A296239 a(n) = distance from n to nearest Fibonacci number.

0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 1, 0, 1, 2, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 16, 15, 14, 13, 12, 11, 10, 9

Offset: 0

Author

Rémy Sigrist, Dec 09 2017

Keywords

• nonn
• easy

Comments

The Fibonacci numbers correspond to sequence A000045.

This sequence is analogous to:

- A051699 (distance to nearest prime),

- A053188 (distance to nearest square),

- A053646 (distance to nearest power of 2),

- A053615 (distance to nearest oblong number),

- A053616 (distance to nearest triangular number),

- A061670 (distance to nearest power),

- A074989 (distance to nearest cube),

- A081134 (distance to nearest power of 3),

The local maxima of the sequence correspond to positive terms of A004695.

a(n) = 0 iff n = A000045(k) for some k >= 0.

a(n) = 1 iff n = A061489(k) for some k > 4.

For any n >= 0, abs(a(n+1) - a(n)) <= 1.

For any n > 0, a(n) < n, and a^k(n) = 0 for some k > 0 (where a^k denotes the k-th iterate of a); k equals A105446(n) for n = 1..80 (and possibly more values).

a(n) > max(a(n-1), a(n+1)) iff n = A001076(k) for some k > 1.

Examples

			For n = 42:
- A000045(9) = 34 <= 42 <= 55 = A000045(10),
- a(42) = min(42 - 34, 55 - 42) = min(8, 13) = 8.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Index entries for sequences related to distance to nearest element of some set

Crossrefs

Cf. A000045, A001076, A004695, A051699, A053188, A053615, A053616, A053646, A061489, A061670, A074989, A081134, A105446.

Programs

Mathematica

fibPi[n_] := 1 + Floor[ Log[ GoldenRatio, 1 + n*Sqrt@5]]; f[n_] := Block[{m = fibPi@ n}, Min[n - Fibonacci[m -1], Fibonacci[m] - n]]; Array[f, 81, 0] (* Robert G. Wilson v, Dec 11 2017 *)
With[{nn=80,fibs=Fibonacci[Range[0,20]]},Table[Abs[n-Nearest[fibs,n]][[1]],{n,0,nn}]] (* Harvey P. Dale, Jul 02 2022 *)

PARI

a(n) = for (i=1, oo, if (n<=fibonacci(i), return (min(n-fibonacci(i-1), fibonacci(i)-n))))

Formula

a(n) = abs(n - Fibonacci(floor(log(sqrt(20)*n)/log((1 + sqrt(5))/2)-1))). - Jon E. Schoenfield, Dec 14 2017