A294549 - OEIS

%I #8 Nov 14 2017 22:19:47

%S 1,2,10,21,42,76,133,226,379,627,1030,1683,2741,4454,7227,11715,18978,

%T 30731,49750,80524,130319,210890,341258,552199,893510,1445764,2339331,

%U 3785154,6124546,9909763,16034374,25944204,41978647,67922922,109901642,177824639

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n + 1, 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) + 1 = 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] + n + 1;

%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}]  (* A294549 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A001622, A294532.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 04 2017