James Beazley - cp's OEIS frontend

User: James Beazley

James Beazley has authored 1 sequences.

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

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

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User: James Beazley

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

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