cp's OEIS Frontend

A294413 Solution of the complementary equation a(n) = a(n-1) + a(n-2) - b(n-1) + 6, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294413 #9 Nov 01 2017 20:53:46
%S A294413 1,3,6,10,15,23,35,53,82,128,202,320,511,819,1317,2122,3424,5530,8936,
%T A294413 14447,23363,37789,61130,98896,160002,258873,418849,677695,1096516,
%U A294413 1774181,2870666,4644815,7515448,12160229,19675642,31835835
%N A294413 Solution of the complementary equation a(n) = a(n-1) + a(n-2) - b(n-1) + 6, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294413 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  See A294414 for a guide to related sequences.
%C A294413 Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
%H A294413 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 A294413 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294413 a(2)  = a(1) + a(0) - b(1) + 6
%e A294413 Complement: (b(n)) = (2, 4, 6, 7, 9, 11, 12, 13, 14, 16, ...)
%t A294413 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294413 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294413 a[n_] := a[n] = a[n - 1] + a[n - 2] - b[n - 1] + 6;
%t A294413 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294413 Table[a[n], {n, 0, 40}]  (* A294413 *)
%t A294413 Table[b[n], {n, 0, 10}]
%Y A294413 Cf. A293076, A293765, A294414.
%K A294413 nonn,easy
%O A294413 0,2
%A A294413 _Clark Kimberling_, Oct 31 2017