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A295754 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-4), 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 A295754 #7 Aug 27 2021 21:04:20
%S A295754 1,2,3,4,12,23,37,61,105,175,284,463,757,1231,1994,3231,5237,8481,
%T A295754 13726,22215,35955,58186,94152,152348,246516,398882,645411,1044305,
%U A295754 1689734,2734059,4423808,7157881,11581709,18739612,30321339,49060968,79382329,128443321
%N A295754 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-4), 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 A295754 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.
%C A295754 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 A295754 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 A295754 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 A295754 b(4) = 9 (least "new number")
%e A295754 a(4) = a(3) + a(1) + a(0) + b(0) = 12
%e A295754 Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)
%t A295754 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295754 a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
%t A295754 b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
%t A295754 a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 4];
%t A295754 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295754 z = 36;  Table[a[n], {n, 0, z}]   (* A295754 *)
%t A295754 Table[b[n], {n, 0, 20}]  (*complement *)
%Y A295754 Cf. A001622, A000045, A293411, A295755.
%K A295754 nonn,easy
%O A295754 0,2
%A A295754 _Clark Kimberling_, Nov 30 2017