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A294542 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294542 #8 Nov 14 2017 22:19:00
%S A294542 1,2,8,16,31,55,96,162,270,445,729,1189,1934,3141,5094,8255,13370,
%T A294542 21647,35040,56711,91776,148513,240316,388857,629202,1018089,1647322,
%U A294542 2665444,4312800,6978279,11291115,18269431,29560584,47830054,77390678,125220773
%N A294542 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294542 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 A294542 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 A294542 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294542 b(1) = 4 (least "new number");
%e A294542 a(2) = a(1) + a(0) + b(1) + 1 = 8.
%e A294542 Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, ...).
%t A294542 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294542 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294542 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 1;
%t A294542 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294542 Table[a[n], {n, 0, 40}]  (* A294542 *)
%t A294542 Table[b[n], {n, 0, 10}]
%Y A294542 Cf. A001622, A294532.
%K A294542 nonn,easy
%O A294542 0,2
%A A294542 _Clark Kimberling_, Nov 04 2017