A293639 a(n) is the greatest integer k such that k/Fibonacci(n) < 2/5.

0, 0, 0, 0, 1, 2, 3, 5, 8, 13, 22, 35, 57, 93, 150, 244, 394, 638, 1033, 1672, 2706, 4378, 7084, 11462, 18547, 30010, 48557, 78567, 127124, 205691, 332816, 538507, 871323, 1409831, 2281154, 3690986, 5972140, 9663126, 15635267, 25298394, 40933662, 66232056

Offset

0,6

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1, 2, -1, -2, 2, 1, -3, -1, 3, 0, -2, 1, 2, -1, -1)

Formula

G.f.: (x^4 (1 + x) (1 - x + x^2) (1 + x - x^2 - 2 x^3 + x^5 + x^6))/((-1 + x) (-1 + x + x^2) (1 + x + x^2 + x^3 + x^4) (1 - x^2 + x^4 - x^6 + x^8)).

a(n) = a(n-1) + 2 a(n-2) - a(n-3) - 2 a(n-4) + 2 a(n-5) + a(n-6) - 3 a(n-7) - a(n-8) + 3 a(n-9) - 2 a(n-11) + a(n-12) + 2 a(n-13) - a(n-14) - a(n-15) for n >= 16.

a(n) = floor(2*Fibonacci(n)/5).

a(n) = A293640(n) - 1 for n > 0.

Mathematica

z = 120; r = 2/5; f[n_] := Fibonacci[n];
Table[Floor[r*f[n]], {n, 0, z}];   (* A293639 *)
Table[Ceiling[r*f[n]], {n, 0, z}]; (* A293640 *)
Table[Round[r*f[n]], {n, 0, z}];   (* A293641 *)
LinearRecurrence[{1, 2, -1, -2, 2, 1, -3, -1, 3, 0, -2, 1, 2, -1, -1}, {0, 0, 0, 0, 1, 2, 3, 5, 8, 13, 22, 35, 57, 93, 150}, 60] (* Harvey P. Dale, May 02 2019 *)

Crossrefs

Cf. A000045, A293640, A293641.

Sequence in context: A053412 A198202 A349840 * A320356 A004697 A245271

Adjacent sequences: A293636 A293637 A293638 * A293640 A293641 A293642

Keyword

nonn,easy

Author

Clark Kimberling, Oct 14 2017

Status

approved