cp's OEIS Frontend

Guenther Rosenbaum showed that the sequence represents the optimal number of guesses in the static Mastermind game with two pegs. Namely, the optimal number of static guesses equals 2k, if the number of colors is either (3k - 1) or 3k and is (2k + 1), if the number of colors is (3k + 1), k >= 1. - Alex Bogomolny, Mar 06 2002

First differences are in A011655. - R. J. Mathar, Mar 19 2008

a(n+1) is the maximum number of wins by a team in a sequence of n basketball games if the team's longest winning streak is 2 games. See example below. In general, floor(k(n+1)/(k+1)) gives the maximum number of wins in n games when the longest winning streak is of length k. - Dennis P. Walsh, Apr 18 2012

Sum_{n>=2} 1/a(n)^k = Sum_{j>=1} Sum_{i=1..2} 1/(i*j)^k = Zeta(k)^2 - Zeta(k)*Zeta(k,3), where Zeta(,) is the generalized Riemann zeta function, for the case k=2 this sum is 5*Pi^2/24. - Enrique Pérez Herrero, Jun 25 2012

a(n) is the pattern of (0+2k, 0+2k, 1+2k), k>=0. a(n) is also the number of odd integers divisible by 3 in ]2(n-1)^2, 2n^2[. - Ralf Steiner, Jun 25 2017

a(n) is also the total domination number of the n-triangular (Johnson) graph for n > 2. - Eric W. Weisstein, Apr 09 2018

a(n) is the maximum total domination number of connected graphs with order n>2. The extremal graphs are "brushes", as defined in the links below. - Allan Bickle, Dec 24 2021

a(n) is the minimal number of ascending or descending staircase walks necessary to cover a chessboard of size n-1, for n > 1. See Ackerman and Pinchasi. - Sela Fried, Jan 16 2023

Watanabe's tag system {00, 1110} maps a word w over {0,1} to w', where if w begins with 0, w' is obtained by appending 00 to w and deleting the first three letters, or if w begins with 1, w' is obtained by appending 1110 to w and deleting the first three letters.

The empty word is included in the count.

Comment from Don Reble, Aug 25 2017: (Start)

The following comment applies to both the 3-shift tag systems {00,1110} (A291068) and {00,0111} (A291069). Number the bits in a binary word w starting at the left with bit 0. For the trajectory of w under the tag system, only bits numbered 0,3,6,9,... are important, the others (the unimportant bits) having no effect on the outcome.

An important 1 bit produces 0111 or 1110, and exactly one of those new 1 bits is important. The number of important 1's never changes. So the number of initial words of length n that terminate (the analog of A289670) is just 2^(number-of-unimportant-bits) = 2^(floor(2*n/3)) = A291778.

The number that end in a cycle is 2^n - 2^(floor(2*n/3)) = A291779.

Furthermore, the number of important zeros is eventually bounded.

Proof. If a word has A important zeros and B important ones, then after A+B steps, there will be at most 2A+4B bits, and at most (2A+4B+2)/3 important bits. B of them are important ones, so at most (2A+B+2)/3 are important zeros.

If A >= B+3, then (2A+B+2)/3 <= (2A+A-1)/3 < A. If A < B+3, then (2A+B+2)/3 < (3B+8)/3 = B+2. The first kind must shrink; the second kind can't grow past A+B+2. QED

Ultimately, a word with B important ones has at most A+B+2 important bits, so can't diverge. So the word "finite" in the definition was unnecessary and has been omitted. (End)

Watanabe's tag system {00, 0111} maps a word w over {0,1} to w', where if w begins with 0, w' is obtained by appending 00 to w and deleting the first three letters, or if w begins with 1, w' is obtained by appending 0111 to w and deleting the first three letters.

The empty word is included in the count.

Comment from Don Reble, Aug 25 2017: (Start)

The following comment applies to both the 3-shift tag systems {00,1110} (A291068) and {00,0111} (A291069). Number the bits in a binary word w starting at the left with bit 0. For the trajectory of w under the tag system, only bits numbered 0,3,6,9,... are important, the others (the unimportant bits) having no effect on the outcome.

An important 1 bit produces 0111 or 1110, and exactly one of those new 1 bits is important. The number of important 1's never changes. So the number of initial words of length n that terminate (the analog of A289670) is just 2^(number-of-unimportant-bits) = 2^(floor(2*n/3)) = A291778.

The number that end in a cycle is 2^n - 2^(floor(2*n/3)) = A291779.

Furthermore, the number of important zeros is eventually bounded.

Proof. If a word has A important zeros and B important ones, then after A+B steps, there will be at most 2A+4B bits, and at most (2A+4B+2)/3 important bits. B of them are important ones, so at most (2A+B+2)/3 are important zeros.

If A >= B+3, then (2A+B+2)/3 <= (2A+A-1)/3 < A. If A < B+3, then (2A+B+2)/3 < (3B+8)/3 = B+2. The first kind must shrink; the second kind can't grow past A+B+2. QED

Ultimately, a word with B important ones has at most A+B+2 important bits, so can't diverge. So the word "finite" in the definition was unnecessary and has been omitted. (End)