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A294415 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294415 #9 Nov 01 2017 20:54:00
%S A294415 1,3,11,24,47,85,148,251,419,693,1138,1859,3027,4918,7979,12933,20950,
%T A294415 33923,54915,88882,143843,232774,376669,609497,986222,1595777,2582059,
%U A294415 4177898,6760021,10937985,17698074,28636129,46334275,74970478,121304829,196275385
%N A294415 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294415 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  See A294414 for a guide to related sequences.
%C A294415 Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
%H A294415 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 A294415 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294415 a(2)  = a(1) + a(0) + b(1) + b(0) + 1 = 11
%e A294415 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14,...)
%t A294415 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294415 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294415 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + 1;
%t A294415 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294415 Table[a[n], {n, 0, 40}]  (* A294415 *)
%t A294415 Table[b[n], {n, 0, 10}]
%Y A294415 Cf. A293076, A293765, A294414.
%K A294415 nonn,easy
%O A294415 0,2
%A A294415 _Clark Kimberling_, Oct 31 2017