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A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.

%I A386950 #25 Aug 24 2025 05:44:51
%S A386950 0,1,0,2,3,0,1,4,0,2,5,1,0,3,0,1,6,2,4,0,1,0,3,2,7,0,5,1,0,2,4,1,3,0,
%T A386950 0,1,6,2,0,3,5,1,4,8,0,2,0,1,0,3,2,1,0,4,7,0,5,1,6,2,3,0,1,0,2,4,0,1,
%U A386950 3,0,2,5,1,0,0,3,4,1,6,2,0,1,0,7,2,3,0
%N A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.
%C A386950 The pdf of the sequence is the probability to see i as the remainder when you divide 1,2,3... with a random number from [1,10].
%C A386950 The respective probabilities are:
%C A386950   p(0) = 7381/25200 = 0.292896...
%C A386950   p(1) = 4861/25200 = 0.192896...
%C A386950   p(2) = 3601/25200 = 0.142896...
%C A386950   p(3) = 2761/25200 = 0.109563...
%C A386950   p(4) = 2131/25200 = 0.084563...
%C A386950   p(5) = 1627/25200 = 0.064563...
%C A386950   p(6) = 1207/25200 = 0.047896...
%C A386950   p(7) = 121/3600   = 0.033611...
%C A386950   p(8) = 19/900     = 0.021111...
%C A386950   p(9) = 1/100      = 0.01
%C A386950 The sequence repeats after 25200 terms. If the random number is picked from [1,n] the period of the sequence is given by n*lcm{1,2,...,n} (A081528).
%C A386950 The arithmetic mean of this sequence is 2.25.
%C A386950 For a sequence with length n, the n-th root of the product of nonzero terms is (2**(5333/12600))*(3**(559/3150))*(5**(1627/25200))*(7**(121/3600)) which is 1.9302557640832...
%H A386950 Jwalin Bhatt, <a href="/A386950/b386950.txt">Table of n, a(n) for n = 1..25200</a>
%e A386950 Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms.
%e A386950 We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
%e A386950 Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
%e A386950 | n | (c(0)+1)/p(0) | (c(1)+1)/p(1) | (c(2)+1)/p(2) | choice |
%e A386950 |---|---------------|---------------|---------------|--------|
%e A386950 | 1 |     3.414     |       -       |       -       |   0    |
%e A386950 | 2 |     6.828     |     5.181     |       -       |   1    |
%e A386950 | 3 |     6.828     |    10.362     |     6.998     |   0    |
%e A386950 | 4 |    10.242     |    10.362     |     6.998     |   2    |
%t A386950 samplePDF[numCoeffs_] := Module[{coeffs, counts},
%t A386950   coeffs = {}; counts = ConstantArray[0, 10];
%t A386950   Do[
%t A386950     minTime = Infinity;
%t A386950     Do[
%t A386950       time = 10*(counts[[i]] + 1)/(HarmonicNumber[10] - HarmonicNumber[i - 1]);
%t A386950       If[time < minTime,minIndex = i;minTime = time],{i, 1, 10}];
%t A386950     counts[[minIndex]] += 1;
%t A386950     coeffs = Append[coeffs, minIndex - 1],
%t A386950     {numCoeffs}
%t A386950    ];
%t A386950   coeffs
%t A386950  ]
%t A386950 A386950 = samplePDF[120]
%o A386950 (Python)
%o A386950 from sympy import harmonic
%o A386950 def sample_pdf(num_coeffs):
%o A386950   coeffs, counts = [], [0]*10
%o A386950   for _ in range(num_coeffs):
%o A386950     min_time = float('inf')
%o A386950     for i, count in enumerate(counts):
%o A386950       time = 10*(count+1) / (harmonic(10)-harmonic(i))
%o A386950       if time < min_time:
%o A386950         min_index, min_time = i, time
%o A386950     counts[min_index] += 1
%o A386950     coeffs.append(min_index)
%o A386950   return coeffs
%o A386950 A386950 = sample_pdf(120)
%Y A386950 Cf. A081528.
%K A386950 nonn,base,new
%O A386950 1,4
%A A386950 _Jwalin Bhatt_, Aug 10 2025