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A294365 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294365 #8 Nov 01 2017 23:04:37
%S A294365 1,3,10,21,41,74,129,219,367,607,997,1629,2653,4311,6995,11339,18369,
%T A294365 29745,48154,77941,126139,204126,330313,534489,864854,1399397,2264307,
%U A294365 3663762,5928129,9591953,15520146,25112165,40632379,65744614,106377065,172121753
%N A294365 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294365 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 A294365 Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.  See A293358 for a guide to related sequences.
%H A294365 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 A294365 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294365 a(2)  = a(1) + a(0) + b(1) + 2 = 10;
%e A294365 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)
%t A294365 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294365 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294365 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + n;
%t A294365 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294365 Table[a[n], {n, 0, 40}]  (* A294365 *)
%t A294365 Table[b[n], {n, 0, 10}]
%Y A294365 Cf. A001622 (golden ratio), A293765.
%K A294365 nonn,easy
%O A294365 0,2
%A A294365 _Clark Kimberling_, Oct 29 2017