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Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-3), 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.
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%I #6 Aug 27 2021 21:04:35

%S 1,2,3,4,13,25,40,66,114,190,308,502,821,1335,2162,3503,5678,9195,

%T 14881,24084,38980,63080,102071,165162,267250,432430,699693,1132136,

%U 1831848,2964004,4795867,7759886,12555774,20315682,32871473,53187172,86058669,139245866

%N Solution of the complementary equation a(n) = a(n-1) + a(n-3) + a(n-4) + b(n-3), 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 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

%C 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 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 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 b(4) = 9 (least "new number")

%e a(4) = a(3) + a(1) + a(0) + b(1) = 13

%e Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 12, 14, 15, ...)

%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

%t a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;

%t b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;

%t a[n_] := a[n] = a[n - 1] + a[n - 3] + a[n - 4] + b[n - 3];

%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

%t z = 36; Table[a[n], {n, 0, z}] (* A295755 *)

%t Table[b[n], {n, 0, 20}] (*complement *)

%Y Cf. A001622, A000045, A293411, A295754.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 30 2017