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 A050181 T(2n+3, n), array T as in A051168; a count of Lyndon words. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 LINKS 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) MAPLE A050181 := proc(n)     A051168(2*n+3, n) ; end proc: # R. J. Mathar, Jul 20 2016 PROG (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 CROSSREFS Cf. A003441. A diagonal of the square array described in A051168. Sequence in context: A089978 A052906 A102898 * A275690 A089931 A148946 Adjacent sequences:  A050178 A050179 A050180 * A050182 A050183 A050184 KEYWORD nonn AUTHOR STATUS approved

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Last modified October 24 01:08 EDT 2018. Contains 316541 sequences. (Running on oeis4.)