A239525 For 0 <= n <= 100, a(n) is smallest integer N such that some positive x/N rounds to n%, with x > 0.

200, 67, 40, 29, 23, 19, 16, 14, 12, 11, 10, 9, 8, 8, 7, 13, 19, 6, 11, 16, 5, 14, 9, 13, 17, 4, 19, 11, 18, 7, 10, 13, 19, 3, 29, 17, 11, 8, 8, 18, 5, 17, 12, 7, 9, 11, 13, 15, 21, 35, 2, 35, 21, 15, 13, 11, 9, 7, 12, 17, 5, 18, 8, 8, 11, 17, 29, 3, 19, 13, 10, 7, 18, 11, 19, 4, 17, 13, 9, 14, 5, 16, 11, 6, 19, 13, 7, 8, 8, 9, 10, 11, 12, 14, 16, 19, 23, 29, 40, 67, 1

Offset: 0

Author

Patrick D McLean, Mar 21 2014

Keywords

• nonn
• base
• easy
• fini
• full

Comments

When the fractional part is 0.5, rounding in either direction is allowed.

From Jonathan Dushoff, Feb 11 2025: (Start)

The sequence arises when considering how to "reverse-engineer" published percentages: A quoted n percent means that proportion was within at least a total of a(n) things.

Having a(n) based on rounding either way when mid-way between two integer percentages makes the terms agnostic as to what rounding was actually used in the published percentage.

(End)

Examples

			a(31)=13 because 4/13 = 0.31 (to two digits after the decimal point).
		

Crossrefs

Cf. A239526 (corresponding x).

Programs

Mathematica

Table[LinearProgramming[{1, 0}, {{-n/100 + 0.005, 1}, {n/100 + 0.005, -1}}, {0, 0}, {1, 1}, Integers], {n, 0, 100}] // Transpose // First

Python

from itertools import count
def A239525(n):
    for y in count(1):
        if not y*((n<<1)+1)%200: return y
        x, z = divmod(y*((n<<1)-1),200)
        if (a:=x+bool(z)) and (200*a+y)//(y<<1) == n:
            return y # Chai Wah Wu, Jun 28 2025

Formula

Find the smallest N such that there is some x > 0 with abs(100*x/N - n) <= 0.5.

Extensions

Edited by Jonathan Dushoff, Apr 22 2022