|
%I #8 Nov 14 2017 22:19:14
%S 1,2,10,20,39,69,119,200,333,548,897,1462,2377,3858,6255,10134,16411,
%T 26569,43005,69600,112632,182260,294921,477211,772163,1249406,2021602,
%U 3271042,5292679,8563757,13856473,22420268,36276780,58697088,94973909,153671040
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C 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 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) = 2, b(0) = 3, so that
%e b(1) = 4 (least "new number");
%e a(2) = a(1) + a(0) + b(1) + 3 = 10.
%e Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...).
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 3; b[0] = 2;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 3;
%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}] (* A294544 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622, A294532.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Nov 04 2017
|
