A294533 - OEIS

%I #4 Nov 03 2017 09:53:42

%S 1,2,7,14,27,48,84,142,237,391,641,1046,1703,2766,4487,7272,11779,

%T 19072,30873,49968,80865,130858,211749,342634,554412,897076,1451519,

%U 2348627,3800179,6148840,9949054,16097930,26047021,42144989,68192049,110337078,178529168

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 1, where a(0) = 1, a(1) = 2, b(0) = 3.

%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(0) + 1 = 7

%e Complement: (b(n)) = (3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, ...)

%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 - 2] + 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}]  (* A294533 *)

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

%Y Cf. A001622, A294532.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 03 2017