cp's OEIS Frontend

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

Also the minimum number of transitive triples in a tournament on n nodes, i.e., a(n) = C(n,3) - A006918(n-2). - Leen Droogendijk, Nov 10 2014

a(n) = the number of binary strings of length n+1 with exactly one pair of adjacent 0's and exactly two pairs of adjacent 1's. - Jeremy Dover, Jul 07 2016

It appears that the row sums give the positive integers of A001629. - Omar E. Pol, Jul 09 2016