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Numbers k such that 3 divides Sum_{j=1..k} binomial(2*j,j).
8

%I #19 Jan 01 2022 08:11:27

%S 0,12,30,36,84,90,108,120,246,252,270,282,324,336,354,360,732,738,756,

%T 768,810,822,840,846,972,984,1002,1008,1056,1062,1080,1092,2190,2196,

%U 2214,2226,2268,2280,2298,2304,2430,2442,2460,2466,2514,2520,2538,2550

%N Numbers k such that 3 divides Sum_{j=1..k} binomial(2*j,j).

%C Apparently a(n)/3 (mod 3) = A010060(n-1), the Thue-Morse sequence.

%F It appears that sequence gives k such that the coefficient of x^k equals 1 in Product_{j>=1} 1-x^(3^j).

%t Reap[ For[n = 0, n <= 3000, n = n + 3, If[ Divisible[ Sum[ Binomial[2 k, k], {k, 1, n}], 3], Print[n]; Sow[n]]]][[2, 1]] (* _Jean-François Alcover_, Jul 01 2013 *)

%t Join[{0},Position[Accumulate[Table[Binomial[2k,k],{k,2600}]],_?( Divisible[ #,3]&)]//Flatten] (* _Harvey P. Dale_, Mar 14 2020 *)

%Y Cf. A066796, A083097, A081601.

%K nonn,easy

%O 1,2

%A _Benoit Cloitre_, Apr 22 2003