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A294383 Solution of the complementary equation a(n) = a(n-1)*b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294383 #11 Nov 06 2018 04:16:25
%S A294383 1,3,7,29,146,877,7017,63154,631541,6946952,83363425,1083724526,
%T A294383 15172143365
%N A294383 Solution of the complementary equation a(n) = a(n-1)*b(n-2) + 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294383 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 A294383 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 A294383 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A294383 a(2)  = a(1)*b(0) + 1 = 7
%e A294383 Complement: (b(n)) = (2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 16, ...)
%t A294383 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294383 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294383 a[n_] := a[n] = a[n - 1]*b[n - 2] + 1;
%t A294383 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294383 Table[a[n], {n, 0, 40}]  (* A294383 *)
%t A294383 Table[b[n], {n, 0, 10}]
%Y A294383 Cf. A293076, A293765, A294381.
%K A294383 nonn,more
%O A294383 0,2
%A A294383 _Clark Kimberling_, Oct 29 2017