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

%I A050181 #26 Sep 17 2019 03:49:39
%S A050181 0,1,3,9,30,99,333,1144,3978,13995,49742,178296,643842,2340135,
%T A050181 8554275,31429026,115997970,429874830,1598952366,5967382200,
%U A050181 22338765540,83859016527,315614844558,1190680751376,4501802223090,17055399281284
%N A050181 T(2n+3, n), array T as in A051168; a count of Lyndon words.
%H A050181 <a href="/index/Lu#Lyndon">Index entries for sequences related to Lyndon words</a>
%F A050181 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 A050181 From _Petros Hadjicostas_, Nov 16 2017: (Start)
%F A050181 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 A050181 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 A050181 Using the above formulae, one can verify _R. J. Mathar_'s conjecture above.
%F A050181 (End)
%p A050181 A050181 := proc(n)
%p A050181     A051168(2*n+3,n) ;
%p A050181 end proc: # _R. J. Mathar_, Jul 20 2016
%t A050181 a[n_] := (1/(2n+3)) Sum[MoebiusMu[d] Binomial[(2n+3)/d, n/d], {d, Divisors[ GCD[n, 3]]}];
%t A050181 a /@ Range[0, 25] (* _Jean-François Alcover_, Sep 17 2019, from PARI *)
%o A050181 (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 A050181 Cf. A003441.
%Y A050181 A diagonal of the square array described in A051168.
%K A050181 nonn
%O A050181 0,3
%A A050181 _Clark Kimberling_