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A294565 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294565 #4 Nov 15 2017 17:39:10
%S A294565 1,2,6,12,25,44,77,130,217,360,590,964,1569,2549,4135,6702,10856,
%T A294565 17578,28455,46055,74533,120614,195173,315814,511015,826858,1337903,
%U A294565 2164792,3502727,5667552,9170313,14837900,24008249,38846186,62854473,101700698,164555211
%N A294565 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294565 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 A294565 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 A294565 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294565 b(1) = 4 (least "new number")
%e A294565 a(2)  = a(1) + a(0) + 2*b(1) - b(0) - 2 = 6
%e A294565 Complement: (b(n)) = (3, 4, 5, 7, 8, 10, 11, 13, 14, 15, 16, ...)
%t A294565 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294565 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294565 a[n_] := a[n] = a[n - 1] + a[n - 2] + 2 b[n - 1] - b[n - 2] - 2;
%t A294565 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294565 Table[a[n], {n, 0, 40}]  (* A294565 *)
%t A294565 Table[b[n], {n, 0, 10}]
%Y A294565 Cf. A001622, A294532.
%K A294565 nonn,easy
%O A294565 0,2
%A A294565 _Clark Kimberling_, Nov 15 2017