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A294482 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n + 1, where a(0) = 1, a(1) = 3, b(0) = 2.

%I A294482 #4 Nov 03 2017 09:53:26
%S A294482 1,3,8,12,19,25,35,43,55,66,80,93,109,124,142,160,180,200,222,244,269,
%T A294482 293,320,346,375,403,434,464,497,530,565,600,637,674,713,752,794,835,
%U A294482 879,922,968,1013,1061,1108,1158,1207,1259,1311,1365,1419,1475,1531
%N A294482 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n + 1, where a(0) = 1, a(1) = 3, b(0) = 2.
%C A294482 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 A294482 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 A294482 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294482 a(2)  = a(0) + b(1) + 3 = 8
%e A294482 Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 13, 15, ...)
%t A294482 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294482 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294482 a[n_] := a[n] = a[n - 2] + b[n - 1] + n + 1;
%t A294482 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294482 Table[a[n], {n, 0, 40}]  (* A294482 *)
%t A294482 Table[b[n], {n, 0, 10}]
%Y A294482 Cf. A293076, A293765, A294476.
%K A294482 nonn,easy
%O A294482 0,2
%A A294482 _Clark Kimberling_, Nov 03 2017