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%I #7 Feb 01 2015 09:37:50
%S 1,0,2,1,1,1,1,0,2,1,2,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,2,1,2,0,1,1,1,1,
%T 2,0,1,0,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,0
%N A014138 (= partial sums of Catalan numbers starting with 1,2,5) mod 3.
%C As usual, "mod 3" means to choose the unique representative in { 0,1,2 } of the equivalence class modulo 3Z.
%C Here the conventions of A014138 are used, but it seems somehow unnatural to start with offset 0 corresponding to the Catalan number A000108(1).
%C For m>1, the length of the m-th block of nonzero elements (and thus the approximate length of the m-th string of consecutive 1's) is given by 2 A137822(m)-1.
%F a(n) = sum( k=1..n+1, C(k) ) (mod 3), where C(k) = binomial(2k,k)/(k+1) = A000108(k).
%F a(n) = 0 <=> n+1 = 2 A137821(m) for some m.
%o (PARI) A137993(n) = lift( sum( k=1,n+1, binomial( 2*k,k )/(k+1), Mod(0,3) ))
%Y Cf. A014138, A000108, A137821-A137824, A107755, A137992, A014137(n+1) = a(n)+1 (mod 3).
%K easy,nonn
%O 0,3
%A _M. F. Hasler_, Mar 16 2008
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