A294536 - OEIS

%I #4 Nov 03 2017 09:54:07

%S 1,2,5,10,20,36,63,107,180,298,490,801,1305,2121,3442,5580,9040,14640,

%T 23701,38363,62087,100474,162586,263086,425699,688813,1114541,1803384,

%U 2917956,4721372,7639361,12360767,20000164,32360968,52361170,84722177,137083387

%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 = 5

%e Complement: (b(n)) = (3, 4, 6, 7, 8, 9, 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] - 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}]  (* A294536 *)

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

%Y Cf. A001622, A294532.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 03 2017