A239525 -id:A239525 - cp's OEIS frontend

A303612 a(n) = min{denominator(r) with r in R} and R = {0 <= r rational <= 1 and [rk] = n}. Here k = 10^(floor(log_10(n))+1) and [x] = floor(x+1/2) + ceiling((2x-1)/4) - floor((2*x-1)/4) - 1.

1, 7, 4, 3, 5, 2, 5, 3, 4, 7, 10, 9, 8, 15, 7, 13, 19, 6, 11, 16, 5, 14, 9, 13, 17, 4, 19, 11, 18, 7, 10, 13, 19, 3, 29, 17, 11, 19, 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, 19, 11, 17, 29, 3, 19, 13, 10

Offset: 0

Luca Petrone, Apr 27 2018

a(n) is the smallest denominator of a fraction that, when rounded to d digits after the decimal point, is equal to 0.n, where d is the number of digits of n, and the rounding convention applied is that a number whose fractional part is 1/2 is rounded to the nearest even integer.
a(k-n) = a(n), where k is the first power of 10 exceeding n.
The sequence [A297367(n)/a(n), n = 10^(k-1)..10^(k)-1] is a subsequence of the Farey sequence A006842/A006843 of order ceiling((2/3)*10^k). For example, the terms a(1)..a(9) are the denominators of {1/7, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 6/7}; this sequence of fractions is a subsequence of the Farey sequence of order ceiling((2/3)*10^1) = 7, i.e., F7 = {0/1, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1/1}.
With the exception of n in {1, 2, 4, 13, 16}, r(n) = A297367(n)/a(n) is in the Farey series of order n (row n of A006842/A006843). - Peter Luschny, May 19 2018

			The table below shows the different rational numbers which satisfy the requirements of the definition. The denominators of the first rational number in each row constitute the sequence. Note that the round function is not implemented uniformly in popular software. For example, Mathematica matches our definition, while Maple's round function would return incorrect values.
.
  |                     |  decimal  |     round(10*r)
n | rational numbers r  |   value   | Mathematica | Maple
--+---------------------+-----------+-------------+------
0 | 0/1                 | 0.0000000 |      0      |   0
1 | 1/7, 1/8, 1/9, 1/10 | 0.1428571 |      1      |   1
2 | 1/4, 1/5, 1/6, 2/9  | 0.2500000 |      2      | * 3 *
3 | 1/3, 2/7, 3/10      | 0.3333333 |      3      |   3
4 | 2/5, 3/7, 3/8, 4/9  | 0.4000000 |      4      |   4
5 | 1/2                 | 0.5000000 |      5      |   5
6 | 3/5, 4/7, 5/8, 5/9  | 0.6000000 |      6      |   6
7 | 2/3, 5/7, 7/10      | 0.6666667 |      7      |   7
8 | 3/4, 4/5, 5/6, 7/9  | 0.7500000 |      8      |   8
9 | 6/7, 7/8, 8/9, 9/10 | 0.8571429 |      9      |   9

C. F. Gauss, Theorematis arithmetici demonstratio nova, Societati regiae scientiarum Gottingensis, Vol. XVI., January 15, 1808, pp. 5-7, section 4-5.
L. Graham and Donald E. Knuth and Oren Patashnik, Concrete mathematics: a foundation for computer science (Second Edition), Addison-Wesley Publishing Company, 1994, pp. 86-101.
Kenneth E. Iverson, A Programming Language, John Wiley And Sons, Inc., 1962 (4th printing 1967), pp. 11-12.
Takeo Kamizawa, Note on the Distance to the Nearest Integer, Faculty of Physics, Astronomy and Informatics, Nicolaus Copernicus University, Toruń, Poland, 2016.
A. M. Legendre, Théorie des nombres (deuxième édition), 1808.
D. Zuras, M. Cowlishaw, R. M. Grow, et al., IEEE Standard for Floating-Point Arithmetic, Std 754(tm)-2008, ISBN 978-0-7381-5753-5, August 28, 2008, p. 16, sections 4.3.1-4.3.3.

Luca Petrone, Table of n, a(n) for n = 0..9999
IEEE, 754-2008 - IEEE Standard for Floating-Point Arithmetic, IEEE Computer Society, August 28, 2008.
Luca Petrone, log-log plot for first 100000 terms
Štefan Porubský, Integer rounding functions, Interactive Information Portal for Algorithmic Mathematics, Institute of Computer Science of the Czech Academy of Sciences, April 1, 2007.
Eric Weisstein's World of Mathematics, Nearest Integer Function
Wikipedia, Nearest integer function

Cf. A239525, A297367, A006842, A006843, A304879.

r := proc(n) local nint, k, p, q; k := 10^(ilog10(n)+1);
nint := m -> floor(m + 1/2) + ceil((2*m-1)/4) - floor((2*m-1)/4) - 1;
for p from 1 to k do for q from p+1 to k do if nint(p*k/q) = n then return p/q fi od od; 0/1 end:
a := n -> denom(r(n)): seq(a(n), n=0..99); # Peter Luschny, May 19 2018

a = {1};
For[i = 1, i <= 100, i++,
nmax = 10^(Floor[Log[10, i]] + 1);
r = i/nmax;
For[n = 1, n <= nmax, n++,
If[Round[Round[n r]/n, 1/nmax] == r,
a = Flatten[Append[a, n]];
Break[];
]]]

A239526 For 0<=n<=100, a(n) is the number of positive responses x such that x/N rounds to n%, minimized over sample size N.

Original entry on oeis.org

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

Offset: 0

Patrick D McLean, Mar 21 2014

			a(31)=4 because 4/13=0.31 (2DP).

Cf. A239525 (Minimal sample sizes).

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

from itertools import count
def A239526(n):
    for y in count(1):
        x, z = divmod(y*((n<<1)+1),200)
        if not z: return x
        x, z = divmod(y*((n<<1)-1),200)
        if (x:=x+bool(z)) and (200*x+y)//(y<<1) == n:
            return x # Chai Wah Wu, Jun 28 2025

A384670 Smallest denominator y for which there exists an integer x with round(100*x/y) = n.

Original entry on oeis.org

1, 67, 41, 29, 23, 19, 16, 14, 12, 11, 10, 9, 17, 8, 7, 13, 19, 6, 11, 16, 5, 14, 9, 13, 17, 4, 19, 11, 18, 7, 10, 13, 19, 3, 29, 17, 11, 19, 8, 18, 5, 17, 12, 7, 9, 11, 13, 15, 21, 35, 2, 35, 21, 15, 13, 11, 9, 7, 12, 17, 5, 18, 13, 8, 11, 17, 29, 3, 19, 13, 10, 7, 18, 11, 19, 4, 17, 13, 9, 14, 5, 16, 11, 6, 19, 13, 7, 15, 8, 9, 10, 11, 12, 14, 16, 19, 23, 29, 40, 67, 1

Offset: 0

James Beazley, Jun 06 2025

We allow x=0 so that a(0)=1 is from round(100*0/1) = 0.

If some published statistic shows n percent, and that percentage was made by rounding to the nearest integer (and 0.5 rounds upwards), then it must have been from a sample of at least a(n) things.

			For n=1, proportion 1/67 = 1.4992...% rounds to n=1 percent and 67 is the smallest denominator allowing that.

Cf. A239525 (round either way).

first(n) = {
    res = vector(n, i, oo);
    todo = 100;
    for(i = 1, 100,
        for(j = 1, i,
            c = round(100*j/i);
            if(0 < c && c <= n,
                if(res[c] == oo,
                    todo--;
                    if(todo == 0,
                        break
                    ));
                    res[c] = min(res[c], i))));
    for(i = 101, n,
        res[i] = res[i-100]);
    concat(1, res)
} \\ David A. Corneth, Jun 23 2025

from itertools import count
def A384670(n):
    for y in count(1):
        x, z = divmod(y*((n<<1)-1),200)
        if (200*(x+bool(z))+y)//(y<<1) == n:
            return y # Chai Wah Wu, Jun 28 2025

cp's OEIS Frontend

A303612 a(n) = min{denominator(r) with r in R} and R = {0 <= r rational <= 1 and [rk] = n}. Here k = 10^(floor(log_10(n))+1) and [x] = floor(x+1/2) + ceiling((2x-1)/4) - floor((2*x-1)/4) - 1.

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Maple

Mathematica

A239526 For 0<=n<=100, a(n) is the number of positive responses x such that x/N rounds to n%, minimized over sample size N.

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Mathematica

Python

A384670 Smallest denominator y for which there exists an integer x with round(100*x/y) = n.

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PARI

Python

cp's OEIS Frontend

A303612 a(n) = min{denominator(r) with r in R} and R = {0 <= r rational <= 1 and [r*k] = n}. Here k = 10^(floor(log_10(n))+1) and [x] = floor(x+1/2) + ceiling((2*x-1)/4) - floor((2*x-1)/4) - 1.

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Maple

Mathematica

A239526 For 0<=n<=100, a(n) is the number of positive responses x such that x/N rounds to n%, minimized over sample size N.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Python

A384670 Smallest denominator y for which there exists an integer x with round(100*x/y) = n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Python

A303612 a(n) = min{denominator(r) with r in R} and R = {0 <= r rational <= 1 and [rk] = n}. Here k = 10^(floor(log_10(n))+1) and [x] = floor(x+1/2) + ceiling((2x-1)/4) - floor((2*x-1)/4) - 1.