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A295362
Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
2
1, 3, 5, 8, 10, 14, 19, 25, 34, 49, 71, 106, 162, 253, 398, 632, 1010, 1621, 2610, 4208, 6793, 10975, 17741, 28688, 46400, 75058, 121428, 196454, 317848, 514267, 832079, 1346309, 2178350, 3524620, 5702930, 9227509, 14930397, 24157863
OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
FORMULA
a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
EXAMPLE
a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = a(2) + a(1) + b(2) - b(1) - b(0) = 8
Complement: (b(n)) = (2, 4, 6, 7, 9, 11, 12, 13, 15, 16, 17, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] - b[n - 2] - b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 32; u = Table[a[n], {n, 0, z}] (* A295362 *)
v = Table[b[n], {n, 0, 10}] (* complement *)
CROSSREFS
Sequence in context: A253081 A088940 A088937 * A162383 A123327 A024679
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 21 2017
STATUS
approved