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A294557 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n - 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294557 #7 Sep 27 2020 18:35:40
%S A294557 1,2,11,24,49,90,159,272,457,759,1250,2046,3336,5425,8807,14281,23140,
%T A294557 37476,60674,98211,158949,257228,416249,673552,1089879,1763512,
%U A294557 2853475,4617074,7470639,12087806,19558541,31646446,51205089,82851640,134056837,216908588
%N A294557 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n - 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294557 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 A294557 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 A294557 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294557 b(1) = 4 (least "new number")
%e A294557 a(2)  = a(1) + a(0) + b(1) + b(0) + 1 = 11
%e A294557 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15, 16, ...)
%t A294557 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294557 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294557 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + n - 1;
%t A294557 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294557 Table[a[n], {n, 0, 40}]  (* A294557 *)
%t A294557 Table[b[n], {n, 0, 10}]
%Y A294557 Cf. A001622, A294532.
%K A294557 nonn,easy
%O A294557 0,2
%A A294557 _Clark Kimberling_, Nov 15 2017
%E A294557 Definition corrected by _Georg Fischer_, Sep 27 2020