OFFSET
0,2
COMMENTS
The pattern in this enumeration must be contiguous (all three values next to each other in one sequence of three letters).
Because A(x) = Sum_{n>=1} a(n)*x^n = 1 - Sum_{n>=1} (phi(n)/n)*log(1-B(x^n)), where B(x) = q*x - x^3 and q = 4, we may find sequence (c(n): n>=1) that satisfies a(n) = (1/n)*Sum_{d|n} phi(n/d)*c(d) for n>=1 by using the formula Sum_{n>=1} c(n)*x^n = C(x) = x*(dB/dx)/(1-B(x)). In our case, C(x) = x*(d(q*x-x^3)/dx)/(1-(q*x-x^3)) = (q*x - 3*x^3)/(1 - q*x + x^3). This implies that c(1) = q, c(2) = q^2, c(3) = q^3 - 3, and c(n) = q*c(n-1) - c(n-3) for n>=4. This comment applies not only to this sequence, but also to sequences A274017, A274018 and A274020 as well (corresponding to cases q=2, 3, and 5, respectively). - Petros Hadjicostas, Jan 31 2018
LINKS
P. Hadjicostas and L. Zhang, On cyclic strings avoiding a pattern, Discrete Mathematics, 341 (2018), 1662-1674.
Math Stackexchange, Marko Riedel et al., Counting circular sequences.
Marko Riedel, Maple code for this sequence.
FORMULA
G.f.: 1 - Sum_{n>=1} (phi(n)/n)*log(x^(3*n)-q*x^n+1), where q=4 is the number of symbols in the alphabet we are using. - Petros Hadjicostas, Sep 12 2017
Define sequence (c(n): n>=1) by c(1) = q, c(2) = q^2, c(3) = q^3-3, and c(n) = q*c(n-1) - c(n-3) for n>=4. Then a(n) = (1/n)*Sum_{d|n} phi(n/d)*c(d) for n>=1. (Here q=4.) - Petros Hadjicostas, Jan 29 2018
EXAMPLE
The following necklace
. 1-1
. / \
. 0 0
. | |
. 1 3
. \ /
. 0-2
contains one instance of the subsequence starting in the upper left corner. Unlike a bracelet, the necklace is oriented.
CROSSREFS
KEYWORD
nonn
AUTHOR
Marko Riedel, Jun 06 2016
STATUS
approved