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A294538 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 2n, where a(0) = 1, a(1) = 2, b(0) = 3.

%I A294538 #4 Nov 03 2017 09:54:20
%S A294538 1,2,10,22,45,83,147,252,424,705,1161,1901,3100,5042,8186,13275,21511,
%T A294538 34839,56406,91304,147773,239143,386985,626200,1013260,1639538,
%U A294538 2652879,4292501,6945467,11238058,18183618,29421772,47605489,77027363,124632957,201660428
%N A294538 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 2n, where a(0) = 1, a(1) = 2, b(0) = 3.
%C A294538 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 A294538 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 A294538 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294538 b(1) = 4 (least "new number")
%e A294538 a(2)  = a(1) + a(0) + b(0) + 4 = 10
%e A294538 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, ...)
%t A294538 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294538 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294538 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2] + 2n;
%t A294538 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294538 Table[a[n], {n, 0, 40}]  (* A294538 *)
%t A294538 Table[b[n], {n, 0, 10}]
%Y A294538 Cf. A001622, A294532.
%K A294538 nonn,easy
%O A294538 0,2
%A A294538 _Clark Kimberling_, Nov 03 2017