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 A085296 Runs of zeros in Catalan sequence modulo 3: consecutive occurrences of binomial(2*k,k)/(k+1) == 0 (mod 3). 6
 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 Antti Karttunen, Table of n, a(n) for n = 1..16384 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 MATHEMATICA Map[If[First@ # == 0, Length@ #, Nothing] &, SplitBy[Array[Mod[CatalanNumber@ #, 3] &, 10^4], # == 0 &]] (* Michael De Vlieger, Nov 02 2018 *) PROG (PARI) A085296(n) = if(n%2, 3, 3*(1+A085296(n/2))); \\ Antti Karttunen, Nov 01 2018 CROSSREFS Cf. A000108, A039969, A085297, A220466. Sequence in context: A069522 A170857 A227106 * A306364 A357819 A357821 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 August 6 06:54 EDT 2024. Contains 374960 sequences. (Running on oeis4.)