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 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
 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 (list; graph; refs; listen; history; text; internal format)
 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 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: A179009 A112757 A219794 * A118492 A079246 A309014 Adjacent sequences:  A286331 A286332 A286333 * A286335 A286336 A286337 KEYWORD nonn,fini,full AUTHOR Matthew Scroggs, May 07 2017 STATUS approved

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Last modified August 8 11:31 EDT 2020. Contains 336298 sequences. (Running on oeis4.)