%I #17 Feb 11 2021 01:41:32
%S 3,9,11,27,29,35,39,81,83,89,93,107,111,117,119,243,245,251,255,269,
%T 273,279,281,323,327,333,335,351,353,359,363,729,731,737,741,755,759,
%U 765,767,809,813,819,821,837,839,845,849,971,975,981,983,999,1001,1007,1011
%N Numbers k such that Sum_{j=1..k} Catalan(j) == 2 (mod 3).
%H Y. More, <a href="http://www.jstor.org/stable/30037533">Problem 11165</a>, Amer. Math. Monthly, 112 (2005), 568.
%p c:=n->binomial(2*n,n)/(n+1): s:=0: for n from 1 to 1500 do s:=s+c(n): a[n]:=s mod 3: od: A:=[seq(a[n],n=1..1500)]: p:=proc(n) if A[n]=2 then n else fi end: seq(p(n),n=1..1500); # _Emeric Deutsch_, Jun 12 2005
%t s0 = s2 = {}; s = 0; Do[s = Mod[s + (2 n)!/n!/(n + 1)!, 3]; Switch[ Mod[s, 3], 0, AppendTo[s0, n], 2, AppendTo[s2, n]], {n, 1055}]; s2 (* _Robert G. Wilson v_, Jun 14 2005 *)
%Y Cf. A000108, A107755, A107756.
%Y Equals A074939 - 1.
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_, Jun 11 2005
%E More terms from _Emeric Deutsch_, Jun 12 2005