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A295357 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295357 #5 Nov 21 2017 21:33:18
%S A295357 1,3,5,20,42,83,149,259,438,730,1204,1973,3219,5237,8504,13792,22350,
%T A295357 36200,58612,94878,153559,248509,402143,650730,1052954,1703768,
%U A295357 2756809,4460667,7217569,11678332,18896000,30574434,49470539
%N A295357 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295357 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. Guide to related sequences:
%C A295357 ***** Part 1:  initial values are a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6
%C A295357 A295357: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3)
%C A295357 A295358: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3)
%C A295357 A295359: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2*b(n-3)
%C A295357 A295360: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 3*b(n-3)
%C A295357 A295361: a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - 3*b(n-3)
%C A295357 A295362: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) - b(n-3)
%C A295357 ***** Part 2: initial values as shown
%C A295357 A295363: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
%C A295357 A295364: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4
%C A295357 A295365: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6
%C A295357 A295366: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6
%C A295357 A295367: a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2); a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
%C A295357 For all of these sequences, a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%H A295357 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 A295357 a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
%e A295357 b(3) = 7 (least "new number")
%e A295357 a(3) = a(1) + a(0) + b(2) + b(1) + b(0) = 20
%e A295357 Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
%t A295357 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295357 a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
%t A295357 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + b[n - 3];
%t A295357 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295357 z = 32; u = Table[a[n], {n, 0, z}]   (* A295357 *)
%t A295357 v = Table[b[n], {n, 0, 10}]  (* complement *)
%Y A295357 Cf. A001622, A293076, A294532.
%K A295357 nonn,easy
%O A295357 0,2
%A A295357 _Clark Kimberling_, Nov 21 2017