%I
%S 1,1,2,5,2,0,0,3,2,2,2,4,4,4,0,3,0,0,0,0,0,0,0,0,0,0,2,2,2,4,4,1,0,0,
%T 0,4,4,4,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,0,0,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,2,2,2,4,4,4,0,0,0,4,4,4,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,4,4,4,2,2,2,0,0,0,2,2,2,4
%N Catalan numbers mod 6.
%C The only odd terms are those with indices n = 2^k  1 (k = 0, 1, 2, 3, ...); see also A038003.
%C It is conjectured that the only k which yield a(2^k1) = 1 are k = 0, 1 and 5. Are there other k than 2 and 8 that yield a(2^k1) = 5? Otherwise said, is a(2^k1) = 3 for all k > 8?
%C The question is equivalent to: does 2^k  1 always contain a digit 2 when converted into base 3 for all k > 8? Similar conjecture has been proposed for 2^k, see A004642.  _Jianing Song_, Sep 04 2018
%H M. Alekseyev, <a href="http://home.gwu.edu/~maxal/gpscripts/">PARI/GP Scripts for Miscellaneous Math Problems</a>, sect. III: Binomial coefficients modulo integers, binomod.gp (v.1.4, 11/2015).
%H V. Reshetnikov, <a href="http://list.seqfan.eu/pipermail/seqfan/2015November/015578.html">A000108(n) ≡ 1 (mod 6)</a>, SeqFan list, Nov. 8, 2015.
%F a(n) = A000108(n) mod 6.
%o (PARI) a(n)=binomial(2*n,n)/(n+1)%6
%o (PARI) A259667(n)=lift(if(n%3!=1,binomod(2*n+1,n,6)/(2*n+1), if(bittest(n,0),binomod(2*n,n1,6)/n,binomod(2*n,n,6)/(n+1)))) \\ using binomod.gp by M. Alekseyev, cf. Links.
%Y Cf. A000108, A004642, A038003.
%K nonn
%O 0,3
%A _M. F. Hasler_, Nov 08 2015
