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%I #20 Jul 23 2023 10:13:10
%S 1,2,3,10,11,28,29,36,37,38,39,82,83,90,91,92,93,108,109,110,111,118,
%T 119,244,245,252,253,254,255,270,271,272,273,280,281,324,325,326,327,
%U 334,335,352,353,360,361,362,363,730,731,738,739,740,741,756,757,758
%N Numbers k such that Sum_{j=1..k} (binomial(2*j,j) mod 3) is even.
%H Robert Israel, <a href="/A081868/b081868.txt">Table of n, a(n) for n = 1..10000</a>
%p N:= 10000: # to get all terms <= N
%p alpha:= 2:
%p beta:= 0:
%p t:= 0:
%p A[1]:= 1:
%p count:= 1:
%p for i from 2 to N do
%p d:= padic:-ordp(4 - 2/i,3);
%p beta:= beta + d;
%p alpha:= alpha * (4-2/i)/3^d mod 3;
%p if beta = 0 then
%p t:= t + alpha mod 2;
%p fi;
%p if t = 0 then
%p count:= count+1;
%p A[count]:= i;
%p fi
%p od:
%p seq(A[i], i=1..count); # _Robert Israel_, May 05 2014
%t Select[Range[800],EvenQ[Sum[Mod[Binomial[2j,j],3],{j,#}]]&] (* _Harvey P. Dale_, Jul 23 2023 *)
%o (PARI) isok(n) = ! (sum(k=1, n, binomial(2*k, k) % 3) % 2); \\ _Michel Marcus_, Dec 04 2013
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 12 2003
%E More terms from _Michel Marcus_, Dec 04 2013
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