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A294397 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294397 #16 Nov 01 2017 23:05:40
%S A294397 1,3,6,11,17,25,34,44,55,68,82,97,113,130,149,169,190,212,235,259,284,
%T A294397 311,339,368,398,429,461,494,528,564,601,639,678,718,759,801,844,888,
%U A294397 934,981,1029,1078,1128,1179,1231,1284,1338,1393,1450,1508,1567,1627
%N A294397 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294397 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A022940 for a guide to related sequences.
%C A294397 Apart from the first two entries this is the same as A081689. - _R. J. Mathar_, Oct 31 2017
%H A294397 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 A294397 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294397 a(2)  = a(1) + b(0) + 1 = 6
%e A294397 Complement: (b(n)) = (2, 4, 5, 7, 8, 10, 11, 12, 13, 14, 16, ...)
%t A294397 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294397 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294397 a[n_] := a[n] = a[n - 1] + b[n - 2] + 1;
%t A294397 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294397 Table[a[n], {n, 0, 40}]  (* A294397 *)
%t A294397 Table[b[n], {n, 0, 10}]
%Y A294397 Cf. A081689, A293076, A293765, A022940.
%K A294397 nonn,easy
%O A294397 0,2
%A A294397 _Clark Kimberling_, Oct 30 2017