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A294401 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294401 #12 Feb 06 2018 13:30:29
%S A294401 1,3,9,19,32,48,67,89,115,144,176,211,249,290,334,381,431,485,542,602,
%T A294401 665,731,800,872,947,1025,1106,1190,1277,1368,1462,1559,1659,1762,
%U A294401 1868,1977,2089,2204,2322,2443,2567,2694,2824,2957,3094,3234,3377,3523,3672
%N A294401 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294401 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 A294401 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 A294401 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294401 a(2)  = a(1) + a(0) + 4 = 9.
%e A294401 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, ...)
%t A294401 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294401 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294401 a[n_] := a[n] = a[n - 1] + b[n - 2] + 2 n;
%t A294401 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294401 Table[a[n], {n, 0, 40}]  (* A294401 *)
%t A294401 Table[b[n], {n, 0, 10}]
%Y A294401 Cf. A293076, A293765, A022940.
%K A294401 nonn,easy
%O A294401 0,2
%A A294401 _Clark Kimberling_, Oct 31 2017