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.

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, 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, 3, 0, 2, 5, 1, 0, 0, 3, 4, 1, 6, 2, 0, 1, 0, 7, 2, 3, 0

Offset: 1

Author

Jwalin Bhatt, Aug 10 2025

Keywords

• nonn
• base
• new

Comments

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].

The respective probabilities are:

p(0) = 7381/25200 = 0.292896...

p(1) = 4861/25200 = 0.192896...

p(2) = 3601/25200 = 0.142896...

p(3) = 2761/25200 = 0.109563...

p(4) = 2131/25200 = 0.084563...

p(5) = 1627/25200 = 0.064563...

p(6) = 1207/25200 = 0.047896...

p(7) = 121/3600 = 0.033611...

p(8) = 19/900 = 0.021111...

p(9) = 1/100 = 0.01

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).

The arithmetic mean of this sequence is 2.25.

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...

Examples

			Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms.
We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
| n | (c(0)+1)/p(0) | (c(1)+1)/p(1) | (c(2)+1)/p(2) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     3.414     |       -       |       -       |   0    |
| 2 |     6.828     |     5.181     |       -       |   1    |
| 3 |     6.828     |    10.362     |     6.998     |   0    |
| 4 |    10.242     |    10.362     |     6.998     |   2    |
		

Links

• Jwalin Bhatt, Table of n, a(n) for n = 1..25200

Crossrefs

Cf. A081528.

Programs

Mathematica

samplePDF[numCoeffs_] := Module[{coeffs, counts},
  coeffs = {}; counts = ConstantArray[0, 10];
  Do[
    minTime = Infinity;
    Do[
      time = 10*(counts[[i]] + 1)/(HarmonicNumber[10] - HarmonicNumber[i - 1]);
      If[time < minTime,minIndex = i;minTime = time],{i, 1, 10}];
    counts[[minIndex]] += 1;
    coeffs = Append[coeffs, minIndex - 1],
    {numCoeffs}
   ];
  coeffs
 ]
A386950 = samplePDF[120]

Python

from sympy import harmonic
def sample_pdf(num_coeffs):
  coeffs, counts = [], [0]*10
  for _ in range(num_coeffs):
    min_time = float('inf')
    for i, count in enumerate(counts):
      time = 10*(count+1) / (harmonic(10)-harmonic(i))
      if time < min_time:
        min_index, min_time = i, time
    counts[min_index] += 1
    coeffs.append(min_index)
  return coeffs
A386950 = sample_pdf(120)