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A298108 Solution b( ) of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 3, 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 A298108 #5 Feb 09 2018 11:10:11
%S A298108 3,4,5,7,8,10,11,12,13,15,17,18,19,20,22,24,25,27,28,30,31,32,33,35,
%T A298108 36,37,39,41,42,44,45,47,48,49,50,52,53,54,56,58,59,60,61,63,65,66,67,
%U A298108 68,70,72,73,75,76,78,79,80,81,83,85,86,88,89,90,91,93,94
%N A298108 Solution b( ) of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 3, 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 A298108 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 A298108 Conjecture:  3/2 < a(n) - n*sqrt(2) < 4 for n >= 1.
%H A298108 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 A298108 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
%t A298108 a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + 2 n - 3;
%t A298108 j = 1; While[j < 80000, k = a[j] - j - 1;
%t A298108  While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
%t A298108 u = Table[a[n], {n, 0, k}]; (* A297833 *)
%t A298108 v = Table[b[n], {n, 0, k}]; (* A298108 *)
%t A298108 Take[u, 50]
%t A298108 Take[v, 50]
%Y A298108 Cf. A297830, A297833.
%K A298108 nonn,easy
%O A298108 0,1
%A A298108 _Clark Kimberling_, Feb 09 2018