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A294533 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 1, where a(0) = 1, a(1) = 2, b(0) = 3.

%I A294533 #4 Nov 03 2017 09:53:42
%S A294533 1,2,7,14,27,48,84,142,237,391,641,1046,1703,2766,4487,7272,11779,
%T A294533 19072,30873,49968,80865,130858,211749,342634,554412,897076,1451519,
%U A294533 2348627,3800179,6148840,9949054,16097930,26047021,42144989,68192049,110337078,178529168
%N A294533 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 1, where a(0) = 1, a(1) = 2, b(0) = 3.
%C A294533 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 A294533 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 A294533 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294533 b(1) = 4 (least "new number")
%e A294533 a(2)  = a(1) + a(0) + b(0) + 1 = 7
%e A294533 Complement: (b(n)) = (3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, ...)
%t A294533 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294533 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294533 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2] + 1;
%t A294533 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294533 Table[a[n], {n, 0, 40}]  (* A294533 *)
%t A294533 Table[b[n], {n, 0, 10}]
%Y A294533 Cf. A001622, A294532.
%K A294533 nonn,easy
%O A294533 0,2
%A A294533 _Clark Kimberling_, Nov 03 2017