|
|
A274017
|
|
Number of n-bead binary necklaces (no turning over allowed) that avoid the subsequence 110.
|
|
5
|
|
|
1, 2, 3, 3, 4, 4, 6, 6, 9, 11, 16, 20, 32, 42, 65, 95, 144, 212, 330, 494, 767, 1171, 1812, 2788, 4342, 6714, 10463, 16275, 25416, 39652, 62076, 97110, 152289, 238839, 375168, 589528, 927556, 1459962, 2300349, 3626243, 5721046, 9030452, 14264310, 22542398, 35646313, 56393863, 89264836, 141358276, 223959712
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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).
The proofs of all my formulas below become evident once it is realized that A001612(n) gives the number of cyclic sequences of length n consisting of zeros and ones that avoid the pattern 001 (or equivalently, the pattern 110) provided the positions of zeros and ones on a circle are fixed. Everything else follows from the material that can be found in A001612. - Petros Hadjicostas, Jan 11 2017
|
|
LINKS
|
|
|
FORMULA
|
For all the formulas below, assume n>=1.
a(n) = 1 + A000358(n). (Notice the different offsets.)
a(n) = 1 + (1/n) * Sum_{d|n} totient(n/d)*(Fibonacci(d-1)+Fibonacci(d+1)).
a(n) = (1/n) * Sum_{d divides n} totient(n/d)*A001612(d).
G.f. = x/(1-x) + Sum_(k>=1} phi(k)/k * log(1/(1-B(x^k))) where B(x)=x*(1+x). (This is a modification of a formula due to Joerg Arndt.)
G.f. = Sum_{k>=1} phi(k)/k * log(1/C(x^k)) where C(x) = (1-x)*(1-B(x)). (End)
|
|
EXAMPLE
|
The following necklace
. 1-1
. / \
. 0 0
. | |
. 1 1
. \ /
. 0-0
contains one instance of the subsequence starting in the upper left corner. Unlike a bracelet, the necklace is oriented.
a(8) = 9: 00000000, 00000001, 00000101, 00001001, 00010001, 00010101, 00100101, 01010101, 11111111.
a(9) = 11: 000000000, 000000001, 000000101, 000001001, 000010001, 000010101, 000100101, 000101001, 001001001, 001010101, 111111111.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|