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A294479 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2.

%I A294479 #8 Nov 02 2017 16:11:43
%S A294479 1,3,7,11,17,24,32,41,52,63,76,89,104,120,137,155,174,194,215,238,261,
%T A294479 286,311,338,365,394,424,455,487,520,554,589,625,662,701,740,781,822,
%U A294479 865,908,953,998,1045,1092,1142,1191,1243,1294,1348,1401,1457,1512
%N A294479 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2.
%C A294479 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  See A294476 for a guide to related sequences.
%H A294479 Clark Kimberling, <a href="/A294479/b294479.txt">Table of n, a(n) for n = 0..1000</a>
%H A294479 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 A294479 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294479 a(2)  = a(0) + b(1) + 2 = 7
%e A294479 Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 10, 12, 13, 14, ...)
%t A294479 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294479 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294479 a[n_] := a[n] = a[n - 2] + b[n - 1] + n;
%t A294479 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294479 Table[a[n], {n, 0, 40}]  (* A294479 *)
%t A294479 Table[b[n], {n, 0, 10}]
%Y A294479 Cf. A293076, A293765, A294476.
%K A294479 nonn,easy
%O A294479 0,2
%A A294479 _Clark Kimberling_, Nov 01 2017