cp's OEIS Frontend

a(n) is also the number of median-reflective (palindrome) symmetric patterns in a top-down equilateral triangular arrangement of closely packed black and white cells satisfying the local matching rule of Pascal's triangle modulo 2, where n is the number of cells in each edge of the arrangement. The matching rule is such that any elementary top-down triangle of three neighboring cells in the arrangement contains either one or three white cells.

The number of possibilities for an n-game (sub)set of tennis with neither player gaining a 2-game advantage. (Motivated by the marathon Isner-Mahut match at Wimbledon, 2010.) - Barry Cipra, Jun 28 2010

Number of achiral rows of n colors using up to two colors. For a(3)=4, the rows are AAA, ABA, BAB, and BBB. - Robert A. Russell, Nov 07 2018

Also the number of walks of length n on the graph x--y--z starting at y. - Sean A. Irvine, May 30 2025

From Petros Hadjicostas, Jul 07 2018: (Start)

Column k of this array is the "BIK" (reversible, indistinct, unlabeled) transform of k,0,0,0,....

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) = BIK(c_k(n): n >= 1) be the output sequence under Bower's BIK transform. It can proved that the g.f. of BIK(c_k(n): n >= 1) is A_k(x) = (1/2)*(C_k(x)/(1-C_k(x)) + (C_k(x^2) + C_k(x))/(1-C_k(x^2))). (See the comments for sequence A001224.)

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 is (1/2)*(C_k(x)/(1-C_k(x)) + (C_k(x^2) + C_k(x))/(1-C_k(x^2))) = (1/2)*(k*x/(1-k*x) + (k*x^2 + k*x)/(1-k*x^2)) = (2 + (1-k)*x - 2*k*x^2)*k*x/(2*(1-k*x^2)*(1-k*x)).

Using the first form the g.f. above and the expansion 1/(1-y) = 1 + y + y^2 + ..., we can easily prove J.-F. Alcover's formula T(n,k) = (k^n + k^((n + mod(n,2))/2))/2.

(End)

Number of palindromes of length n using a maximum of four different symbols.

Number of achiral rows of n colors using up to four colors. - Robert A. Russell, Nov 09 2018

Interleaving of A000302 and 4*A000302.

Unsigned version of A141125.

Binomial transform is A164907. Second binomial transform is A164908. Third binomial transform is A057651. Fourth binomial transform is A016129.

One followed by powers of 3 with positive exponent, repeated. - Omar E. Pol, Jul 27 2009

Number of achiral rows of n colors using up to three colors. E.g., for a(3) = 9, the rows are AAA, ABA, ACA, BAB, BBB, BCB, CAC, CBC, and CCC. - Robert A. Russell, Nov 07 2018

Number of achiral rows of n colors using up to five colors. For a(3) = 25, the rows are AAA, ABA, ACA, ADA, AEA, BAB, BBB, BCB, BDB, BEB, CAC, CBC, CCC, CDC, CEC, DAD, DBD, DCD, DDD, DED, EAE, EBE, ECE, EDE, and EEE. - Robert A. Russell, Nov 09 2018

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)

Number of achiral rows of length n using up to six different colors. For a(3) = 36, the rows are AAA, ABA, ACA, ADA, AEA, AFA, BAB, BBB, BCB, BDB, BEB, BFB, CAC, CBC, CCC, CDC, CEC, CFC, DAD, DBD, DCD, DDD, DED, DFD, EAE, EBE, ECE, EDE, EEE, EFE, FAF, FBF, FCF, FDF, FEF, and FFF. - Robert A. Russell, Nov 08 2018

Also: a(n) is the number of palindromes with n digits using a maximum of six different symbols. - David A. Corneth, Nov 09 2018