cp's OEIS Frontend

a(n) is 1/6 the number of colorings of a 2 X 2 hexagonal array with n+2 colors. - R. H. Hardin, Feb 23 2002

a(n) is the sum of all numbers that cannot be written as t*(n+1) + u*(n+2) for nonnegative integers t,u (see Schuh). - Floor van Lamoen, Oct 09 2002

a(n) is the total number of rectangles (including squares) contained in a stepped pyramid of n rows (or of base 2n-1) of squares. A stepped pyramid of squares of base 2*6 - 1 = 11, for instance, has the following vertices:

..........X.X

........X.X.X.X

......X.X.X.X.X.X

....X.X.X.X.X.X.X.X

..X.X.X.X.X.X.X.X.X.X

X.X.X.X.X.X.X.X.X.X.X.X

X.X.X.X.X.X.X.X.X.X.X.X - Lekraj Beedassy, Sep 02 2003

Partial sums of A002412. - Jonathan Vos Post, Mar 16 2006

a(n) equals -1 times the coefficient of x^3 of the characteristic polynomial of the (n + 2) X (n + 2) matrix with 2's along the main diagonal and 1's everywhere else (see Mathematica code below). - John M. Campbell, May 28 2011

a(n) is the n-th antidiagonal sum of the convolution array A213750. - Clark Kimberling, Jun 20 2012

Convolution of A000027 with A000384 (excluding 0). - Bruno Berselli, Dec 06 2012

The sequence is the binomial transform of (1, 7, 15, 13, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 31 2015

Also the number of 3-cycles in the (n+2)-triangular graph. - Eric W. Weisstein, Aug 14 2017

A general formula for polygonal numbers is P(n,k) = (n-2)*(k-1)*k/2 + k, where P(n,k) is the k-th n-gonal number.

The triangle sums, see A180662 for their definitions, link this square array read by antidiagonals with twelve different sequences, see the crossrefs. Most triangle sums are linear sums of shifted combinations of a sequence, see e.g. A189374. - Johannes W. Meijer, Apr 29 2011

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

A(n,k) is the number of non-isomorphic set multipartitions (multiset of sets) with k parts each part has size n.

Fourth column (m=3) of triangle A060098.

Partial sums of A038163.

Equals the tetrahedral numbers, [1, 4, 10, 20, ...] convolved with the aerated triangular numbers, [1, 0, 3, 0, 6, 0, 10, ...]. [Gary W. Adamson, Jun 11 2009]

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is (1/(1-x*z/((1-z^2)*(1-z))))/(1-z).

Row sums give A052534. Column sequences (without leading zeros) give A000012 (powers of 1), A002620(n+1), A002624, A060099-A060101 for m=0..5.

The bisections of the column sequences give triangles A060102 and A060556.

Riordan array (1/(1-x), x/((1-x)*(1-x^2))). - Paul Barry, Mar 28 2011

Convolution of A006918 with A001399, or of A002625 with A059841 (A000035 if offsets are respected),

or of A038163 with A022003 or of A057524 with A027656 or of A014125 with the aerated version of A000217,

or of A002624 with A103221, or of A002623 with A008731, or of other combinations of splitting the signature -/3,3,1 into two components.

If we apply the enumeration of Molien series as described in A139672,

this is row 45=9*5 of a table of values related to Molien series, i.e., the

product of the sequence on row 9 (A006918) with the sequence on row 5 (A001399).

This is associated with the root system E6, and can be described using the additive function on the affine E6 diagram:

1

|

2

|

1--2--3--2--1

With a(0) = 0 nontrivial integer solutions of (x + y)^3 = (x - y)^4. If x = a(n) then y = a(n + (-1)^n). - Thomas Scheuerle, Mar 22 2023

Equals triangle A152205 as an infinite lower triangular matrix * the triangular numbers: [1, 3, 6, ...]. - Gary W. Adamson, Feb 14 2010

a(n) is the number of partitions of n into four kinds of parts 1 and two kinds of parts 2. - Joerg Arndt, Mar 09 2016

This is row 21 of a table of values related to Molien series. It is the product of the sequence on row 3 (A008619) with the sequence on row 7 (A001400).

This table may be constructed by moving the rows of table A008284 to prime locations and generating the composite locations by multiplication in a manner similar to the calculation illustrated in the present sequence.

Rows 1 thru 20 and 22 thru 25 are as follows:

A000012 A008619 A000027 A001399 A002620 A001400 A000217 A006918 A000601 A001401 A002623 A001402 A002621 A097701 A000292 A008630 A002624 A008631 A057524 A002622 A008632 A001752 A117485.