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A295757 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-1), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295757 #6 Aug 27 2021 21:05:02
%S A295757 1,2,3,4,15,29,46,76,132,220,356,580,949,1543,2498,4047,6560,10623,
%T A295757 17191,27822,45030,72870,117910,190790,308720,499531,808263,1307806,
%U A295757 2116091,3423920,5540025,8963959,14504008,23467992,37972016,61440024,99412066,160852117
%N A295757 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-1), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295757 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.
%C A295757 a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).
%H A295757 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 A295757 a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
%e A295757 b(4) = 9 (least "new number")
%e A295757 a(4) = a(3) + a(1) + a(0) + b(3) = 15
%e A295757 Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)
%t A295757 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295757 a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
%t A295757 b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
%t A295757 a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 1];
%t A295757 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295757 z = 36;  Table[a[n], {n, 0, z}]   (* A295757 *)
%t A295757 Table[b[n], {n, 0, 20}]  (*complement *)
%Y A295757 Cf. A001622, A000045, A293411, A295754.
%K A295757 nonn,easy
%O A295757 0,2
%A A295757 _Clark Kimberling_, Dec 01 2017