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%I #11 Nov 06 2018 04:16:45
%S 1,3,4,13,61,361,2521,20161,181441,1814401,19958401,239500801,
%T 3353011202,50295168017
%N Solution of the complementary equation a(n) = a(n-1)*b(n-2) - n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294381 for a guide to related sequences.
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.pdf">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%e a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e a(2) = a(1)*b(0) - 2 = 4
%e Complement: (b(n)) = (2, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, ...)
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t a[n_] := a[n] = a[n - 1]*b[n - 2] - n;
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 40}] (* A294384 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A293076, A293765, A294381.
%K nonn,more
%O 0,2
%A _Clark Kimberling_, Oct 29 2017
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