cp's OEIS Frontend

a(floor(n/2)) = A002378(n).

Sum of the even numbers among the smallest parts in the partitions of 2n into two parts (see example). - Wesley Ivan Hurt, Jul 25 2014

For n > 0, a(n-1) is the sum of the smallest parts of the partitions of 2n into two distinct even parts. - Wesley Ivan Hurt, Dec 06 2017

Also, number of 4-element subsets of the set {1,...,n-1} whose elements sum up to an odd integer, i.e., 4th column of the triangle A159916, cf. there. - M. F. Hasler, May 01 2009

Also, if the offset is changed to 3, so that a(3)=2, a(n) = number of non-equivalent (mod D_3) ways to place 2 indistinguishable points on a triangular grid of side n so that they are not adjacent. - Heinrich Ludwig, Mar 23 2014

Also, the number of binary strings of length n with exactly one pair of consecutive 0s and exactly three pairs of consecutive 1s. - Jeremy Dover, Jul 07 2016

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 = 5, and (according to C. G. Bower) a(n) = a_{k=5}(n) is the number of reversible non-palindromic compositions of n with 5 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 is such a composition of n (with b_i >= 1), then it is equivalent to the composition n = 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[5] 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 4 black balls and n-5 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 = b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 4 black balls and n-5 white balls that are mirror images of each other.

Hence, for n >= 2, a(n) = a_{k=5}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 4 black balls and n-k = n-5 white balls. (Clearly, a(n) = a_{k=5}(n) > 0 only for n >= 6. For n=5, the composition 1+1+1+1+1, which corresponds to string BBBB, is discarded because it is palindromic.)

For k = 3 (an odd integer) we have a_k(n) = A002620(n-2) (for n >= 4), while for k = 7 (also an odd integer), we have a_k(n) = A032093(n) (for n >= 8).

For k = 4 (which is even), we have a_k(n) = A006584(n-2) (for n >= 5), while for k = 6 and k = 8 (which are also even integers), we get sequences A032092 and A032094, respectively. When k is even, the g.f. in these cases is A_k(x) = (C(x)^k - C(x^2)^(k/2))/2, where C(x) = x/(1-x). The formula for a_k(n) (given above) needs to be modified as well.

(End)

The formula for a(n) for this sequence was Ralf Stephan's conjecture 73. It was solved by Elizabeth Wilmer (see Proposition 2 in one of the links below). There is a minor typo in the original conjecture. - Petros Hadjicostas, Jul 04 2018

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)

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)

In other words, the number of 2-dissections of an n-gon modulo the cyclic action.

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)

Antidiagonal sums: (2,4,10,16,28,40,60,80,...)=A006584 (conjectured).

Diagonals: A007590, A000217.

a(n) is the number of binary Lyndon words of length n+3 having 3 blocks of 0's, see Math.SE. - Andrey Zabolotskiy, Nov 16 2021

Multiplication of the lower triangular matrix A157898 from the left by A000012 means: these are partial column sums of A157898.

The same as A006584 (apart from the offset). - R. J. Mathar, Jan 15 2017

There are two cases: n is odd and n is even.

Let n be an odd integer and n > 3, the sum of 3 integers is odd when all of them are odd or one is odd and the others are even. Number of ways to choose 3 odd numbers: C((n+1)/2, 3). Number of ways to choose 2 even numbers and 1 odd: C((n-1)/2, 2)*C((n+1)/2, 1). Total number of ways: C((n+1)/2, 3) + C((n-1)/2, 2)*C((n+1)/2,1).

Let n be an even integer and n > 3. Number of ways to choose 3 odd numbers: C(n/2, 3). Number of ways to choose 2 even numbers and 1 odd: C(n/2, 2)*C(n/2, 1). Total number of ways: C(n/2, 3) + C(n/2, 2)*C(n/2, 1).

Take a chessboard of n X n unit squares in which the a1 square is black. a(n) is the number of composite squares having white unit squares on their vertices. For the number of composite squares having black unit squares on their vertices see A005993. - Ivan N. Ianakiev, Aug 19 2018