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%I #18 Sep 23 2020 04:11:23
%S 1,3,6,13,24,44,76,129,215,355,582,951,1548,2515,4080,6613,10712,
%T 17345,28078,45445,73546,119016,192588,311631,504247,815907,1320184,
%U 2136122,3456338,5592493,9048865,14641393,23690294,38331724,62022056,100353819,162375915
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C The complementary sequences a() and b() are uniquely determined by the titular equation and initial values, which for each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
%C A293076: a(n) = a(n-1) + a(n-2) + b(n-2)
%C A293316: a(n) = a(n-1) + a(n-2) + b(n-2) + 1
%C A293057: a(n) = a(n-1) + a(n-2) + b(n-2) + 2
%C A293058: a(n) = a(n-1) + a(n-2) + b(n-2) + 3
%C A293317: a(n) = a(n-1) + a(n-2) + b(n-2) - 1
%C A293349: a(n) = a(n-1) + a(n-2) + b(n-2) + n
%C A293350: a(n) = a(n-1) + a(n-2) + b(n-2) + 2*n
%C A293351: a(n) = a(n-1) + a(n-2) + b(n-2) + n - 1
%C A293357: a(n) = a(n-1) + a(n-2) + b(n-2) + n + 1
%C Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
%H 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 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e a(2) = a(1) + a(0) + b(0) = 3 + 1 + 2 = 6;
%e a(3) = a(2) + a(1) + b(1) = 6 + 3 + 4 = 13.
%e Complement: (b(n)) = (2,4,5,7,8,9,10,11,12,14,...)
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2];
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 40}] (* A293076 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622 (golden ratio), A293358.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 28 2017
%E Comments corrected by _Georg Fischer_, Sep 23 2020
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