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A294535 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 3, where a(0) = 1, a(1) = 2, b(0) = 3.

%I A294535 #4 Nov 03 2017 09:53:59
%S A294535 1,2,9,18,35,62,107,180,300,494,809,1319,2145,3482,5646,9148,14816,
%T A294535 23987,38827,62839,101692,164558,266278,430865,697173,1128069,1825274,
%U A294535 2953376,4778684,7732095,12510815,20242947,32753801,52996788,85750630,138747460
%N A294535 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 3, where a(0) = 1, a(1) = 2, b(0) = 3.
%C A294535 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  See A294532 for a guide to related sequences.  Conjecture:  a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622)..
%H A294535 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 A294535 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294535 b(1) = 4 (least "new number")
%e A294535 a(2)  = a(1) + a(0) + b(0) + 3 = 9
%e A294535 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, ...)
%t A294535 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294535 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294535 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2] + 3;
%t A294535 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294535 Table[a[n], {n, 0, 40}]  (* A294535 *)
%t A294535 Table[b[n], {n, 0, 10}]
%Y A294535 Cf. A001622, A294532.
%K A294535 nonn,easy
%O A294535 0,2
%A A294535 _Clark Kimberling_, Nov 03 2017