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

%I A384670 #20 Jun 28 2025 20:19:37
%S A384670 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,
%T A384670 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,
%U A384670 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
%N A384670 Smallest denominator y for which there exists an integer x with round(100*x/y) = n.
%C A384670 We allow x=0 so that a(0)=1 is from round(100*0/1) = 0.
%C A384670 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.
%e A384670 For n=1, proportion 1/67 = 1.4992...% rounds to n=1 percent and 67 is the smallest denominator allowing that.
%o A384670 (PARI)
%o A384670 first(n) = {
%o A384670     res = vector(n, i, oo);
%o A384670     todo = 100;
%o A384670     for(i = 1, 100,
%o A384670         for(j = 1, i,
%o A384670             c = round(100*j/i);
%o A384670             if(0 < c && c <= n,
%o A384670                 if(res[c] == oo,
%o A384670                     todo--;
%o A384670                     if(todo == 0,
%o A384670                         break
%o A384670                     ));
%o A384670                     res[c] = min(res[c], i))));
%o A384670     for(i = 101, n,
%o A384670         res[i] = res[i-100]);
%o A384670     concat(1, res)
%o A384670 } \\ _David A. Corneth_, Jun 23 2025
%o A384670 (Python)
%o A384670 from itertools import count
%o A384670 def A384670(n):
%o A384670     for y in count(1):
%o A384670         x, z = divmod(y*((n<<1)-1),200)
%o A384670         if (200*(x+bool(z))+y)//(y<<1) == n:
%o A384670             return y # _Chai Wah Wu_, Jun 28 2025
%Y A384670 Cf. A239525 (round either way).
%K A384670 nonn,easy
%O A384670 0,2
%A A384670 _James Beazley_, Jun 06 2025