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A294412 Solution of the complementary equation a(n) = a(n-1) + 2*b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294412 #12 Feb 06 2018 13:30:58
%S A294412 1,3,8,17,28,41,56,75,96,119,144,171,200,231,264,301,340,381,424,469,
%T A294412 516,565,616,669,724,783,844,907,972,1039,1108,1179,1252,1327,1404,
%U A294412 1483,1564,1649,1736,1825,1916,2009,2104,2201,2300,2401,2504,2609,2716,2825
%N A294412 Solution of the complementary equation a(n) = a(n-1) + 2*b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294412 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 A294412 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 A294412 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294412 a(2)  = a(1) + 2*b(0) + 1 = 8
%e A294412 Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15, 16, ...)
%t A294412 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294412 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294412 a[n_] := a[n] = a[n - 1] + 2b[n - 2] + 1;
%t A294412 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294412 Table[a[n], {n, 0, 40}]  (* A294412 *)
%t A294412 Table[b[n], {n, 0, 10}]
%Y A294412 Cf. A293076, A293765, A022940.
%K A294412 nonn,easy
%O A294412 0,2
%A A294412 _Clark Kimberling_, Oct 31 2017