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A294539 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + n - 1, where a(0) = 1, a(1) = 2, b(0) = 3.

%I A294539 #4 Nov 03 2017 09:54:26
%S A294539 1,2,7,15,30,55,98,168,283,470,774,1267,2066,3361,5457,8850,14341,
%T A294539 23227,37606,60873,98521,159438,258005,417491,675546,1093089,1768689,
%U A294539 2861835,4630583,7492479,12123125,19615669,31738861,51354599,83093531,134448203
%N A294539 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + n - 1, where a(0) = 1, a(1) = 2, b(0) = 3.
%C A294539 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 A294539 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 A294539 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294539 b(1) = 4 (least "new number")
%e A294539 a(2)  = a(1) + a(0) + b(0) + 1 = 7
%e A294539 Complement: (b(n)) = (3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 16, ...)
%t A294539 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294539 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294539 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2] + n - 1;
%t A294539 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294539 Table[a[n], {n, 0, 40}]  (* A294539 *)
%t A294539 Table[b[n], {n, 0, 10}]
%Y A294539 Cf. A001622, A294532.
%K A294539 nonn,easy
%O A294539 0,2
%A A294539 _Clark Kimberling_, Nov 03 2017