%I #32 Jul 23 2017 21:55:12

%S 0,3,9,12,27,30,36,39,81,84,90,93,108,111,117,120,243,246,252,255,270,

%T 273,279,282,324,327,333,336,351,354,360,363,729,732,738,741,756,759,

%U 765,768,810,813,819,822,837,840,846,849,972,975,981,984,999,1002,1008,1011

%N Numbers n such that 3 does not divide Sum_{k=0..n} binomial(2k,k) = A006134(n).

%C Apparently a(n)/3 mod 2 = A010060(n-1), the Thue-Morse sequence.

%C a(n+1) is the smallest number with exactly n+1 partitions into distinct powers of 2 or of 3: A131996(a(n+1)) = n+1 and A131996(m) < n+1 for m < a(n+1). - _Reinhard Zumkeller_, Aug 06 2007

%H R. Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences ...</a>

%H R. Stephan, <a href="/A079944/a079944.ps">Table of generating functions</a>

%F Apparently a(n) = 3*A005836(n).

%F G.f.: (x/(1 - x))*Sum_{k>=0} 3^(k+1)*x^(2^k)/(1 + x^(2^k)) (conjecture). - _Ilya Gutkovskiy_, Jul 23 2017

%e For n=0, A006134(0) = 1, hence 0 is a term.

%t Select[Range[0, 1020], Mod[Sum[Binomial[2 k, k], {k, 0, #}], 3] != 0 &] (* _Michael De Vlieger_, Nov 28 2015 *)

%o (PARI) for(n=0, 1e3, if(sum(k=0, n, binomial(2*k, k)) % 3 > 0, print1(n,", "))) \\ _Altug Alkan_, Nov 26 2015

%Y Cf. A005836, A006134, A010060, A083096, A131996.

%Y Equals A089118(n-2) + 1, n > 1.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Apr 22 2003

%E Zero prepended to the sequence and formulas modified accordingly by _L. Edson Jeffery_, Nov 25 2015