cp's OEIS Frontend

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Zero together with the partial sum of the even terms of A016116. - Omar E. Pol, Sep 14 2021

For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 1. - David desJardins, Oct 27 2022

Column 2 of A209727.

From Richard Locke Peterson, Apr 26 2020: (Start)

The formula a(n) = 2*a(n-2)-1 also fits empirically. With the given initial numbers a(1)=3, a(2)=4, a(3)=5, this new formula implies the old empirical formula. (But it is not established that the old empirical formula is true, so it is not established that the new formula is true either.) Furthermore, if the initial numbers had somehow, for example, been 3,4,6 instead, the new formula no longer implies the old formula.

If the new formula actually is true, it follows that a(n) is the number of distinct integer triangles that can be formed with sides of length a(n-1) and a(n-2), since the greatest length the third side can have is a(n-1)+a(n-2)-1, and the least length would be a(n-1)-a(n-2)+1. (End)

Conjectures: a(n) = A029744(n+1)+1. Also, a(n) = positions of the zeros in A309019(n+2) - A002487(n+2). - George Beck, Mar 26 2022

Column 3 of A209727.

Column 4 of A209727.

Number of edges along the boundary of the graph G(n) described in A342759.

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple incremental action, bundle existing output as an integrated product (which requires p=-4 steps), or implement the current bundled action (which requires q=5 steps). The first use of a novel bundle erases (or makes obsolete) all prior actions. How large an output can be generated in n time steps?"

1. A production scheme with replacement R(p,q) eliminates existing output followinging a bundling action, while an additive scheme A(p,q) retains the output. The schemes correspond according to A(p,q)=R(p-q,q), with the replacement scheme serving as the default presentation.

2. This problem is structurally similar to the Copy and Paste Keyboard problem: Existing sequences (A178715 and A193286) should be regarded as Paradigm-Shift Sequences with production schemes R(1,1) and R(2,1) with replacement, respectively.

3. The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 2.

4. All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

5. The paradigm shift sequence for the R(-4,5) scheme is also the solution to the R(-2,4) scheme.

From John P. McSorley, Aug 24 2010: (Start)

A combinatorial interpretation of this triangle is as follows:

Ignore the first column of 1's of the above triangle and the call the (n,k) entry of the new triangle formed RE(n,k).

Hence row 8 of the 'RE(n,k)' triangle is 1 4 3 6 3 4 1 1.

Now see sequence A180171 for the definition of a k-reverse of n.

Briefly, a k-reverse of n is a k-composition of n which is cyclically equivalent to its reverse.

Sequence A180171 is the 'R(n,k)' triangle read by rows where R(n,k) is the total number of k-reverses of n.

Then RE(n,k) is the number of k-reverses of n up to cyclic equivalence.

In sequence A180171 we have R(8,3)=9 because there are 9 3-reverses of 8.

In cyclically equivalent classes: {116,611,161} {224,422,242}, and {233,323,332}; since there are 3 such classes we have RE(8,3)=3.

Similarly, in A180171, we have R(8,6)=21 because all 21 6-compositions of 8 are 6-reverses of 8, but they come in 4 cyclically equivalent classes (with representatives 111113, 111122, 111212, and 112112) hence RE(8,6)=4.

There is another (equivalent) interpretation for RE(n,k) involving k-subsets of Z_n, the integers modulo n, and the multiplier -1. See the McSorley/Schoen paper below for more details.

In this case it is convenient to count k-subsets up to dihedral equivalence, rather than cyclic equivalence.

The counts are the same. The row sums of the 'RE(n,k)' triangle give sequence A052955.

(End)

From Petros Hadjicostas, Oct 12 2017: (Start)

When 1 <= k <= n, each cyclically equivalence class of k-reverses of n is a "Sommerville symmetrical cyclic composition," which was introduced by Sommerville (1909). On pp. 301-304 of his paper, he proves that the number of such (equivalence classes of) compositions of n with length k is exactly T(n,k) = RE(n,k).

The equivalence class of a Sommerville symmetrical cyclic composition contains at least one palindromic composition (type I) or a composition that becomes a palindromic composition if we remove the first part (type II). A composition with only one part is a palindromic composition of both types. Hadjicostas and Zhang (2017) have proved that each equivalence class of k-reverses of n contains exactly two compositions that are either of type I or type II (except in the case when k | n and all the parts are the same).

For example, consider the case with n=8 and k=3, where RE(8,3)=3. As pointed above by J. P. McSorley, in cyclically equivalent classes we have {116,611,161} {224,422,242}, and {233,323,332}. The first class contains one composition of type I (161) and one of type II (611); the second class contains one composition of type I (242) and one of type II (422); and the last class contains one composition of type I (323) and one of type II (233).

When n = 6 and k = 4, the class of 4-reverses {1221, 2211, 2112, 1122} contains two compositions of type I (1221 and 2112).

If A is a set of positive integers and 1 <= k <= n, let RE_A(n,k) be the total number of Sommerville symmetrical cyclic compositions of n with length k and parts only in A (= number of cyclically equivalence classes of k-reverses of n with parts only in A). Then the g.f. of RE_A(n,k) is Sum_{n,k >= 1} RE_A(n,k) * x^n * y^k = (-1/2) + (1 + y * f_A(x))^2/(2 * (1 - y^2 * f_A(x^2)), where f_A(x) = Sum_{m in A} x^m. (For this sequence, A = all positive integers.)

Sequence A292200 contains the total number of Sommerville symmetrical cyclic compositions of n that are Carlitz (compositions that have length one, or have length >= 1 and adjacent parts of the composition on a circle are distinct).

(End)