%I #8 May 05 2018 04:18:04
%S 1,2,10,31,35,95,99,108,112,289,293,302,306,330,335,343,348,875,880,
%T 888,893,916,921,929,934,1002,1007,1018,1023,1043,1048,1059,1064,2641,
%U 2646,2657,2662,2682,2687,2698,2703,2768,2773,2784,2789,2809,2814,2825,2830
%N Solution (a(n)) of the system of 2 complementary equations in Comments.
%C Define sequences a(n), b(n), c(n) recursively, starting with a(0) = 1, a(1) = 1, b(0) = 3; for n >= 1,
%C a(2n) = 3*a(n) + b(n);
%C a(2n+1) = 3*a(n-1) + n;
%C b(n) = least new;
%C where "least new k" means the least positive integer not yet placed. The sequences (a(n)) and (b(n)) are complementary.
%H Clark Kimberling, <a href="/A297467/b297467.txt">Table of n, a(n) for n = 0..2000</a>
%e n: 0 1 2 3 4 5 6 7 8
%e a: 1 2 10 31 35 95 99 108 112
%e b: 3 4 5 6 7 8 9 11 12
%t z = 300;
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t a = {1, 2}; b = {3};
%t Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]];
%t AppendTo[a, 3 a[[#/2 + 1]] + b[[#/2 + 1]]] &[Length[a]];
%t AppendTo[a, 3 a[[(# + 3)/2]] + (# - 1)/2] &[Length[a]], {z}]
%t Take[a, 100] (* A297467 *)
%t Take[b, 100] (* A297468 *)
%t (* _Peter J. C. Moses_, Apr 22 2018 *)
%Y Cf. A299634, A297468.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Apr 24 2018