%I #8 Nov 14 2017 22:19:20
%S 1,2,6,12,24,43,75,127,212,351,576,941,1532,2489,4038,6545,10602,
%T 17167,27790,44979,72793,117797,190616,308440,499084,807553,1306667,
%U 2114251,3420950,5535234,8956218,14491487,23447741,37939265,61387044,99326348
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294532 for a guide to related sequences. Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%e a(0) = 1, a(1) = 2, b(0) = 3, so that
%e b(1) = 4 (least "new number");
%e a(2) = a(1) + a(0) + b(1) -1 = 6.
%e Complement: (b(n)) = (3, 4, 5, 7, 8, 9, 11, 13, 14, 15, 16, ...).
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 3; b[0] = 2;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] - 1;
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 40}] (* A294545 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622, A294532.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Nov 04 2017