%I #5 Nov 01 2017 12:26:46
%S 1,3,8,15,28,49,86,144,240,395,647,1055,1718,2789,4524,7331,11874,
%T 19225,31120,50367,81510,131901,213436,345363,558828,904220,1463078,
%U 2367329,3830439,6197801,10028274,16226110,26254420,42480567,68735025,111215631,179950696
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 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. See A294414 for a guide to related sequences.
%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) + 2*b(1) - b(0) - 2 = 8
%e Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 16,...)
%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] + 2*b[n - 1] - b[n - 2] - 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}] (* A294426 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A293076, A293765, A294414.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Nov 01 2017