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A294381 Solution of the complementary equation a(n) = a(n-1)*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A294381 #24 Apr 29 2018 13:08:37
%S A294381 1,3,6,24,120,840,6720,60480,604800,6652800,79833600,1037836800,
%T A294381 14529715200,217945728000,3487131648000,59281238016000,
%U A294381 1067062284288000,20274183401472000,405483668029440000,8515157028618240000,187333454629601280000,4308669456480829440000,107716736412020736000000
%N A294381 Solution of the complementary equation a(n) = a(n-1)*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A294381 The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
%C A294381 A294381:  a(n) = a(n-1)*b(n-2)
%C A294381 A294382:  a(n) = a(n-1)*b(n-2) - 1
%C A294381 A294383:  a(n) = a(n-1)*b(n-2) + 1
%C A294381 A294384:  a(n) = a(n-1)*b(n-2) - n
%C A294381 A294385:  a(n) = a(n-1)*b(n-2) + n
%H A294381 Jack W Grahl, <a href="/A294381/b294381.txt">Table of n, a(n) for n = 0..49</a>
%H A294381 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%e A294381 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that a(2) = a(1)*b(0) = 6.
%e A294381 Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...).
%t A294381 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294381 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A294381 a[n_] := a[n] = a[n - 1]*b[n - 2];
%t A294381 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294381 Table[a[n], {n, 0, 40}]  (* A294381 *)
%t A294381 Table[b[n], {n, 0, 10}]
%Y A294381 Cf. A293076, A293765.
%K A294381 nonn,easy
%O A294381 0,2
%A A294381 _Clark Kimberling_, Oct 29 2017
%E A294381 More terms from _Jack W Grahl_, Apr 26 2018