A286334 The smallest number a such that there exists an integer b such that a/b is equal to n% rounded to the nearest percent.

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 2, 3, 1, 3, 2, 3, 4, 1, 5, 3, 5, 2, 3, 4, 6, 1, 10, 6, 4, 7, 3, 7, 2, 7, 5, 3, 4, 5, 6, 7, 10, 17, 1, 18, 11, 8, 7, 6, 5, 4, 7, 10, 3, 11, 8, 5, 7, 11, 19, 2, 13, 9, 7, 5, 13, 8, 14, 3, 13, 10, 7, 11, 4, 13, 9, 5, 16, 11, 6, 13, 7, 8, 9, 10, 11, 13, 15, 18, 22, 28, 39, 66, 1

Offset

0,13

Comments

a(n) is the smallest number of marks that gives you a mark of n% when rounded to the nearest percent.

a(n) is the smallest number a such that there exists an integer b such that a/b is equal to n% rounded to the nearest percent.

References

D. Griller, Elastic Numbers, Rational Falcon, 37.

Links

Table of n, a(n) for n=0..100.

Example

A mark of 1/20 is 5%, so a(5)=1.
A mark of 2/17 is 12% and 1/m doesn't give 12% for any m, so a(12)=2.
A mark of 3/19 is 16% and 1/m and 2/m don't give 16% for any m, so a(16)=3.

Mathematica

r[n_] := If[EvenQ@ Floor[n], Round[n + 1] - 1, Round[n]]; {0}~Join~Table[Module[{a = 1, b = 2, m}, While[While[100 a/b > n, b++]; !MemberQ[Set[m, Map[r, 100 a/Range@ b]], n], a++]; {a, Position[m, n][[1, 1]]}], {n, 100}][[All, 1]] (* Michael De Vlieger, May 09 2017 *)

Prog

(Python) #
from __future__ import division
from math import floor
least = [None] * 101
i = 1
while None in least.values():
....for j in range(i+1):
........p = int(floor(100*j/i+.5))
........if least[p] is None:
............least[p] = j
....i += 1
print(least)

Crossrefs

Sequence in context: A112757 A219794 A351469 * A118492 A079246 A309014

Adjacent sequences: A286331 A286332 A286333 * A286335 A286336 A286337

Keyword

nonn,fini,full

Author

Matthew Scroggs, May 07 2017

Status

approved