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A294367 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294367 #7 Nov 01 2017 23:04:51
%S A294367 1,3,9,19,37,67,117,200,335,555,912,1491,2429,3948,6407,10387,16829,
%T A294367 27253,44121,71415,115579,187039,302665,489753,792469,1282275,2074799,
%U A294367 3357131,5431989,8789181,14221233,23010479,37231779,60242328,97474179,157716581
%N A294367 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294367 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
%C A294367 Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.  See A293358 for a guide to related sequences.
%H A294367 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 A294367 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294367 a(2)  = a(1) + a(0) + b(1) + 1 = 12;
%e A294367 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, ...)
%t A294367 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294367 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294367 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + n - 1;
%t A294367 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294367 Table[a[n], {n, 0, 40}]  (* A294367 *)
%t A294367 Table[b[n], {n, 0, 10}]
%Y A294367 Cf. A001622 (golden ratio), A293765.
%K A294367 nonn,easy
%O A294367 0,2
%A A294367 _Clark Kimberling_, Oct 29 2017