login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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
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.
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
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 24 2003
STATUS
approved