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A294555 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294555 #7 Sep 27 2020 18:39:17
%S A294555 1,2,13,27,54,97,169,286,477,787,1290,2106,3428,5568,9032,14638,23710,
%T A294555 38390,62144,100580,162772,263402,426226,689682,1115965,1805707,
%U A294555 2921734,4727505,7649305,12376878,20026253,32403203,52429530,84832809,137262417,222095306
%N A294555 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294555 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 A294555 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 A294555 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294555 b(1) = 4 (least "new number")
%e A294555 a(2)  = a(1) + a(0) + b(1) + b(0) + 3 = 13
%e A294555 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 14, 15, 16, ...)
%t A294555 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294555 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294555 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + 3;
%t A294555 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294555 Table[a[n], {n, 0, 40}]  (* A294555 *)
%t A294555 Table[b[n], {n, 0, 10}]
%Y A294555 Cf. A001622, A294532.
%K A294555 nonn,easy
%O A294555 0,2
%A A294555 _Clark Kimberling_, Nov 15 2017
%E A294555 Definition corrected by _Georg Fischer_, Sep 27 2020