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A298004 Solution b( ) of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 3*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

%I A298004 #4 Feb 09 2018 11:09:28
%S A298004 3,4,5,6,7,8,9,10,12,13,14,16,17,18,20,21,22,24,25,26,28,29,30,32,33,
%T A298004 34,36,37,38,39,40,42,43,45,46,47,49,50,51,52,53,55,56,58,59,60,62,63,
%U A298004 64,65,66,68,69,71,72,73,75,76,77,78,79,81,82,84,85,86
%N A298004 Solution b( ) of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 3*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)).  See Comments.
%C A298004 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  The solution a( ) is given at A297836.  See A297830 for a guide to related sequences.
%C A298004 Conjecture: 7/10 < a(n) - n*L < 3 for n >= 1, where L = (-1 + sqrt(13))/2.
%H A298004 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%t A298004 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
%t A298004 a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + 3 n;
%t A298004 j = 1; While[j < 80000, k = a[j] - j - 1;
%t A298004 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
%t A298004 u = Table[a[n], {n, 0, k}]; (* A297836 *)
%t A298004 v = Table[b[n], {n, 0, k}]; (* A298004 *)
%t A298004 Take[u, 50]
%t A298004 Take[v, 50]
%Y A298004 Cf. A297830, A297836.
%K A298004 nonn,easy
%O A298004 0,1
%A A298004 _Clark Kimberling_, Feb 09 2018