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A294481 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 A294481 #5 Nov 01 2017 12:27:30
%S A294481 1,3,6,10,16,22,30,39,49,60,72,85,100,115,132,149,168,188,209,231,254,
%T A294481 278,303,329,357,385,415,445,477,509,543,577,614,650,689,727,768,808,
%U A294481 851,893,938,983,1030,1077,1126,1175,1226,1277,1330,1383,1438,1494
%N A294481 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 A294481 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 A294481 Clark Kimberling, <a href="/A294481/b294481.txt">Table of n, a(n) for n = 0..1000</a>
%H A294481 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 A294481 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294481 a(2)  = a(0) + b(1) + 1 = 6
%e A294481 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 12, 13, 15, ...)
%t A294481 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294481 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294481 a[n_] := a[n] = a[n - 2] + b[n - 1] + n - 1;
%t A294481 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294481 Table[a[n], {n, 0, 40}]  (* A294481 *)
%t A294481 Table[b[n], {n, 0, 10}]
%Y A294481 Cf. A293076, A293765, A294476.
%K A294481 nonn,easy
%O A294481 0,2
%A A294481 _Clark Kimberling_, Nov 01 2017