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

Author

Paul D. Hanna, Jun 24 2003

Keywords

• nonn

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

Crossrefs

Cf. A000108, A039969, A085297, A220466.

Programs

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 *)

PARI

A085296(n) = if(n%2,3,3*(1+A085296(n/2))); \\ Antti Karttunen, Nov 01 2018

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)