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A298112 Solution b( ) of the complementary equation a(n) = a(1)*b(n) - a(0)*b(n-1) + 3*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

%I A298112 #8 Feb 11 2018 03:08:50
%S A298112 3,4,5,6,7,8,9,10,11,13,14,15,17,18,19,21,22,23,25,26,27,29,30,31,33,
%T A298112 34,35,37,38,39,40,41,43,44,46,47,48,50,51,52,53,54,56,57,59,60,61,63,
%U A298112 64,65,66,67,69,70,72,73,74,76,77,78,79,80,82,83,85,86
%N A298112 Solution b( ) of the complementary equation a(n) = a(1)*b(n) - a(0)*b(n-1) + 3*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (b(n)) is the increasing sequence of positive integers not in (a(n)).  See Comments.
%C A298112 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  The solution a( ) is given at A298001.  See A297830 for a guide to related sequences.
%C A298112 Conjecture:  2/5 < a(n) - n*sqrt(2) < 3 for n >= 1.
%H A298112 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 A298112 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5;
%t A298112 a[n_] := a[n] = a[1]*b[n] - a[0]*b[n - 1] + 3 n;
%t A298112 j = 1; While[j < 80000, k = a[j] - j - 1;
%t A298112 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
%t A298112 u = Table[a[n], {n, 0, k}]; (* A298001 *)
%t A298112 v = Table[b[n], {n, 0, k}]; (* A298112 *)
%t A298112 Take[u, 50]
%t A298112 Take[v, 50]
%Y A298112 Cf. A297830, A298001.
%K A298112 nonn,easy
%O A298112 0,1
%A A298112 _Clark Kimberling_, Feb 09 2018