A050181 - OEIS

%I #26 Sep 17 2019 03:49:39

%S 0,1,3,9,30,99,333,1144,3978,13995,49742,178296,643842,2340135,

%T 8554275,31429026,115997970,429874830,1598952366,5967382200,

%U 22338765540,83859016527,315614844558,1190680751376,4501802223090,17055399281284

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

%H <a href="/index/Lu#Lyndon">Index entries for sequences related to Lyndon words</a>

%F 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

%F From _Petros Hadjicostas_, Nov 16 2017: (Start)

%F 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.)

%F 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.

%F Using the above formulae, one can verify _R. J. Mathar_'s conjecture above.

%F (End)

%p A050181 := proc(n)

%p     A051168(2*n+3,n) ;

%p end proc: # _R. J. Mathar_, Jul 20 2016

%t a[n_] := (1/(2n+3)) Sum[MoebiusMu[d] Binomial[(2n+3)/d, n/d], {d, Divisors[ GCD[n, 3]]}];

%t a /@ Range[0, 25] (* _Jean-François Alcover_, Sep 17 2019, from PARI *)

%o (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

%Y Cf. A003441.

%Y A diagonal of the square array described in A051168.

%K nonn

%O 0,3

%A _Clark Kimberling_