A293672 a(n) is the least integer k such that k/Fibonacci(n) > 4/5.

0, 1, 1, 2, 3, 4, 7, 11, 17, 28, 44, 72, 116, 187, 302, 488, 790, 1278, 2068, 3345, 5412, 8757, 14169, 22926, 37095, 60020, 97115, 157135, 254249, 411384, 665632, 1077016, 1742648, 2819663, 4562310, 7381972, 11944282, 19326254, 31270536, 50596789, 81867324

Offset

0,4

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 (-1 + x^2 + x^5 - x^10 + x^12 + x^13))/((-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(4*Fibonacci(n)/5).

a(n) = A293671(n) + 1 for n > 0.

Mathematica

z = 120; r = 4/5; f[n_] := Fibonacci[n];
Table[Floor[r*f[n]], {n, 0, z}];   (* A293671 *)
Table[Ceiling[r*f[n]], {n, 0, z}]; (* A293672 *)
Table[Round[r*f[n]], {n, 0, z}];   (* A293673 *)

Crossrefs

Cf. A000045, A293671, A293673.

Sequence in context: A222024 A222025 A222026 * A050193 A173173 A303025

Adjacent sequences: A293669 A293670 A293671 * A293673 A293674 A293675

Keyword

nonn,easy

Author

Clark Kimberling, Oct 15 2017

Status

approved