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A294543 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294543 #8 Nov 14 2017 22:19:09
%S A294543 1,2,9,18,35,62,107,181,301,496,812,1324,2153,3495,5667,9183,14872,
%T A294543 24078,38974,63077,102077,165181,267286,432496,699812,1132339,1832183,
%U A294543 2964555,4796772,7761362,12558170,20319570,32877779,53197389,86075209,139272640
%N A294543 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294543 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 A294543 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 A294543 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294543 b(1) = 4 (least "new number");
%e A294543 a(2) = a(1) + a(0) + b(1) + 2 = 9.
%e A294543 Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, ...).
%t A294543 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294543 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294543 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 2;
%t A294543 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294543 Table[a[n], {n, 0, 40}]  (* A294543 *)
%t A294543 Table[b[n], {n, 0, 10}]
%Y A294543 Cf. A001622, A294532.
%K A294543 nonn,easy
%O A294543 0,2
%A A294543 _Clark Kimberling_, Nov 04 2017