A050181 T(2n+3, n), array T as in A051168; a count of Lyndon words.

0, 1, 3, 9, 30, 99, 333, 1144, 3978, 13995, 49742, 178296, 643842, 2340135, 8554275, 31429026, 115997970, 429874830, 1598952366, 5967382200, 22338765540, 83859016527, 315614844558, 1190680751376, 4501802223090, 17055399281284

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Links

• Index entries for sequences related to Lyndon words

Crossrefs

Cf. A003441.

A diagonal of the square array described in A051168.

Programs

Maple

A050181 := proc(n)
    A051168(2*n+3,n) ;
end proc: # R. J. Mathar, Jul 20 2016

Mathematica

a[n_] := (1/(2n+3)) Sum[MoebiusMu[d] Binomial[(2n+3)/d, n/d], {d, Divisors[ GCD[n, 3]]}];
a /@ Range[0, 25] (* Jean-François Alcover, Sep 17 2019, from PARI *)

PARI

a(n) = (1/(2*n+3))*sumdiv(gcd(n,3), d, moebius(d)*binomial((2*n+3)/d, n/d)); \\ Michel Marcus, Nov 18 2017

Formula

Conjecture: -(n-1)*(n+3)*(n+2)*a(n) + 2*(3*n-4)*(n+2)*(n+1)*a(n-1) - 4*n*(n+1)*(2*n-5)*a(n-2) + 2*(n-1)*(n+2)*(2*n-3)*a(n-3) - 4*(2*n-5)*(3*n-4)*(n+1)*a(n-4) + 8*n*(2*n-5)*(2*n-7)*a(n-5) = 0. - R. J. Mathar, Jul 20 2016

From Petros Hadjicostas, Nov 16 2017: (Start)

a(n) = (1/(2*n+3))*Sum_{d|gcd(n,3)} mu(d)*binomial((2*n+3)/d, n/d). (This is a special case of A. Howroyd's formula for double array A051168.)

a(n) = (1/(2*n+3))*(binomial(2*n+3, n) - binomial((2*n/3)+1, n/3)) if 3|n; = (1/(2*n+3))*binomial(2*n+3, n) otherwise.

Using the above formulae, one can verify R. J. Mathar's conjecture above.

(End)