A293640 a(n) is the least integer k such that k/Fibonacci(n) > 2/5.

0, 1, 1, 1, 2, 2, 4, 6, 9, 14, 22, 36, 58, 94, 151, 244, 395, 639, 1034, 1673, 2706, 4379, 7085, 11463, 18548, 30010, 48558, 78568, 127125, 205692, 332816, 538508, 871324, 1409832, 2281155, 3690986, 5972141, 9663127, 15635268, 25298395, 40933662, 66232057

Offset

0,5

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) = ceiling(2*Fibonacci(n)/5).

a(n) = A293639(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, 1, 1, 1, 2, 2, 4, 6, 9, 14, 22, 36, 58, 94, 151}, 60] (* Harvey P. Dale, Jul 13 2025 *)

Crossrefs

Cf. A000045, A004698, A293642, A293671, A293639, A293641.

Sequence in context: A058518 A393225 A018139 * A025049 A293552 A293546

Adjacent sequences: A293637 A293638 A293639 * A293641 A293642 A293643

Keyword

nonn,easy

Author

Clark Kimberling, Oct 14 2017

Status

approved