OFFSET
1,1
COMMENTS
A Lyndon word is the aperiodic necklace representative which is lexicographically earliest among its cyclic shifts. Thus we can apply the fixed density formulas: L_k(n,d)=sum L(n-d, n_1,..., n_(k-1)); n_1+...+n_(k-1)=d where L(n_0, n_1,...,n_(k-1))=(1/n)sum mu(j)*[(n/j)!/((n_0/j)!(n_1/j)!...(n_(k-1)/j)!)]; j|gcd(n_0, n_1,...,n_(k-1)). For this sequence, sum over n_0=even. Alternatively, a(n)=(sum mu(d)*3^(n/d)/n; d|n) - (sum mu(d)*(3^(n/d)-1)/(2n); d|n, d odd).
REFERENCES
M. Lothaire, Combinatorics on Words, Addison-Wesley, Reading, MA, 1983.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..650
E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.
F. Ruskey and J. Sawada, An Efficient Algorithm for Generating Necklaces with Fixed Density, SIAM J. Computing, 29 (1999) 671-684.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
M. Zabrocki, Course website
FORMULA
a(1)=2; for n>1, if n=2^k for some k, then a(n)=((3^(n/2)-1)^2)/(2n). Otherwise, if n=even then a(n)=sum mu(d)*(3^(n/d)-2*3^(n/(2d))/(2n); d|n, d odd. If n=odd then a(n)=sum mu(d)*(3^(n/d)-1)/(2n); d|n, d odd.
EXAMPLE
For n=3, out of 8 possible Lyndon words: 112, 113, 122, 123, 132, 133, 223, 233, only the first two and the last two have an even number of 1's. Thus a(3) = 4.
MAPLE
with(numtheory): a:= n-> add(mobius(d) *3^(n/d), d=divisors(n))/n -add(mobius(d) *(3^(n/d)-1), d=select(x-> irem(x, 2)=1, divisors(n)))/ (2*n): seq(a(n), n=1..30); # Alois P. Heinz, Jul 29 2011
MATHEMATICA
a[n_] := DivisorSum[n, MoebiusMu[#]*(3^(n/#) - (1/2)*Boole[OddQ[#]]*(3^(n/#)-1))&]/n; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, Mar 21 2017, after Alois P. Heinz *)
PROG
(PARI) a133267(n) = sumdiv(n, d, moebius(d)*3^(n/d)/n - if (d%2, moebius(d)*(3^(n/d)-1)/(2*n))); \\ Michel Marcus, May 17 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Jennifer Woodcock (jennifer.woodcock(AT)ugdsb.on.ca), Jan 03 2008
STATUS
approved