A085296 Runs of zeros in Catalan sequence modulo 3: consecutive occurrences of binomial(2*k,k)/(k+1) == 0 (mod 3).

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

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