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A136704 Number of Lyndon words on {1,2,3} with an odd number of 1's and an odd number of 2's. +0
2
0, 1, 2, 5, 12, 30, 78, 205, 546, 1476, 4026, 11070, 30660, 85410, 239144, 672605, 1899120, 5380830, 15292914, 43584804, 124527988, 356602950, 1023295422, 2941974270, 8472886092, 24441017580, 70607383938 (list; graph; listen)
OFFSET

1,3

COMMENT

1) 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. 2) Also, a Lyndon word is the aperiodic necklace representative which is lexicographically smallest 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

E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.

M. Lothaire, Combinatorics on Words, Addison-Wesley, Reading, MA, 1983.

F. Ruskey and J. Sawada, "An Efficient Algorithm for Generating Necklaces with Fixed Density", SIAM J. Computing, 29 (1999) 671-684.

LINKS

F. Ruskey, Counting Necklaces

M. Zabrocki, MATH5020 York University Course Website

FORMULA

a(1)=0; for n>1, if n=odd then a(n)= sum(mu(d)*3^(n/d))/(4n); d|n. If n=even, then a(n)= sum(mu(d)*(3^(n/d)-1))/(4n); d|n, d odd.

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.

CROSSREFS

Cf. A006575; A027376; A133267; A136703.

Sequence in context: A024851 A145267 A103287 this_sequence A120895 A101785 A003089

Adjacent sequences: A136701 A136702 A136703 this_sequence A136705 A136706 A136707

KEYWORD

nonn

AUTHOR

Jennifer Woodcock (jennifer.woodcock(AT)ugdsb.on.ca), Jan 16 2008

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Last modified December 16 13:01 EST 2009. Contains 170825 sequences.


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