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A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2.

%I A294476 #7 Nov 01 2017 12:26:54
%S A294476 1,3,6,9,14,18,25,30,38,44,54,61,72,81,93,103,116,127,141,154,169,183,
%T A294476 199,215,232,249,267,285,304,323,344,364,386,407,430,453,477,501,526,
%U A294476 551,577,603,630,657,686,714,744,773,804,834,867,898,932,964,999
%N A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2.
%C A294476 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2:
%C A294476 A294476:  a(n) = a(n-2) + b(n-1) + 1
%C A294476 A294477:  a(n) = a(n-2) + b(n-1) + 2
%C A294476 A294478:  a(n) = a(n-2) + b(n-1) + 3
%C A294476 A294479:  a(n) = a(n-2) + b(n-1) + n
%C A294476 A294480:  a(n) = a(n-2) + b(n-1) + 2n
%C A294476 A294481:  a(n) = a(n-2) + b(n-1) + n - 1
%C A294476 A294482:  a(n) = a(n-2) + b(n-1) + n + 1
%C A294476 For a(n-2) + b(n-1), with offset 1 instead of 0, see A022942.
%H A294476 Clark Kimberling, <a href="/A294476/b294476.txt">Table of n, a(n) for n = 0..1000</a>
%H A294476 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 A294476 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294476 a(2)  = a(0) + b(1) + 1 = 6
%e A294476 Complement: (b(n)) = (2, 4, 5, 7, 8, 10, 11, 12, 13, 15,...)
%t A294476 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294476 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294476 a[n_] := a[n] = a[n - 2] + b[n - 1] + 1;
%t A294476 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294476 Table[a[n], {n, 0, 40}]  (* A294476 *)
%t A294476 Table[b[n], {n, 0, 10}]
%Y A294476 Cf. A293076, A293765, A293358, A294414.
%K A294476 nonn,easy
%O A294476 0,2
%A A294476 _Clark Kimberling_, Nov 01 2017