%I #8 Nov 02 2017 09:19:37
%S 1,3,8,18,35,64,112,192,322,534,878,1436,2340,3804,6174,10010,16219,
%T 26266,42524,68831,111398,180274,291719,472042,763812,1235907,1999774,
%U 3235738,5235571,8471370,13707004,22178439,35885511,58064020,93949603,152013697
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + n, 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 A293076 for a guide to related sequences.
%C Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
%e a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e a(2) = a(1) + a(0) + b(0) + 2 = 8;
%e a(3) = a(2) + a(1) + b(1) + 3 = 18.
%e Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 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] + n;
%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}] (* A293349 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622 (golden ratio), A293076.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 28 2017