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A294399 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294399 #9 Nov 01 2017 23:05:55
%S A294399 1,3,8,15,23,32,42,54,67,81,96,112,129,148,168,189,211,234,258,283,
%T A294399 310,338,367,397,428,460,493,527,563,600,638,677,717,758,800,843,887,
%U A294399 933,980,1028,1077,1127,1178,1230,1283,1337,1392,1448,1506,1565,1625,1686
%N A294399 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294399 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A022940 for a guide to related sequences.
%H A294399 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.pdf">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%e A294399 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294399 a(2)  = a(1) + b(0) + 3 = 8
%e A294399 Complement: (b(n)) = (2, 4, 5, 6, 7, 10, 11, 12, 13, 14, 16, ...)
%t A294399 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294399 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294399 a[n_] := a[n] = a[n - 1] + b[n - 2] + 3;
%t A294399 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294399 Table[a[n], {n, 0, 40}]  (* A294399 *)
%t A294399 Table[b[n], {n, 0, 10}]
%Y A294399 Cf. A293076, A293765, A022940.
%K A294399 nonn,easy
%O A294399 0,2
%A A294399 _Clark Kimberling_, Oct 30 2017