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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

%I

%S 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,

%T 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,

%U 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

%N Runs of zeros in Catalan sequence modulo 3: consecutive occurrences of binomial(2*k,k)/(k+1) == 0 (mod 3).

%C 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.

%H Antti Karttunen, <a href="/A085296/b085296.txt">Table of n, a(n) for n = 1..16384</a>

%H R. Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences ...</a>

%H R. Stephan, <a href="/A079944/a079944.ps">Table of generating functions</a>

%F a(2*n-1) = 3, a(2*n) = 3*(a(n)+1), for n >= 1.

%F a(n) = (9 * 3^A007814(n) - 1) / 2 - 1. - _Ralf Stephan_, Oct 10 2003

%F From _Johannes W. Meijer_, Feb 11 2013: (Start)

%F a((2*n-1)*2^p) = (3^(p+2)-3)/2, p >= 0 and n >= 1. Observe that a(2^p) = A029858(p+2).

%F 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)

%p 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

%t Map[If[First@ # == 0, Length@ #, Nothing] &, SplitBy[Array[Mod[CatalanNumber@ #, 3] &, 10^4], # == 0 &]] (* _Michael De Vlieger_, Nov 02 2018 *)

%o (PARI) A085296(n) = if(n%2,3,3*(1+A085296(n/2))); \\ _Antti Karttunen_, Nov 01 2018

%Y Cf. A000108, A039969, A085297, A220466.

%K nonn

%O 1,1

%A _Paul D. Hanna_, Jun 24 2003

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Last modified June 25 12:55 EDT 2019. Contains 324352 sequences. (Running on oeis4.)