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 A085296 Runs of zeros in Catalan sequence modulo 3: consecutive occurrences of binomial(2*k,k)/(k+1) (Mod 3) = 0. 5
 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 363, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 1092, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 363, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 3279, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS When we prepend a '1' to the Catalan sequence modulo 3, the only nonzero digit strings are {1,1,1,2,2,2} and {2,2,2,1,1,1}; see A085297 for the occurrences of these digit strings. LINKS R. Stephan, Some divide-and-conquer sequences ... R. Stephan, Table of generating functions FORMULA a(2*n-1) = 3, a(2*n) = 3*(a(n)+1), for n >= 1. a(n) = (9 * 3^A007814(n) - 1) / 2 - 1. - Ralf Stephan, Oct 10 2003 From Johannes W. Meijer, Feb 11 2013: (Start) a((2*n-1)*2^p) = (3^(p+2)-3)/2, p >= 0 and n >= 1. Observe that a(2^p) = A029858(p+2). a(2^(p+3)*n + 2^(p+2) - 1) = a(2^(p+2)*n + 2^(p+1) - 1) for p >= 0 and n >= 1. (End) MAPLE nmax:=79: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (3^(p+2)-3)/2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 11 2013 CROSSREFS Cf. A000108, A039969, A085297, A220466. Sequence in context: A069522 A170857 A227106 * A214401 A009781 A266913 Adjacent sequences:  A085293 A085294 A085295 * A085297 A085298 A085299 KEYWORD nonn AUTHOR Paul D. Hanna, Jun 24 2003 STATUS approved

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Last modified October 23 20:52 EDT 2018. Contains 316530 sequences. (Running on oeis4.)