A293637 a(n) is the least integer k such that k/Fibonacci(n) > 1/5.

0, 1, 1, 1, 1, 1, 2, 3, 5, 7, 11, 18, 29, 47, 76, 122, 198, 320, 517, 837, 1353, 2190, 3543, 5732, 9274, 15005, 24279, 39284, 63563, 102846, 166408, 269254, 435662, 704916, 1140578, 1845493, 2986071, 4831564, 7817634, 12649198, 20466831, 33116029, 53582860

Offset

0,7

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) (-1 + x + x^2 - x^4 + x^5 - x^7 + x^9 - x^10 + x^12))/((-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(Fibonacci(n)/5).

a(n) = A004698(n) + 1 for n > 0.

Mathematica

z = 120; r = 1/5; f[n_] := Fibonacci[n];
Table[Floor[r*f[n]], {n, 0, z}];   (* A004698 *)
Table[Ceiling[r*f[n]], {n, 0, z}]; (* A293637 *)
Table[Round[r*f[n]], {n, 0, z}];   (* A293638 *)

Crossrefs

Cf. A000045, A004698, A293638.

Sequence in context: A240566 A067904 A018146 * A284250 A291660 A069749

Adjacent sequences: A293634 A293635 A293636 * A293638 A293639 A293640

Keyword

nonn,easy

Author

Clark Kimberling, Oct 14 2017

Status

approved