceiling[n/2]))/2][m - k + 1], {m, 11}, {k, m, 1, -1}] // Flatten (* Michael De Vlieger, Oct 11 2017 *)

PARI

T(n,k) = (k^n - k^(ceil(n/2)))/2;

Formula

T(n,k) = (k^n - k^(ceiling(n/2)))/2.

G.f. for column k: (1/2)*(k - 1)*k*x^2/((k*x^2 - 1)*(k*x - 1)). - Petros Hadjicostas, Jul 07 2018

From Robert A. Russell, Nov 16 2018: (Start)

T(n,k) = (A003992(k,n) - A321391(n,k)) / 2.

T(n,k) = = A003992(k,n) - A277504(n,k) = A277504(n,k) - A321391(n,k).

G.f. for row n: (Sum_{j=0..n} S2(n,j)*j!*x^j/(1-x)^(j+1) - Sum_{j=0..ceiling(n/2)} S2(ceiling(n/2),j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.

G.f. for row n>1: x * Sum_{k=1..n-1} A145883(n,k) * x^k / (1-x)^(n+1).

E.g.f. for row n: (Sum_{k=0..n} S2(n,k)*x^k - Sum_{k=0..ceiling(n/2)} S2(ceiling(n/2),k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.

T(0,k) = T(1,k) = 0; T(2,k) = binomial(k,2); for n>2, T(n,k) = k*(T(n-3,k)+T(n-2,k)-k*T(n-1,k)).

For k>n, T(n,k) = Sum_{j=1..n+1} -binomial(j-n-2,j) * T(n,k-j). (End)