A294535 - OEIS

%I #4 Nov 03 2017 09:53:59

%S 1,2,9,18,35,62,107,180,300,494,809,1319,2145,3482,5646,9148,14816,

%T 23987,38827,62839,101692,164558,266278,430865,697173,1128069,1825274,

%U 2953376,4778684,7732095,12510815,20242947,32753801,52996788,85750630,138747460

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 3, 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) + 3 = 9

%e Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 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] + 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}]  (* A294535 *)

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

%Y Cf. A001622, A294532.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 03 2017