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A051841
Number of binary Lyndon words with an even number of 1's.
15
1, 0, 1, 1, 3, 4, 9, 14, 28, 48, 93, 165, 315, 576, 1091, 2032, 3855, 7252, 13797, 26163, 49929, 95232, 182361, 349350, 671088, 1290240, 2485504, 4792905, 9256395, 17894588, 34636833, 67106816, 130150493, 252641280, 490853403, 954429840, 1857283155, 3616800768, 7048151355, 13743869130, 26817356775
OFFSET
1,5
COMMENTS
Also number of trace 0 irreducible polynomials over GF(2).
Also number of trace 0 Lyndon words over GF(2).
REFERENCES
May, Robert M. "Simple mathematical models with very complicated dynamics." Nature, Vol. 261, June 10, 1976, pp. 459-467; reprinted in The Theory of Chaotic Attractors, pp. 85-93. Springer, New York, NY, 2004. The sequences listed in Table 2 are A000079, A027375, A000031, A001037, A000048, A051841. - N. J. A. Sloane, Mar 17 2019
FORMULA
a(n) = 1/(2*n)*Sum_{d|n} gcd(d,2)*mu(d)*2^(n/d).
a(n) ~ 2^(n-1) / n. - Vaclav Kotesovec, May 31 2019
From Richard L. Ollerton, May 10 2021: (Start)
a(n) = 1/(2*n)*Sum_{k=1..n} gcd(gcd(n,k),2)*mu(gcd(n,k))*2^(n/gcd(n,k))/phi(n/gcd(n,k)).
a(n) = (1/n)*Sum_{k=1..n} gcd(n/gcd(n,k),2)*mu(n/gcd(n,k))*2^gcd(n,k)/phi(n/gcd(n,k)). (End)
EXAMPLE
a(5) = 3 = |{ 00011, 00101, 01111 }|.
MATHEMATICA
a[n_] := Sum[GCD[d, 2]*MoebiusMu[d]*2^(n/d), {d, Divisors[n]}]/(2n);
Table[a[n], {n, 1, 32}]
(* Jean-François Alcover, May 14 2012, from formula *)
PROG
(PARI)
L(n, k) = sumdiv(gcd(n, k), d, moebius(d) * binomial(n/d, k/d) );
a(n) = sum(k=0, n, if( (n+k)%2==0, L(n, k), 0 ) ) / n;
vector(33, n, a(n))
/* Joerg Arndt, Jun 28 2012 */
(Haskell)
a051841 n = (sum $ zipWith (\u v -> gcd 2 u * a008683 u * 2 ^ v)
ds $ reverse ds) `div` (2 * n) where ds = a027750_row n
-- Reinhard Zumkeller, Mar 17 2013
CROSSREFS
Same as A001037 - A000048. Same as A042980 + A042979.
Cf. A000010.
Sequence in context: A002823 A109509 A006053 * A096081 A054162 A174783
KEYWORD
nonn,easy,nice
AUTHOR
Frank Ruskey, Dec 13 1999
STATUS
approved