OFFSET
1,3
COMMENTS
This sequence is also the number of Lyndon words on {1,2,3} with an even number of 1's and an odd number of 2's except that a(1) = 1 in this case.
Also, 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, n_1 = odd.
REFERENCES
M. Lothaire, Combinatorics on Words, Addison-Wesley, Reading, MA, 1983.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..1000
E. N. Gilbert and John Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.
Frank Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
Frank Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
Frank Ruskey and Joe Sawada, An Efficient Algorithm for Generating Necklaces with Fixed Density, SIAM J. Computing, 29 (1999), 671-684.
Mike Zabrocki, MATH5020 York University Course Website.
FORMULA
a(1) = 0; for n>1, if n = odd then a(n) = Sum_{d|n} (mu(d)*3^(n/d))/(4n). If n = even, then a(n) = Sum_{d|n, d odd} (mu(d)*(3^(n/d)-1))/(4n).
EXAMPLE
For n = 3, out of 8 possible Lyndon words: 112, 113, 122, 123, 132, 133, 223, 233, only 123 and 132 have an odd number of both 1's and 2's. Thus a(3) = 2.
MATHEMATICA
a[1] = 0;
a[n_] := If[OddQ[n], Sum[MoebiusMu[d] * 3^(n/d), {d, Divisors[n]}], Sum[Boole[OddQ[d]] MoebiusMu[d] * (3^(n/d)-1), {d, Divisors[n]}]]/(4n);
Array[a, 27] (* Jean-François Alcover, Aug 26 2019 *)
PROG
(PARI) a(n) = if (n==1, 0, if (n % 2, sumdiv(n, d, moebius(d)*3^(n/d))/(4*n), sumdiv(n, d, if (d%2, moebius(d)*(3^(n/d)-1)))/(4*n))); \\ Michel Marcus, Aug 26 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Jennifer Woodcock (jennifer.woodcock(AT)ugdsb.on.ca), Jan 16 2008
STATUS
approved