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A294558 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 2*n, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294558 #7 Sep 27 2020 18:36:47
%S A294558 1,2,14,31,64,118,209,358,602,999,1644,2690,4386,7133,11580,18778,
%T A294558 30427,49278,79782,129141,209008,338238,547339,885674,1433114,2318893,
%U A294558 3752116,6071122,9823356,15894601,25718084,41612816,67331035,108943990,176275168,285219305
%N A294558 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 2*n, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294558 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 A294558 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 A294558 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294558 b(1) = 4 (least "new number")
%e A294558 a(2) = a(1) + a(0) + b(1) + b(0) + 1 = 11
%e A294558 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, ...)
%t A294558 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294558 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294558 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + 2*n;
%t A294558 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294558 Table[a[n], {n, 0, 18}]  (* A294558 *)
%t A294558 Table[b[n], {n, 0, 10}]
%Y A294558 Cf. A001622, A294532.
%K A294558 nonn,easy
%O A294558 0,2
%A A294558 _Clark Kimberling_, Nov 16 2017
%E A294558 Definition corrected by _Georg Fischer_, Sep 27 2020