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A298007 Solution b( ) of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 2, 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 A298007 #4 Feb 09 2018 11:10:05
%S A298007 3,4,5,6,8,9,11,12,14,15,16,17,19,21,22,23,24,26,28,29,30,31,33,35,36,
%T A298007 38,39,41,42,43,44,46,47,48,50,52,53,55,56,58,59,60,61,63,64,65,67,69,
%U A298007 70,72,73,75,76,77,78,80,81,82,84,86,87,88,89,91,93,94
%N A298007 Solution b( ) of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 2, 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 A298007 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  The solution a( ) is given at A297832.  See A297830 for a guide to related sequences.
%C A298007 Conjecture:  1 < a(n) - n*sqrt(2) < 3 for n >= 1.
%H A298007 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 A298007 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
%t A298007 a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + 2 n - 2;
%t A298007 j = 1; While[j < 80000, k = a[j] - j - 1;
%t A298007  While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
%t A298007 u = Table[a[n], {n, 0, k}]; (* A297832 *)
%t A298007 v = Table[b[n], {n, 0, k}]; (* A298007 *)
%t A298007 Take[u, 50]
%t A298007 Take[v, 50]
%Y A298007 Cf. A297830, A297832.
%K A298007 nonn,easy
%O A298007 0,1
%A A298007 _Clark Kimberling_, Feb 09 2018