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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. 7

%I #24 Apr 29 2018 13:08:37

%S 1,3,6,24,120,840,6720,60480,604800,6652800,79833600,1037836800,

%T 14529715200,217945728000,3487131648000,59281238016000,

%U 1067062284288000,20274183401472000,405483668029440000,8515157028618240000,187333454629601280000,4308669456480829440000,107716736412020736000000

%N 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 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: a(n) = a(n-1)*b(n-2)

%C A294382: a(n) = a(n-1)*b(n-2) - 1

%C A294383: a(n) = a(n-1)*b(n-2) + 1

%C A294384: a(n) = a(n-1)*b(n-2) - n

%C A294385: a(n) = a(n-1)*b(n-2) + n

%H Jack W Grahl, <a href="/A294381/b294381.txt">Table of n, a(n) for n = 0..49</a>

%H 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 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that a(2) = a(1)*b(0) = 6.

%e Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 12, 13, 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];

%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}] (* A294381 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A293076, A293765.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 29 2017

%E More terms from _Jack W Grahl_, Apr 26 2018

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)