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A294477 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 3, b(0) = 2.

%I A294477 #5 Nov 01 2017 12:27:01
%S A294477 1,3,7,10,15,20,26,33,40,48,56,66,75,86,96,109,120,134,146,161,175,
%T A294477 191,206,223,239,257,275,294,313,333,353,374,396,418,441,464,488,512,
%U A294477 537,563,589,616,643,671,699,728,758,788,819,850,882,914,947,980,1014
%N A294477 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 2, where a(0) = 1, a(1) = 3, b(0) = 2.
%C A294477 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  See A294476 for a guide to related sequences.
%H A294477 Clark Kimberling, <a href="/A294477/b294477.txt">Table of n, a(n) for n = 0..1000</a>
%H A294477 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 A294477 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294477 a(2) = a(0) + b(1) + 2 = 7
%e A294477 Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 11, 12, 13, 14, 16,...)
%t A294477 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294477 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294477 a[n_] := a[n] = a[n - 2] + b[n - 1] + 2;
%t A294477 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294477 Table[a[n], {n, 0, 40}]  (* A294477 *)
%t A294477 Table[b[n], {n, 0, 10}]
%Y A294477 Cf. A293076, A293765, A294476.
%K A294477 nonn
%O A294477 0,2
%A A294477 _Clark Kimberling_, Nov 01 2017