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Search: a294382 -id:a294382
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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.
+10
7
1, 3, 6, 24, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000, 107716736412020736000000
OFFSET
0,2
COMMENTS
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:
A294381: a(n) = a(n-1)*b(n-2)
A294382: a(n) = a(n-1)*b(n-2) - 1
A294383: a(n) = a(n-1)*b(n-2) + 1
A294384: a(n) = a(n-1)*b(n-2) - n
A294385: a(n) = a(n-1)*b(n-2) + n
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that a(2) = a(1)*b(0) = 6.
Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...).
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1]*b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294381 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 29 2017
EXTENSIONS
More terms from Jack W Grahl, Apr 26 2018
STATUS
approved

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