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A295756 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-2), 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 A295756 #6 Aug 27 2021 21:04:49
%S A295756 1,2,3,4,14,27,43,71,123,205,332,541,885,1439,2330,3775,6119,9909,
%T A295756 16036,25953,42005,67975,109990,177976,287985,465980,753977,1219970,
%U A295756 1973968,3193959,5167941,8361915,13529879,21891817,35421712,57313546,92735283,150048854
%N A295756 Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-2), 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 A295756 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.
%C A295756 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 A295756 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 A295756 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 A295756 b(4) = 9 (least "new number")
%e A295756 a(4) = a(3) + a(1) + a(0) + b(2) = 14
%e A295756 Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 13, 15, ...)
%t A295756 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295756 a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
%t A295756 b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
%t A295756 a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 2];
%t A295756 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295756 z = 36;  Table[a[n], {n, 0, z}]   (* A295756 *)
%t A295756 Table[b[n], {n, 0, 20}]  (*complement *)
%Y A295756 Cf. A001622, A000045, A293411, A295754.
%K A295756 nonn,easy
%O A295756 0,2
%A A295756 _Clark Kimberling_, Dec 01 2017