cp's OEIS Frontend

Triangle T(n, k) is irregularly shaped: 1 <= k <= A239438(n). First row corresponds to n = 1.

The maximal number of points that can be placed on a triangular grid of side n so that no two of them are adjacent is given by A239438(n).

Without the restriction "non-equivalent (mod D_3)" numbers are given by A239567.

If the offset is changed to 3, this is the 2nd Witt transform of A000217 [Moree]. - R. J. Mathar, Nov 08 2008

From Petros Hadjicostas, May 19 2018: (Start)

Let k be an integer >= 2. The g.f. of the BHK[k] transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n>=1} c(n)*x^n, is A_k(x) = (C(x)^k - C(x^2)^(k/2))/2 if k is even, and A_k(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. This follows easily from the formulae in C. G. Bower's web link below about transforms.

When k is even and c(n) = 1 for all n >= 1, we get C(x) = x/(1-x) and A_k(x) = (1/2)*((x/(1-x))^k - (x^2/(1-x^2))^{k/2}). If (a_k(n): n >= 1) is the output sequence (with g.f. A_k(x)), then it can be proved (using Taylor expansions) that a_k(n) = (1/2)*(binomial(n-1, n-k) - binomial((n/2)-1, (n-k)/2)) for even n >= k+1 and a_k(n) = (1/2)*binomial(n-1, n-k) for odd n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)

In this sequence, k = 6, and (according to C. G. Bower) a(n) = a_{k=6}(n) is the number of reversible non-palindromic compositions of n with 6 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 is such a composition of n (with b_i >= 1), then it is equivalent to the composition n = b_6 + b_5 + b_4 + b_3 + b_2 + b_1, and each equivalent class has two elements because here linear palindromes are not allowed as compositions of n.

The fact that we are finding the BHK[6] transform of 1, 1, 1, ... means that each part of each composition of n can have exactly one color (see Bower's link below about transforms).

In each such composition replace each b_i with one black (B) ball followed by b_i - 1 white (W) balls. Then drop the first black (B) ball. We then get a reversible non-palindromic string of length n-1 that has 5 black balls and n-6 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 = b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 5 black balls and n-6 white balls that are mirror images of each other.

Hence, for n >= 2, a(n) = a_{k=6}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 5 black balls and n-k = n-6 white balls. (Clearly, a(n) = a_{k=6}(n) > 0 only for n >= 7.)

(End)

If the offset is changed to 3, this is the 2nd Witt transform of A000292 [Moree]. - R. J. Mathar, Nov 08 2008

Also 7th column of A159916, i.e., number of 7-element subsets of {1,...,n-1} whose elements add up to an odd integer. - M. F. Hasler, May 02 2009

From Petros Hadjicostas, May 19 2018: (Start)

Let k be an integer >= 2. The g.f. of the BHK[k] transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n>=1} c(n)*x^n, is A_k(x) = (C(x)^k - C(x^2)^(k/2))/2 if k is even, and A_k(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. This follows easily from the formulae in C. G. Bower's web link below about transforms.

When k is even and c(n) = 1 for all n >= 1, we get C(x) = x/(1-x) and A_k(x) = (1/2)*((x/(1-x))^k - (x^2/(1-x^2))^{k/2}). If (a_k(n): n >= 1) is the output sequence (with g.f. A_k(x)), then it can be proved (using Taylor expansions) that a_k(n) = (1/2)*(binomial(n-1, n-k) - binomial((n/2)-1, (n-k)/2)) for even n >= k+1 and a_k(n) = (1/2)*binomial(n-1, n-k) for odd n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)

In this sequence, k = 8, and (according to C. G. Bower) a(n) = a_{k=8}(n) is the number of reversible non-palindromic compositions of n with 8 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 + b_8 is such a composition of n (with b_i >= 1), then it is equivalent to the composition n = b_8 + b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, and each equivalent class has two elements because here linear palindromes are not allowed as compositions of n.

The fact that we are finding the BHK[8] transform of 1, 1, 1, ... means that each part of each composition of n can have exactly one color (see Bower's link below about transforms).

In each such composition replace each b_i with one black (B) ball followed by b_i - 1 white (W) balls. Then drop the first black (B) ball. We then get a reversible non-palindromic string of length n-1 that has k-1 = 7 black balls and n-k = n-8 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 + b_8 = b_8 + b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 7 black balls and n-8 white balls that are mirror images of each other.

Hence, for n >= 2, a(n) = a_{k=8}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 7 black balls and n-k = n-8 white balls. (Clearly, a(n) = a_{k=8}(n) > 0 only for n >= 9.)

(End)

Rotations and reflections of placements are not counted. If they are to be counted see A239569.

Rotations and reflections of placements are not counted. If they are to be counted see A239570.

Rotations and reflections of placements are not counted. If they are to be counted see A239571.

From Petros Hadjicostas, May 19 2018: (Start)

Let k be an integer >= 2. The g.f. of the BHK[k] transform of the sequence (c(n): n>=1), with g.f. C(x) = Sum_{n>=1} c(n)*x^n, is A_k(x) = (C(x)^k - C(x^2)^(k/2))/2 if k is even, and A_k(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. This follows easily from the formulae in C. G. Bower's web link below about transforms.

When k is odd and c(n) = 1 for all n>=1, we get C(x) = x/(1-x) and A_k(x) = (1/2)*(x/(1-x))*((x/(1-x))^{k-1} - (x^2/(1-x^2))^{(k-1)/2}). If (a_k(n): n>=1) is the output sequence (with g.f. A_k(x)), then it can be proved (using Taylor expansions) that a_k(n) = (1/2)*(binomial(n-1, n-k) - binomial(floor((n-1)/2), floor((n-k)/2))) for n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)

In this sequence, k = 7, and (according to C. G. Bower) a(n) = a_{k=7}(n) is the number of reversible non-palindromic compositions of n with 7 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 is such a composition of n (with b_i >=1), then it is equivalent to the composition n = b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, and each equivalent class has two elements because here linear palindromes are not allowed as compositions of n.

The fact that we are finding the BHK[7] transform of 1, 1, 1, ... means that each part of each composition of n can have exactly one color (see Bower's link below about transforms).

In each such composition replace each b_i with one black (B) ball followed by b_i - 1 white (W) balls. Then drop the first black (B) ball. We then get a reversible non-palindromic string of length n-1 that has 6 black balls and n-7 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 = b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 6 black balls and n-7 white balls that are mirror images of each other.

Hence, for n>=2, a(n) = a_{k=7}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 6 black balls and n-k = n-7 white balls. (Clearly, a(n) = a_{k=7}(n) > 0 only for n >= 8. For n=7, the composition 1+1+1+1+1+1+1, which corresponds to string BBBBBB, is discarded because it is palindromic.)

(End)

Rotations and reflections of placements are not counted. For numbers if they are to be counted see A282998.

Also number of linear unbranched n-4-catafusenes of C_{2v} symmetry.

Number of n-bead black-white reversible strings with k black beads; also binary grids; string is not palindromic. - Yosu Yurramendi, Aug 08 2008

The first seven columns are A004526, A002620, A006584, A032091, A032092, A032093, A032094. Row sums give essentially A032085. - Yosu Yurramendi, Aug 08 2008

From Álvar Ibeas, Jun 01 2020: (Start)

T(n, k) is the sum of odd-degree coefficients of the Gaussian polynomial [n, k]_q. The area below a NE lattice path between (0,0) and (k, n-k) is even for A034851(n, k) paths and odd for T(n, k) of them.

For a (non-reversible) string of k black and n-k white beads, consider the minimum number of bead transpositions needed to place the black ones to the left and the white ones to the right (in other words, the number of inversions of the permutation obtained by labeling the black beads by integers 1,...,k and the white ones by k+1,...,n, in the same order they take on the string). It is even for A034851(n, k) strings and odd for T(n, k) cases.

(End)