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Column 4 of A209727.

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

A stopping problem: begin with n and at each stage if even divide by 2 or if odd subtract 1. That is, iterate A029578 while nonzero.

From Peter Kagey, Jul 16 2015: (Start)

The number of appearances of n in this sequence is identically A000045(n). Proof:

By application of the formula,

"a(0) = 0, a(2*n+1) = a(2*n) + 1 and a(2*n) = a(n) + 1",

it can be seen that:

{i: a(i) = n} = {2*i: a(i) = n-1, n>0} U {2*i+1: a(i) = n-2, n>1}.

Because the two sets on the left hand side share no elements:

|{i: a(i) = n}| = |{i: a(i) = n-1, n>0}| + |{i: a(i) = n-2, n>1}|

Notice that

|{i : a(i) = 1}| = |{1}| = 1 = A000045(1) and

|{i : a(i) = 2}| = |{2}| = 1 = A000045(2).

Therefore the number of appearances of n in this sequence is A000045(n). (End)

The folds are always made so the longer side becomes the shorter side.

We could have counted not only the holes but also all the notches: 4, 6, 9, 15, 25, 45, 81, 153, 289, ... which has the formula a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) and appears to match the sequence A183978. - Philippe Gibone, Jul 06 2016

The same sequence (0,0,1,3,9,21,49,...) turns up when you start with an isosceles right triangular piece of paper and repeatedly fold it in half, snipping corners as you go. Is there an easy way to see why the two questions have the same answer? - James Propp, Jul 05 2016

Reply from Tom Karzes, Jul 05 2016: (Start)

This case seems a little more complicated than the rectangular case, since with the triangle you alternate between horizontal/vertical folds vs. diagonal folds, and the resulting fold pattern is more complex, but I think the basic argument is essentially the same.

Note that with the triangle, the first hole doesn't appear until after you've made 3 folds, so if you start counting at zero folds, you have three leading zeros in the sequence: 0,0,0,1,3,9,21,... (End)

Also the number of subsets of {1,2,...,n} that contain both even and odd numbers. For example, a(3)=3 and the 3 subsets are {1,2}, {2,3}, {1,2,3}; a(4)=9 and the 9 subsets are {1,2}, {1,4}, {2,3}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}. (See comments in A052551 for the number of subsets of {1,2,...,n} that contain only odd and even numbers.) - Enrique Navarrete, Mar 26 2018

Also the number of integer compositions of n + 1 with an odd part other than the first or last. The complementary compositions are counted by A052955(n>0) = A027383(n) + 1. - Gus Wiseman, Feb 05 2022

Also the number of unit squares in the (n+1)-st iteration in the version of the dragon curve where the rotation directions alternate, so that any clockwise rotation is followed by a counterclockwise rotation, and vice versa (see image link below). - Talmon Silver, May 09 2023

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)

We define a pattern to be a finite sequence covering an initial interval of positive integers. Patterns are counted by A000670 and ranked by A333217.

We define a sequence to be weakly alternating if it is alternately weakly increasing and weakly decreasing, starting with either.

In other words, number of steps to reach 1 starting from n and using the process: x -> x-1 if n is odd and x -> x/2 otherwise.

a(n) = number of 0's + twice number of 1's (disregarding the leading digit 1) in the binary expansion of n, i.e., A007088(n). - Lekraj Beedassy, May 28 2010

From Daniel Forgues, Jul 31 2012: (Start)

For the binary Fibonacci rabbits sequence (A036299) (cf. OEIS Wiki link below) we have the substitution/concatenation rule: a(n), n >= 3, may be obtained by the concatenation of a(n-1) and a(n-2), with a(1) = 0, a(2) = 1. Thus, using . (dot) as the concatenation operator, we have the recursive substitution/concatenation

a(n) = a(n-0)

a(n) = a(n-1).a(n-2)

a(n) = a(n-2).a(n-3).a(n-3).a(n-4)

a(n) = a(n-3).a(n-4).a(n-4).a(n-5).a(n-4).a(n-5).a(n-5).a(n-6)

which suggests the sequence

{0}

{1, 2}

{2, 3, 3, 4}

{3, 4, 4, 5, 4, 5, 5, 6}

whose concatenation gives A014701 (this sequence).

Number of multiplications to compute n-th power by the Chandah-sutra method, also called left-to-right binary exponentiation:

x^1 = x^( 1_2) = (x) (0 prod)

x^2 = x^( 10_2) = (x^2) (1 prod)

x^3 = x^( 11_2) = (x^2) * (x) (2 prod)

x^4 = x^( 100_2) = (x^2)^2 (2 prod)

x^5 = x^( 101_2) = (x^2)^2 * (x) (3 prod)

x^6 = x^( 110_2) = (x^2)^2 * (x^2) (3 prod)

x^7 = x^( 111_2) = (x^2)^2 * (x^2) * (x) (4 prod)

x^8 = x^(1000_2) = ((x^2)^2)^2 (3 prod) (End)

From Ya-Ping Lu, Mar 03 2021: (Start)

Index at which record m occurs is A052955(m).

First appearance of m in the sequence (or the record value m) is at n = 2^(m/2 + 1) - 1 for even m, and at n = 3*2^((m - 1)/2) - 1 for odd m.

The last appearance of m in the sequence is at n = 2^m. (End)

a(n) is the digit sum of n-1 in bijective base-2. Since the Fibonacci number F(m) can be defined as the number of ways to compose m as the sum of 1s and 2s, we get that m appears F(m) times in the sequence. - Oscar Cunningham, Apr 14 2024

Conjecture: a(n+1) is the minimal number of steps to go from 0 to n, by choosing before each step, after the first step, whether to keep the same step length or double it. The initial step length is 1. - Jean-Marc Rebert, May 15 2025

Interleaving of A000079 without initial 1 and A007283.

Agrees from a(2) onward with A145751 for all terms listed there (up to 65536). Apparently equal to 2, 3 followed by A090989. Equals 2 followed by A163978.

Binomial transform is A000129 without first two terms, second binomial transform is A020727, third binomial transform is A164033, fourth binomial transform is A164034, fifth binomial transform is A164035.

Number of achiral necklaces or bracelets with n beads using up to 2 colors. For n=5, the eight achiral necklaces or bracelets are AAAAA, AAAAB, AAABB, AABAB, AABBB, ABABB, ABBBB, and BBBBB. - Robert A. Russell, Sep 22 2018